3.2 worked examples for case one

3.2.1 Example 1

Let $$y^{\prime \prime }=\frac{4x^6-8x^5+12x^4+4x^3+7x^2-20x+4}{4x^4}y$$ Therefore $$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{4x^6-8x^5+12x^4+4x^3+7x^2-20x+4}{4x^4}\\ \text{(1)}& & =x^2-2x+3+\frac{1}{x}+\frac{7}{x^2}-\frac{5}{x^3}+\frac{1}{x^4}\end{array}$$

Step 1 In this we find all ${\left[\sqrt{r}\right]}_c$ and associated ${\alpha }_c^{\pm }$ for each pole. There is only one pole at $x=0$ of order 4. Hence $2v=4$. And $v=2$. This is step (C3) now in the paper (1).

We need now to find Laurent series of $\sqrt{r}$ expanded around $x=c=0$. This is given by (using series command on the computer) $$\begin{array}{}\text{(2)}& \frac{1}{x^2}-\frac{5}{2}\frac{1}{x}-\frac{9}{4}-\frac{41}{8}x-\frac{443}{32}{x}^2+\cdots \end{array}$$ We need to add all terms in the Laurent series expansion of $\sqrt{r}$ from $v=2$ down to $2$. Hence $$\begin{array}{}\text{(3)}& {\left[\sqrt{r}\right]}_c=\frac{1}{{(x-0)}^2}\end{array}$$ Is only term from 2. Comparing the above to $\frac{a}{{(x-0)}^2}$ shows that $$\begin{array}{}\text{(4)}& a=1\end{array}$$ Hence$$\begin{array}{rl}\text{(5)}& {\left[\sqrt{r}\right]}_c& =\frac{1}{x^2}{\alpha }_c^{+}& =\frac{1}{2}(\frac{b}{a}+v)\\ \text{(6)}& {\alpha }_c^{-}& =\frac{1}{2}(-\frac{b}{a}+v)\end{array}$$

Where $v=2$ and $a=1$. We still need to find $b$. But $b$ is the coefficient of the term $\frac{1}{{(x-0)}^{v+1}}$ in $r$ minus the coefficient of $\frac{1}{{(x-0)}^{v+1}}$ in ${\left[\sqrt{r}\right]}_c$ which we just found above. Looking at $r$ from Eq (1) we see that the term $\frac{1}{{(x-0)}^{v+1}}=\frac{1}{{(x-0)}^3}$ has coefficient $-5$. And looking at Eq (3) we see that there is no term $\frac{1}{{(x-0)}^3}$ in it. Hence $$\begin{array}{rl}b& =-5-0\\ \text{(7)}& & =-5\end{array}$$

Now we found $a,b$, then (5,6) becomes (since $v=2$)$$\begin{array}{}\text{(8)}& {\alpha }_c^{+}& =\frac{1}{2}(-5+2)=\frac{-3}{2}\text{(9)}& {\alpha }_c^{-}& =\frac{1}{2}(5+2)=\frac{7}{2}\end{array}$$

We are done with all the poles.

Now we consider $O\left(\mathrm{\infty }\right)$ which is $\mathrm{deg}\left(t\right)-\mathrm{deg}\left(s\right)=4-6=-2$. Since this is even order and negative then $-2v=-2$ or $$v=1$$ We need the Laurent series of $\sqrt{r}$ around $\mathrm{\infty }$. Using the computer this is $${\left[\sqrt{r}\right]}_{\mathrm{\infty }}=x-1+\frac{1}{x}+\frac{3}{2}\frac{1}{x^2}+\frac{15}{8x^3}+\cdots$$ Now we only want the terms $x^i$ where $0\le i\le v$. This implies the above is reduced to $${\left[\sqrt{r}\right]}_{\mathrm{\infty }}=x-1$$ The $a$ is the coefficient of $x^v=x$ which is$$a=1$$ Now we need to find ${\alpha }_{\mathrm{\infty }}^{\pm }$ associated with ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$. For this we need to first find $b$. Recall from above that $b$ is the coefficient of $x^{v-1}$ or $x^0$ in $r$ minus the coefficient of $x^{v-1}=x^0$ in ${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2$. Since $v=1$ then we want the coefficient of $x^0$ in $r$ and subtract from it the coefficient of $x^0$ in ${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2$. But$$\begin{array}{rl}{\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2& ={(x-1)}^2\\ & =x^2+1-2x\end{array}$$

Hence the coefficient of $x^0$ in ${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2$ is $1$.  To find the coefficient of $x^0$ in $r$ long division is done$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{4x^6-8x^5+12x^4+4x^3+7x^2-20x+4}{4x^4}\\ & =Q+\frac{R}{4x^2}\end{array}$$

Where $Q$ is the quotient and $R$ is the remainder. This gives$$r=(x^2-2x+3)+\frac{4x^3+7x^2-20x+4}{4x^2}$$ For the case of $v\ne 0$ then the coefficient is read from $Q$ above. Which is $3$. Hence$$\begin{array}{rl}b& =3-1\\ & =2\end{array}$$

For the other case of $v=0$ then the coefficient of $x^{-1}$ in $r$ is found using $\frac{lcoeff\left(R\right)}{lcoeff\left(t\right)}$ which will give $1$ in this case. (More examples below).

Now that we found $a,b$, then from the above section describing the algorithm, we see in this case that$$\begin{array}{rl}{\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)=\frac{1}{2}(\frac{2}{1}-1)=\frac{1}{2}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)=\frac{1}{2}(-\frac{2}{1}-1)=-\frac{3}{2}\end{array}$$

This completes step 1 of the solution. We have found ${\left[\sqrt{r}\right]}_c$ and its associated ${\alpha }_c^{\pm }$ and found ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$. Now we go to step 2 which is to find the $d^{\prime }s$.

step 2 Since we have a pole at zero, and we have one $O\left(\mathrm{\infty }\right)$, each with $\pm$ signs, then we set up this table to make it easier to work with. This implements$$d={\alpha }_{\mathrm{\infty }}^{\pm }-\sum _{i=1}^1{\alpha }_{c_i}^{\pm }$$ Therefore we obtain 4 possible $d$ values.

pole $c$	${\alpha }_c$ value	$O\left(\mathrm{\infty }\right)$ value	$d$	$d$ value
$x=0$	${\alpha }_c^{+}=\frac{-3}{2}$	${\alpha }_{\mathrm{\infty }}^{+}=\frac{1}{2}$	${\alpha }_{\mathrm{\infty }}^{+}-\left({\alpha }_c^{+}\right)=\frac{1}{2}-\left(\frac{-3}{2}\right)$	$2$
$x=0$	${\alpha }_c^{+}=\frac{-3}{2}$	${\alpha }_{\mathrm{\infty }}^{-}=-\frac{3}{2}$	${\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_c^{+}\right)=-\frac{3}{2}-\left(\frac{-3}{2}\right)$	$0$
$x=0$	${\alpha }_c^{-}=\frac{7}{2}$	${\alpha }_{\mathrm{\infty }}^{+}=\frac{1}{2}$	${\alpha }_{\mathrm{\infty }}^{+}-\left({\alpha }_c^{-}\right)=\frac{1}{2}-\left(\frac{7}{2}\right)$	$-3$
$x=0$	${\alpha }_c^{-}=\frac{7}{2}$	${\alpha }_{\mathrm{\infty }}^{-}=-\frac{3}{2}$	${\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_c^{-}\right)=-\frac{3}{2}-\left(\frac{7}{2}\right)$	$-5$

We see from the above that we took each pole in this problem (there is only one pole here at $x=0$) and its associated ${\alpha }_c^{\pm }$ with each ${\alpha }_{\mathrm{\infty }}^{\pm }$ and generated all possible $d$ values from all the combinations. Hence we obtain 4 possible $d$ values in this case. If we had 2 poles, then we would have 8 possible $d$ values. Hence the maximum possible $d$ values we can get is $2^{p+1}$ where $p$ is number of poles. Now we remove all negative $d$ values. Hence the trial $d$ values remaining is $$d=\{0,2\}$$ Now for each $d$ value, we generate $\omega$ using$$\omega =(\sum _{c}s\left(c\right){\left[\sqrt{r}\right]}_c+\frac{{\alpha }_c}{x-c})+s\left(\mathrm{\infty }\right){\left[\sqrt{r}\right]}_{\mathrm{\infty }}$$ To apply the above, we update the table above, now using only $d=0,d=2$ values, but also add columns for ${\left[\sqrt{r}\right]}_c,{\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ to make the computation easier. Here is the new table

pole $c$	${\alpha }_c$ value	$s\left(c\right)$	${\left[\sqrt{r}\right]}_c$	$O\left(\mathrm{\infty }\right)$ value	$s\left(\mathrm{\infty }\right)$	${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$	$d$ value	$\omega$ value $(\sum _{c}s\left(c\right){\left[\sqrt{r}\right]}_c+\frac{{\alpha }_c}{x-c})+s\left(\mathrm{\infty }\right){\left[\sqrt{r}\right]}_{\mathrm{\infty }}$
$x=0$	${\alpha }_c^{+}=\frac{-3}{2}$	$+$	$\frac{1}{x^2}$	${\alpha }_{\mathrm{\infty }}^{-}=-\frac{3}{2}$	$-$	$x-1$	$0$	$(+\left(\frac{1}{x^2}\right)+\frac{\frac{-3}{2}}{x-0})+(-)(x-1)=\frac{1}{x^2}-\frac{3}{2x}-x+1$
$x=0$	${\alpha }_c^{+}=\frac{-3}{2}$	$+$	$\frac{1}{x^2}$	${\alpha }_{\mathrm{\infty }}^{+}=\frac{1}{2}$	$+$	$x-1$	$2$	$(+\left(\frac{1}{x^2}\right)+\frac{\frac{-3}{2}}{x-0})+(+)(x-1)=x-\frac{3}{2x}+\frac{1}{x^2}-1$

The above are the two candidate $\omega$ values. For each $\omega$ we need to find polynomial $P$ by solving $$\begin{array}{}\text{(8)}& P^{\prime \prime }+2\omega P^{\prime }+({\omega }^{\prime }+{\omega }^2-r)P=0\end{array}$$ If we are able to find $P$, then we stop and the ode $y^{\prime \prime }=ry$ is solved. If we try all candidate $\omega$ and can not find $P$ then this case is not successful and we go to the next case.

step 3 Now for each candidate $\omega$ we solve the above Eq (8). Starting with $\omega =\frac{1}{x^2}-\frac{3}{2x}-x+1$ associated with $d=0$ in the table, then (8) becomes

$$\begin{array}{rl}{P}^{\prime \prime }+2(\frac{1}{x^2}-\frac{3}{2x}-x+1)P^{\prime }+((\frac{3}{2x^2}-\frac{2}{x^3}-1)+{(\frac{1}{x^2}-\frac{3}{2x}-x+1)}^2-\left(\frac{4x^6-8x^5+12x^4+4x^3+7x^2-20x+4}{4x^4}\right))P& =0\\ P^{\prime \prime }+(\frac{2}{x^2}-\frac{3}{2x}-2x+2)P^{\prime }+(\frac{4}{x^2}-\frac{6}{x})P& =0\end{array}$$

Since this the case for $d=0$, then $P$ has zero degree, Hence $P$ is constant. Therefore the above simplifies to$$(\frac{4}{x^2}-\frac{6}{x})P=0$$ Which means$$\frac{4}{x^2}-\frac{6}{x}=0$$ Which is not possible for all $x$. Hence $d=0$ do not generate valid $P$ polynomial. We now try the case of $d=2$. Since $d=2$, it means the polynomial $d$ is of degree two. Let $$P=x^2+ax+b$$ Substituting this in (8) using $\omega =x-\frac{3}{2x}+\frac{1}{x^2}-1$. This gives

$$\begin{array}{rl}{P}^{\prime \prime }+2(x-\frac{3}{2x}+\frac{1}{x^2}-1)P^{\prime }+((\frac{3}{2x^2}-\frac{2}{x^3}+1)+{(x-\frac{3}{2x}+\frac{1}{x^2}-1)}^2-\left(\frac{4x^6-8x^5+12x^4+4x^3+7x^2-20x+4}{4x^4}\right))P& =0\\ P^{\prime \prime }+2(x-\frac{3}{2x}+\frac{1}{x^2}-1)P^{\prime }+(\frac{4}{x}-4)P& =0\end{array}$$

Using $P=x^2+ax+b$ the above becomes$$\begin{array}{rl}2+2(x-\frac{3}{2x}+\frac{1}{x^2}-1)(2x+a)+(\frac{4}{x}-4)(x^2+ax+b)& =0\\ 2a-4b-3\frac{a}{x}+2\frac{a}{x^2}+4\frac{b}{x}-2ax+\frac{4}{x}-4& =0\\ (2a-4b-4)+\frac{1}{x}(-\frac{3}{4}a+4b+4)+\frac{1}{x^2}\left(2a\right)-2ax& =0\end{array}$$

Therefore$$\begin{array}{rl}2a-4b-4& =0\\ -\frac{3}{4}a+4b+4& =0\\ 2a& =0\\ 2a& =0\end{array}$$

hence $a=0$ from last equation. Using first or second equation gives $b=-1$. Therefore a solution is found. Hence $$p\left(x\right)=x^2-1$$ Therefore the solution to $y^{\prime \prime }=ry$ is$$\begin{array}{rl}y& =p\left(x\right)e^{\int \omega dx}\\ & =(x^2-1)e^{\int x-\frac{3}{2x}+\frac{1}{x^2}-1dx}\\ & =(x^2-1)e^{-\frac{1}{x}-\frac{3}{2}\ln x+\frac{x^2}{2}-x}\\ & =(x^2-1)x^{\frac{-3}{2}}{e}^{-\frac{1}{x}+\frac{x^2}{2}-x}\end{array}$$

The second solution can be found by reduction of order.  

3.2.2 Example 2

Let $$y^{\prime \prime }=(\frac{2}{x^2}-1)y$$ Therefore $$\begin{array}{rl}r& =\frac{s}{t}\\ \text{(1)}& & =\frac{2-x^2}{x^2}\end{array}$$

Step 1 In this we find all ${\left[\sqrt{r}\right]}_c$ and associated ${\alpha }_c^{\pm }$ for each pole. There is one pole at $x=0$ of order 2. In this case, from the description of the algorithm earlier, we write$$\begin{array}{rl}{\left[\sqrt{r}\right]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4b}\\ {\alpha }_c^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4b}\end{array}$$

Where $b$ is the coefficient of $\frac{1}{{(x-0)}^2}$ in the partial fraction decomposition of $r$ which is $\frac{2}{x^2}-1$. Hence $b=2$. Therefore$$\begin{array}{rl}{\left[\sqrt{r}\right]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+8}=\frac{1}{2}+\frac{3}{2}=2\\ {\alpha }_c^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+8}=\frac{1}{2}-\frac{3}{2}=-1\end{array}$$

We are done with all the poles. Now we consider $O\left(\mathrm{\infty }\right)$ which is $\mathrm{deg}\left(t\right)-\mathrm{deg}\left(s\right)=2-2=0$. This falls in the case $-2v\le 0$. Hence $$v=0$$ We need the Laurent series of $\sqrt{r}$ around $\mathrm{\infty }$. Using the computer this is $i-\frac{i}{x^2}-\frac{1}{2x^4}i+\cdots$. Hence we need the coefficient of $x^0$ in this series ($0$ because that is value of $v$).$${\left[\sqrt{r}\right]}_{\mathrm{\infty }}=i{x}^0$$ Recall that ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ is the sum of terms of $x^j$ for $0\le j\le v$ or for $j=0$ since $v=0$. Looking at the series above, we see that $$a=i$$ Which is the coefficient of the term $x^0$. Now we need to find ${\alpha }_{\mathrm{\infty }}^{\pm }$ associated with ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$. For this we need to first find $b$ which is the coefficient of $x^{v-1}=\frac{1}{x}$ in $r$ minus the coefficient of $x^{v-1}=\frac{1}{x}$ in ${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2$. But$${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2=i^2=-1$$ Hence the coefficient of $x^{-1}$ in ${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2$ is $0$. To find the coefficient of $x^{-1}$ in $r$ long division is done$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{2-x^2}{x^2}\\ & =Q+\frac{R}{x^2}\end{array}$$

Where $Q$ is the quotient and $R$ is the remainder. This gives$$r=-1+\frac{2}{x^2}$$ For the other case of $v=0$ then the coefficient of $x^{-1}$ is found by looking up the coefficient in $R$ of $x$ to the degree of of $t$ then subtracting one and dividing result by $lcoeff\left(t\right)$. But degree of $t$ is $2$. Therefore we want the coefficient of $x^{2-1}$ or $x$ in $R$ which is zero. Hence $b=0-0=0$.

Now that we found $a,b$, then from the above section describing the algorithm, we see in this case that$$\begin{array}{rl}{\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)=\frac{1}{2}(0-0)=0\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)=\frac{1}{2}(-0-0)=0\end{array}$$

This completes step 1 of the solution. We have found ${\left[\sqrt{r}\right]}_c$ and its associated ${\alpha }_c^{\pm }$ and found ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$. Now we go to step 2 which is to find the $d^{\prime }s$.

step 2 Since we have a pole at zero, and we have one $O\left(\mathrm{\infty }\right)$, each with $\pm$ signs, then we set up this table to make it easier to work with. This implements$$d={\alpha }_{\mathrm{\infty }}^{\pm }-\sum _{i=1}^1{\alpha }_{c_i}^{\pm }$$ Therefore we obtain 4 possible $d$ values.

pole $c$	${\alpha }_c$ value	$O\left(\mathrm{\infty }\right)$ value	$d$	$d$ value
$x=0$	${\alpha }_c^{+}=2$	${\alpha }_{\mathrm{\infty }}^{+}=0$	${\alpha }_{\mathrm{\infty }}^{+}-\left({\alpha }_c^{+}\right)=0-\left(2\right)$	$-2$
$x=0$	${\alpha }_c^{+}=2$	${\alpha }_{\mathrm{\infty }}^{-}=0$	${\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_c^{+}\right)=0-\left(2\right)$	$-2$
$x=0$	${\alpha }_c^{-}=-1$	${\alpha }_{\mathrm{\infty }}^{+}=0$	${\alpha }_{\mathrm{\infty }}^{+}-\left({\alpha }_c^{-}\right)=0-(-1)$	$1$
$x=0$	${\alpha }_c^{-}=-1$	${\alpha }_{\mathrm{\infty }}^{-}=0$	${\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_c^{-}\right)=0-(-1)$	$1$

We see from the above that we took each pole in this problem (there is only one pole here at $x=0$) and its associated ${\alpha }_c^{\pm }$ with each ${\alpha }_{\mathrm{\infty }}^{\pm }$ and generated all possible $d$ values from all the combinations. Hence we obtain 4 possible $d$ values in this case. If we had 2 poles, then we would have 8 possible $d$ values. Hence the maximum possible $d$ values we can get is $2^{p+1}$ where $p$ is number of poles. Now we remove all negative $d$ values. Hence the trial $d$ values remaining is $$d=\left\{1\right\}$$ There is one $d$ value to try. We can pick any one of the two values of $d$ generated since there are both $d=1$. Both will give same solution. We generate $\omega$ using$$\omega =(\sum _{c}s\left(c\right){\left[\sqrt{r}\right]}_c+\frac{{\alpha }_c}{x-c})+s\left(\mathrm{\infty }\right){\left[\sqrt{r}\right]}_{\mathrm{\infty }}$$ To apply the above, we update the table above, now using only the first $d=1$ value in the above table. (selecting the second $d=1$ row, will not change the final solution). but we also add columns for ${\left[\sqrt{r}\right]}_c,{\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ to make the computation easier. Here is the new table

pole $c$	${\alpha }_c$ value	$s\left(c\right)$	${\left[\sqrt{r}\right]}_c$	$O\left(\mathrm{\infty }\right)$ value	$s\left(\mathrm{\infty }\right)$	${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$	$d$ value	$\omega$ value $(\sum _{c}s\left(c\right){\left[\sqrt{r}\right]}_c+\frac{{\alpha }_c}{x-c})+s\left(\mathrm{\infty }\right){\left[\sqrt{r}\right]}_{\mathrm{\infty }}$
$x=0$	${\alpha }_c^{-}=-1$	$-$	$0$	${\alpha }_{\mathrm{\infty }}^{+}=0$	$+$	$i$	$1$	$(-\left(0\right)+\frac{-1}{x-0})+(+)\left(i\right)=\frac{-1}{x}+i$

The above gives one candidate $\omega$ value to try. For this $\omega$ we need to find polynomial $P$ by solving $$\begin{array}{}\text{(8)}& P^{\prime \prime }+2\omega P^{\prime }+({\omega }^{\prime }+{\omega }^2-r)P=0\end{array}$$ If we are able to find $P$, then we stop and the ode $y^{\prime \prime }=ry$ is solved. If we try all candidate $\omega$ and can not find $P$ then this case is not successful and we go to the next case.

step 3 Now for each candidate $\omega$ we solve the above Eq (8). Starting with $\omega =\frac{-1}{x}+i$ associated with first $d=1$ in the table, then (8) becomes$$\begin{array}{rl}{P}^{\prime \prime }+2(\frac{-1}{x}+i)P^{\prime }+({(\frac{-1}{x}+i)}^{\prime }+{(\frac{-1}{x}+i)}^2-(\frac{2}{x^2}-1))P& =0\\ P^{\prime \prime }+2(\frac{-1}{x}+i)P^{\prime }+\left(\frac{-2i}{x}\right)P& =0\end{array}$$

This needs to be solved for $P$. Since degree of $p\left(x\right)$ is $d=1$. Let $p=x+a$. The above becomes$$\begin{array}{rl}2(\frac{-1}{x}+i)+\left(\frac{-2i}{x}\right)(x+a)& =0\\ \frac{-2}{x}+2i-2i-\frac{2ia}{x}& =0\\ \frac{-2}{x}-\frac{2ia}{x}& =0\end{array}$$

Which means$$a=i$$ Hence we found the polynomial $$p\left(x\right)=x+i$$ Therefore the solution to $y^{\prime \prime }=ry$ is$$\begin{array}{rl}y& =p{e}^{\int \omega dx}\\ & =(x+i)e^{\int \frac{-1}{x}+idx}\\ & =(x+i)\frac{1}{x}{e}^{ix}\\ & =\frac{x+i}{x}(\cos x+i\sin x)\end{array}$$

The second solution can be found by reduction of order. The full general solution to $y^{\prime \prime }=(\frac{2}{x^2}-1)y$ is$$y\left(x\right)=\frac{c_1}{x}(x\cos x-\sin x)+\frac{c_2}{x}(\cos x+x\sin x)$$

3.2.3 Example 3

Let $$y^{\prime \prime }=(x^2+3)y$$ Therefore $$\begin{array}{rl}r& =\frac{s}{t}\\ \text{(1)}& & =\frac{x^2+3}{1}\end{array}$$

Step 1 In this step we find all ${\left[\sqrt{r}\right]}_c$ and associated ${\alpha }_c^{\pm }$ for each pole. There are no poles. In this case${\alpha }_c^{\pm }=0$ (paper was not explicit in saying this, but from example 3 in paper this can be inferred). Hence the value of $d$ is decided by ${\alpha }_{\mathrm{\infty }}^{\pm }$ only in this case.

Now we consider $O\left(\mathrm{\infty }\right)$ which is $\mathrm{deg}\left(t\right)-\mathrm{deg}\left(s\right)=0-2=-2$. This falls in the case $-2v\le 0$. Hence $2v=-2$ or $$v=1$$ We need the Laurent series of $\sqrt{r}$ around $\mathrm{\infty }$. Using the computer this is $$x+\frac{3}{2x}-\frac{9}{8x^3}+\cdots$$ Hence we need the coefficient of $x^1$ in this series ($1$ because that is value of $v$). Recall that ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ is the sum of terms of $x^j$ for $0\le j\le v$ or for $j=0,1$ since $v=1$. Looking at the series above, we see that $$a=1$$ Which is the coefficient of the term $x$. There is no term $x^0$. Hence $${\left[\sqrt{r}\right]}_{\mathrm{\infty }}=x$$ Now we need to find ${\alpha }_{\mathrm{\infty }}^{\pm }$ associated with ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$. For this we need to first find $b$. Recall from above that $b$ is the coefficient of $x^{v-1}$ or $x^0$ in $r$ minus the coefficient of $x^{v-1}=x^0$ in ${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2$.  But ${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2=x^2$. Hence the coefficient of $x^0$ is zero. To find the coefficient of $x^0$ in $r$ long division is done$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{x^2+3}{1}\\ & =Q+\frac{R}{1}\end{array}$$

Where $Q$ is the quotient and $R$ is the remainder. This gives$$r=x^2+3+\frac{0}{1}$$ For the case of $v\ne 0$ then the coefficient is read from $Q$ above. Which is $3$. Hence$$\begin{array}{rl}b& =3-0\\ & =3\end{array}$$

Now that we found $a,b$, then from the above section describing the algorithm, we see in this case that$$\begin{array}{rl}{\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)=\frac{1}{2}(3-1)=1\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)=\frac{1}{2}(-3-1)=-2\end{array}$$

This completes step 1 of the solution. We have found ${\left[\sqrt{r}\right]}_c$ and its associated ${\alpha }_c^{\pm }$ (these are zero, in this example, since there are no poles) and found ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$. Now we go to step 2 which is to find the $d^{\prime }s$.

step 2 We set up this table to make it easier to work with. This implements$$d={\alpha }_{\mathrm{\infty }}^{\pm }-\sum _{i=1}^0{\alpha }_{c_i}^{\pm }$$ Therefore we obtain 2 possible $d$ values.

pole $c$	${\alpha }_c$ values (all zero)	$O\left(\mathrm{\infty }\right)$ value	$d$	$d$ value
$x=$N/A	${\alpha }_c=0$	${\alpha }_{\mathrm{\infty }}^{+}=1$	${\alpha }_{\mathrm{\infty }}^{+}=1$	$1$
$x=$N/A	${\alpha }_c=0$	${\alpha }_{\mathrm{\infty }}^{-}=-2$	${\alpha }_{\mathrm{\infty }}^{-}=-2$	$-2$

Picking the positive $d$ integers, this gives$$d=\left\{1\right\}$$ There is one $d$ value to try. We can pick any one of the two values of $d$ generated since there are both $d=1$. Both will give same solution. We generate $\omega$ using$$\omega =(\sum _{c}s\left(c\right){\left[\sqrt{r}\right]}_c+\frac{{\alpha }_c}{x-c})+s\left(\mathrm{\infty }\right){\left[\sqrt{r}\right]}_{\mathrm{\infty }}$$ To apply the above, we update the table above, now using only the first $d=1$ value in the above table. (selecting the first $d=1$ row). but we also add columns for ${\left[\sqrt{r}\right]}_c,{\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ to make the computation easier. Here is the new table

pole $c$	${\alpha }_c$ value	$s\left(c\right)$	${\left[\sqrt{r}\right]}_c$	$O\left(\mathrm{\infty }\right)$ value	$s\left(\mathrm{\infty }\right)$	${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$	$d$ value	$\omega$ value $(\sum _{c}s\left(c\right){\left[\sqrt{r}\right]}_c+\frac{{\alpha }_c}{x-c})+s\left(\mathrm{\infty }\right){\left[\sqrt{r}\right]}_{\mathrm{\infty }}$
$x=0$	${\alpha }_c=0$	$+$	$0$	${\alpha }_{\mathrm{\infty }}^{+}=1$	$+$	$x$	$1$	$(+\left(0\right)+0)+(+)\left(x\right)=x$

The above gives one candidate $\omega$ value to try. For this $\omega$ we need to find polynomial $P$ by solving $$\begin{array}{}\text{(8)}& P^{\prime \prime }+2\omega P^{\prime }+({\omega }^{\prime }+{\omega }^2-r)P=0\end{array}$$ If we are able to find $P$, then we stop and the ode $y^{\prime \prime }=ry$ is solved. If we try all candidate $\omega$ and can not find $P$ then this case is not successful and we go to the next case.

step 3 Now for each candidate $\omega$ (there is only one in this example) we solve the above Eq (8). Starting with $\omega =x$ associated with first $d=1$ in the table, then (8) becomes$$\begin{array}{rl}{P}^{\prime \prime }+2\left(x\right)P^{\prime }+({\left(x\right)}^{\prime }+{\left(x\right)}^2-(x^2+3))P& =0\\ P^{\prime \prime }+2x{P}^{\prime }+(1+x^2-x^2-3)P& =0\\ P^{\prime \prime }+2x{P}^{\prime }-2P& =0\end{array}$$

This needs to be solved for $P$. Since degree of $p\left(x\right)$ is $d=1$. Let $p=x+a$. The above becomes$$\begin{array}{rl}2x-2(x+a)& =0\\ 2x-2x-2a& =0\\ 2a& =0\end{array}$$

Which means$$a=0$$ Hence we found the polynomial $$p\left(x\right)=x$$ Therefore the solution to $y^{\prime \prime }=ry$ is$$\begin{array}{rl}y& =p{e}^{\int \omega dx}\\ & =x{e}^{\int xdx}\\ & =x{e}^{\frac{x^2}{2}}\end{array}$$

The second solution can be found by reduction of order. The full general solution to $y^{\prime \prime }=(x^2+3)y$ is$$y\left(x\right)=c_{1}x{e}^{\frac{x^2}{2}}+c_2(x{e}^{\frac{x^2}{2}}\sqrt{\pi }\mathrm{erf}\left(x\right)+e^{\frac{-x^2}{2}})$$

3.2.4 Example 4

Let $$y^{\prime \prime }=\frac{2}{x^2}y$$ Therefore $$\begin{array}{rl}r& =\frac{s}{t}\\ \text{(1)}& & =\frac{2}{x^2}\end{array}$$

Step 1 In this we find all ${\left[\sqrt{r}\right]}_c$ and associated ${\alpha }_c^{\pm }$ for each pole. There is one pole at $x=0$ of order 2. In this case, from the description of the algorithm earlier, we write$$\begin{array}{rl}{\left[\sqrt{r}\right]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4b}\\ {\alpha }_c^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4b}\end{array}$$

Where $b$ is the coefficient of $\frac{1}{{(x-0)}^2}$ in the partial fraction decomposition of $r$ which is $\frac{2}{x^2}$. Hence $b=2$. Therefore$$\begin{array}{rl}{\left[\sqrt{r}\right]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+8}=\frac{1}{2}+\frac{3}{2}=2\\ {\alpha }_c^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+8}=\frac{1}{2}-\frac{3}{2}=-1\end{array}$$

We are done with all the poles. Now we consider $O\left(\mathrm{\infty }\right)$ which is $\mathrm{deg}\left(t\right)-\mathrm{deg}\left(s\right)=2-0=2$. Since $O\left(\mathrm{\infty }\right)=2$ then from the algorithm above $${\left[\sqrt{r}\right]}_{\mathrm{\infty }}=0$$ Now we calculate $b$ for this case. This is given by the leading coefficient of $s$ divided by the leading coefficient of $t$ when $gcd(s,t)=1$. In this case $r=\frac{2}{x^2}$ , hence $b=\frac{2}{1}=2$. Therefore$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{\mathrm{\infty }}& =0\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4b}=\frac{1}{2}+\frac{1}{2}\sqrt{1+8}=2\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4b}=\frac{1}{2}-\frac{1}{2}\sqrt{1+8}=-1\end{array}$$

This completes step 1 of the solution. We have found ${\left[\sqrt{r}\right]}_c$ and its associated ${\alpha }_c^{\pm }$ and found ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$. Now we go to step 2 which is to find the $d^{\prime }s$.

step 2 Since we have a pole at zero, and we have one $O\left(\mathrm{\infty }\right)$, each with $\pm$ signs, then we set up this table to make it easier to work with. This implements$$d={\alpha }_{\mathrm{\infty }}^{\pm }-\sum _{i=1}^1{\alpha }_{c_i}^{\pm }$$ Therefore we obtain 4 possible $d$ values.

pole $c$	${\alpha }_c$ value	$O\left(\mathrm{\infty }\right)$ value	$d$	$d$ value
$x=0$	${\alpha }_c^{+}=2$	${\alpha }_{\mathrm{\infty }}^{+}=2$	${\alpha }_{\mathrm{\infty }}^{+}-\left({\alpha }_c^{+}\right)=2-\left(2\right)$	$0$
$x=0$	${\alpha }_c^{+}=2$	${\alpha }_{\mathrm{\infty }}^{-}=-1$	${\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_c^{+}\right)=-1-\left(2\right)$	$-3$
$x=0$	${\alpha }_c^{-}=-1$	${\alpha }_{\mathrm{\infty }}^{+}=2$	${\alpha }_{\mathrm{\infty }}^{+}-\left({\alpha }_c^{-}\right)=2-(-1)$	$3$
$x=0$	${\alpha }_c^{-}=-1$	${\alpha }_{\mathrm{\infty }}^{-}=-1$	${\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_c^{-}\right)=-1-(-1)$	$0$

Hence the trial $d$ values which are not negative are $$d=\{0,3\}$$ For $d=0$, since it shows in two rows, we take the first row. Now we generate $\omega$ for each $d$ using$$\omega =(\sum _{c}s\left(c\right){\left[\sqrt{r}\right]}_c+\frac{{\alpha }_c}{x-c})+s\left(\mathrm{\infty }\right){\left[\sqrt{r}\right]}_{\mathrm{\infty }}$$ To apply the above, we update the table above, but we also add columns for ${\left[\sqrt{r}\right]}_c,{\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ to make the computation easier. Here is the new table

pole $c$	${\alpha }_c$ value	$s\left(c\right)$	${\left[\sqrt{r}\right]}_c$	$O\left(\mathrm{\infty }\right)$ value	$s\left(\mathrm{\infty }\right)$	${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$	$d$ value	$\omega$ value $(\sum _{c}s\left(c\right){\left[\sqrt{r}\right]}_c+\frac{{\alpha }_c}{x-c})+s\left(\mathrm{\infty }\right){\left[\sqrt{r}\right]}_{\mathrm{\infty }}$
$x=0$	${\alpha }_c^{+}=2$	$+$	$0$	${\alpha }_{\mathrm{\infty }}^{+}=2$	$+$	$0$	$0$	$(+\left(0\right)+\frac{2}{x-0})+(+)\left(0\right)=\frac{2}{x}$
$x=0$	${\alpha }_c^{-}=-1$	$-$	$0$	${\alpha }_{\mathrm{\infty }}^{+}=2$	$+$	$0$	$3$	$(-\left(0\right)+\frac{-1}{x-0})+(+)\left(0\right)=\frac{-1}{x}$

The above gives two candidate $\omega =\{\frac{2}{x},\frac{-1}{x}\}$ value to try. For this $\omega$ we need to find polynomial $P$ by solving $$\begin{array}{}\text{(8)}& P^{\prime \prime }+2\omega P^{\prime }+({\omega }^{\prime }+{\omega }^2-r)P=0\end{array}$$ If we are able to find $P$, then we stop and the ode $y^{\prime \prime }=ry$ is solved. If we try all candidate $\omega$ and can not find $P$ then this case is not successful and we go to the next case.

step 3 Now for each candidate $\omega$ we solve the above Eq (8). Starting with $\omega =\frac{2}{x}$ associated with first $d=0$ in the table, then (8) becomes$$\begin{array}{rl}{P}^{\prime \prime }+2\left(\frac{2}{x}\right)P^{\prime }+({\left(\frac{2}{x}\right)}^{\prime }+{\left(\frac{2}{x}\right)}^2-\left(\frac{2}{x^2}\right))P& =0\\ P^{\prime \prime }+\frac{4}{x}{P}^{\prime }+(-\frac{2}{x^2}+\frac{4}{x^2}-\frac{2}{x^2})P& =0\\ P^{\prime \prime }+\frac{4}{x}{P}^{\prime }& =0\end{array}$$

This needs to be solved for $P$. Since degree of $p\left(x\right)$ is $d=0$. Let $p=a$. The above becomes$$0=0$$ No unique solution. Hence $d=0$ did not work. Now we try the second $\omega =\frac{-1}{x}$ associated with $d=3$. Substituting in 8 gives$$\begin{array}{rl}{P}^{\prime \prime }+2\left(\frac{-1}{x}\right)P^{\prime }+({\left(\frac{-1}{x}\right)}^{\prime }+{\left(\frac{-1}{x}\right)}^2-\left(\frac{2}{x^2}\right))P& =0\\ P^{\prime \prime }+\frac{-2}{x}{P}^{\prime }+(\frac{1}{x^2}+\frac{1}{x^2}-\frac{2}{x^2})P& =0\\ P^{\prime \prime }-\frac{2}{x}{P}^{\prime }& =0\end{array}$$

Since $d=3$, let $$p\left(x\right)=x^3+a{x}^2+bx+c$$ Then $P^{\prime \prime }-\frac{2}{x}{P}^{\prime }=0$ becomes$$\begin{array}{rl}(6x+2a)-\frac{2}{x}(3x^2+2ax+b)& =0\\ -2a-2\frac{b}{x}& =0\end{array}$$

Hence $a=0,b=0$ is solution. $c$ is arbitrary. Taking $c=0$ then$$p\left(x\right)=x^3$$ Therefore the solution to $y^{\prime \prime }=ry$ is$$\begin{array}{rl}y& =p\left(x\right)e^{\int \omega dx}\\ & =x^3{e}^{\int \frac{-1}{x}dx}\\ & =x^3{e}^{-\ln x}\\ & =x^2\end{array}$$

The second solution can be found by reduction of order. The full general solution to $y^{\prime \prime }=\frac{2}{x^2}y$ is$$y\left(x\right)=c_1\frac{1}{x}+c_2{x}^2$$

3.2.5 Example 5

Let $$y^{\prime \prime }=\frac{32x^2-27x+27}{144x^4-288x^3+144x^2}y$$ Therefore $$\begin{array}{rl}r& =\frac{s}{t}\\ \text{(1)}& & =\frac{32x^2-27x+27}{144x^4-288x^3+144x^2}\text{(2)}& & =\frac{3}{16x}-\frac{3}{16}\frac{1}{(x-1)}+\frac{3}{16}\frac{1}{x^2}+\frac{2}{9}\frac{1}{{(x-1)}^2}\end{array}$$

The necessary conditions for case 1 are met.

Step 1 In this we find all ${\left[\sqrt{r}\right]}_c$ and associated ${\alpha }_c^{\pm }$ for each pole. There is one pole at $x=0$ of order 2 and one pole at $x=1$ of order 2. For the pole at $x=0$ since order is $2$ then$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c=0}& =0\\ {\alpha }_{c=0}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4b}\\ {\alpha }_{c=0}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4b}\end{array}$$

Where $b$ is the coefficient of $\frac{1}{{(x-0)}^2}$ in the partial fraction decomposition of $r$ given in Eq (2) which is $\frac{3}{16}$. Hence $b=\frac{3}{16}$. Therefore$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c=0}& =0\\ {\alpha }_{c=0}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4\left(\frac{3}{16}\right)}=\frac{1}{4}\sqrt{7}+\frac{1}{2}\\ {\alpha }_{c=0}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4\left(\frac{3}{16}\right)}=\frac{1}{2}-\frac{1}{4}\sqrt{7}\end{array}$$

And for the pole at $x=1$ which is order 2, $$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c=1}& =0\\ {\alpha }_{c=1}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4b}\\ {\alpha }_{c=1}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4b}\end{array}$$

Where $b$ is the coefficient of $\frac{1}{{(x-1)}^2}$ in the partial fraction decomposition of $r$ given in Eq (2) which is $\frac{2}{9}$. Hence $b=\frac{2}{9}$. Therefore the above becomes$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c=1}& =0\\ {\alpha }_{c=1}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4\left(\frac{2}{9}\right)}=\frac{1}{6}\sqrt{17}+\frac{1}{2}\\ {\alpha }_{c=1}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4\left(\frac{2}{9}\right)}=\frac{1}{2}-\frac{1}{6}\sqrt{17}\end{array}$$

We are done with all the poles. Now we consider $O\left(\mathrm{\infty }\right)$ which is $\mathrm{deg}\left(t\right)-\mathrm{deg}\left(s\right)=4-2=2$. Since $O\left(\mathrm{\infty }\right)=2$ then from the algorithm above $${\left[\sqrt{r}\right]}_{\mathrm{\infty }}=0$$ Now we calculate $b$ for this case. This is given by the leading coefficient of $s$ divided by the leading coefficient of $t$ when $gcd(s,t)=1$. In this case $r=\frac{32x^2-27x+27}{144x^4-288x^3+144x^2}$ from Eq (1) , hence $b=\frac{32}{144}=\frac{2}{9}$. Therefore$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{\mathrm{\infty }}& =0\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4b}=\frac{1}{2}+\frac{1}{2}\sqrt{1+4\left(\frac{2}{9}\right)}=\frac{1}{6}\sqrt{17}+\frac{1}{2}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4b}=\frac{1}{2}-\frac{1}{2}\sqrt{1+4\left(\frac{2}{9}\right)}=\frac{1}{2}-\frac{1}{6}\sqrt{17}\end{array}$$

This completes step 1 of the solution. We have found ${\left[\sqrt{r}\right]}_c$ and its associated ${\alpha }_c^{\pm }$ and found ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$. Now we go to step 2 which is to find the $d^{\prime }s$.

step 2 Since we have a pole at $x=c_1=0$ and pole at $x=c_2=1$, and we have one $O\left(\mathrm{\infty }\right)$, each with $\pm$ signs. The following now implements $$d={\alpha }_{\mathrm{\infty }}^{\pm }-\sum _{i=1}^2{\alpha }_{c_i}^{\pm }$$ By trying all possible combinations. There are 8 possible $d$ values. This gives$$\begin{array}{rl}{d}_1& ={\alpha }_{\mathrm{\infty }}^{+}-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{+})=(\frac{1}{6}\sqrt{17}+\frac{1}{2})-(\frac{1}{4}\sqrt{7}+\frac{1}{2})-(\frac{1}{6}\sqrt{17}+\frac{1}{2})=-\frac{1}{4}\sqrt{7}-\frac{1}{2}\\ d_2& ={\alpha }_{\mathrm{\infty }}^{+}-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{-})=(\frac{1}{6}\sqrt{17}+\frac{1}{2})-(\frac{1}{4}\sqrt{7}+\frac{1}{2})-(\frac{1}{2}-\frac{1}{6}\sqrt{17})=\frac{1}{3}\sqrt{17}-\frac{1}{4}\sqrt{7}-\frac{1}{2}\\ d_3& ={\alpha }_{\mathrm{\infty }}^{+}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{+})=(\frac{1}{6}\sqrt{17}+\frac{1}{2})-(\frac{1}{2}-\frac{1}{4}\sqrt{7})-(\frac{1}{6}\sqrt{17}+\frac{1}{2})=\frac{1}{4}\sqrt{7}-\frac{1}{2}\\ d_4& ={\alpha }_{\mathrm{\infty }}^{+}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{-})=(\frac{1}{6}\sqrt{17}+\frac{1}{2})-(\frac{1}{2}-\frac{1}{4}\sqrt{7})-(\frac{1}{2}-\frac{1}{6}\sqrt{17})=\frac{1}{4}\sqrt{7}+\frac{1}{3}\sqrt{17}-\frac{1}{2}\\ d_5& ={\alpha }_{\mathrm{\infty }}^{-}-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{+})=(\frac{1}{2}-\frac{1}{6}\sqrt{17})-(\frac{1}{4}\sqrt{7}+\frac{1}{2})-(\frac{1}{6}\sqrt{17}+\frac{1}{2})=-\frac{1}{4}\sqrt{7}-\frac{1}{3}\sqrt{17}-\frac{1}{2}\\ d_6& ={\alpha }_{\mathrm{\infty }}^{-}-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{-})=(\frac{1}{2}-\frac{1}{6}\sqrt{17})-(\frac{1}{4}\sqrt{7}+\frac{1}{2})-(\frac{1}{2}-\frac{1}{6}\sqrt{17})=-\frac{1}{4}\sqrt{7}-\frac{1}{2}\\ d_7& ={\alpha }_{\mathrm{\infty }}^{-}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{+})=(\frac{1}{2}-\frac{1}{6}\sqrt{17})-(\frac{1}{2}-\frac{1}{4}\sqrt{7})-(\frac{1}{6}\sqrt{17}+\frac{1}{2})=\frac{1}{4}\sqrt{7}-\frac{1}{3}\sqrt{17}-\frac{1}{2}\\ d_8& ={\alpha }_{\mathrm{\infty }}^{-}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{-})=(\frac{1}{2}-\frac{1}{6}\sqrt{17})-(\frac{1}{2}-\frac{1}{4}\sqrt{7})-(\frac{1}{2}-\frac{1}{6}\sqrt{17})=\frac{1}{4}\sqrt{7}-\frac{1}{2}\end{array}$$

None of the $d$ found are integer. Hence case 1 did not work we need to try case 2 and if that also fail, try case 3. We will find all three cases fail on this ode..

3.2.6 Example 6

Let $$y^{\prime \prime }=y$$ Therefore $$r=\frac{s}{t}=1$$ The necessary conditions for case 1 are met since zero order pole and $O\left(\mathrm{\infty }\right)=0$.

Step 1 In this we find all ${\left[\sqrt{r}\right]}_c$ and associated ${\alpha }_c^{\pm }$ for each pole. There are no poles. In this case${\left[\sqrt{r}\right]}_c=0$ and ${\alpha }_c^{\pm }=0$. Since $O\left(\mathrm{\infty }\right)=0$, we are in case $2v\le 0$. Hence $v=0$. Then now ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ is the sum of all terms $x^i$ for $0\le i\le v$ in the Laurent series expansion of $\sqrt{r}$ at $\mathrm{\infty }$.$${\left[\sqrt{r}\right]}_{\mathrm{\infty }}=1$$ Hence $a=1$.  And $b$ is the coefficient of $x^{v-1}=x^{-1}$ in $r$ minus the coefficient of $x^{v-1}=x^{-1}$ in ${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2$. Hence $b=0$. Then $$\begin{array}{rl}{\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)=0\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)=0\end{array}$$

This completes step 1 of the solution. We have found ${\left[\sqrt{r}\right]}_c$ and its associated ${\alpha }_c^{\pm }$ and found ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$. Now we go to step 2 which is to find the $d^{\prime }s$.

step 2 Since we have a pole at zero and pole at $x=1$, and we have one $O\left(\mathrm{\infty }\right)$, each with $\pm$ signs, then we set up this table to make it easier to work with. This implements$$d={\alpha }_{\mathrm{\infty }}^{\pm }-\sum _{i=1}^0{\alpha }_{c_i}^{\pm }$$ Therefore we obtain 2 possible $d$ values.

pole $c$	${\alpha }_c$ value	$O\left(\mathrm{\infty }\right)$ value	$d$	$d$ value
$x=NA$	${\alpha }_c=0$	${\alpha }_{\mathrm{\infty }}^{+}=0$	${\alpha }_{\mathrm{\infty }}^{+}=0$	$0$
$x=NA$	${\alpha }_c=0$	${\alpha }_{\mathrm{\infty }}^{-}=0$	${\alpha }_{\mathrm{\infty }}^{+}=0$	$0$

Hence the trial $d$ values which are not negative integers are $$d=\left\{0\right\}$$ For $d=0$, since it shows in two rows, we take the first row. Now we generate $\omega$ for each $d$ using$$\omega =(\sum _{c}s\left(c\right){\left[\sqrt{r}\right]}_c+\frac{{\alpha }_c}{x-c})+s\left(\mathrm{\infty }\right){\left[\sqrt{r}\right]}_{\mathrm{\infty }}$$ To apply the above, we update the table above, but we also add columns for ${\left[\sqrt{r}\right]}_c,{\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ to make the computation easier. Here is the new table

pole $c$	${\alpha }_c$ value	$s\left(c\right)$	${\left[\sqrt{r}\right]}_c$	$O\left(\mathrm{\infty }\right)$ value	$s\left(\mathrm{\infty }\right)$	${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$	$d$ value	$\omega$ value $(\sum _{c}s\left(c\right){\left[\sqrt{r}\right]}_c+\frac{{\alpha }_c}{x-c})+s\left(\mathrm{\infty }\right){\left[\sqrt{r}\right]}_{\mathrm{\infty }}$
$x=NA$	${\alpha }_c=0$	$+$	$0$	${\alpha }_{\mathrm{\infty }}^{+}=0$	$+$	$0$	$0$	$(+\left(0\right)+0)+(+)\left(1\right)=1$

The above gives candidate $\omega =1$ value to try. For this $\omega$ we need to find polynomial $P$ by solving $$\begin{array}{}\text{(8)}& P^{\prime \prime }+2\omega P^{\prime }+({\omega }^{\prime }+{\omega }^2-r)P=0\end{array}$$ If we are able to find $P$, then we stop and the ode $y^{\prime \prime }=ry$ is solved.

step 3 Now for each candidate $\omega$ we solve the above Eq (8). Starting with $\omega =1$ associated with $d=0$ in the table. Let $p\left(x\right)=1$ since degree is zero, then (8) becomes$$\begin{array}{rl}{P}^{\prime \prime }+2\left(1\right)P^{\prime }+({\left(1\right)}^{\prime }+{\left(1\right)}^2-\left(1\right))P& =0\\ (0+1-1)& =0\\ 0& =0\end{array}$$

Hence $p\left(x\right)=1$ is valid solution. Therefore the solution to $y^{\prime \prime }=y$ is$$\begin{array}{rl}y& =p\left(x\right)e^{\int \omega dx}\\ & =e^{\int 1dx}\\ & =e^x\end{array}$$

The second solution can be found by reduction of order. The full general solution to $y^{\prime \prime }=y$ is$$y\left(x\right)=c_1{e}^x+c_2{e}^{-x}$$

3.2.7 Example 7

Let $$(x^2-2x)y^{\prime \prime }+(2-x^2)y^{\prime }+(2x-2)y=0$$ Normalizing so that coefficient of $y^{\prime \prime }$ is one gives$$\begin{array}{rl}{y}^{\prime \prime }+\frac{(2-x^2)}{(x^2-2x)}{y}^{\prime }+\frac{(2x-2)}{(x^2-2x)}y& =0\\ \text{(1)}& y^{\prime \prime }+a{y}^{\prime }\left(x\right)+by& =0\end{array}$$

Hence $$\begin{array}{rl}a& =\frac{(2-x^2)}{(x^2-2x)}\\ b& =\frac{(2x-2)}{(x^2-2x)}\end{array}$$

It is first transformed to the following ode by eliminating the first derivative $$\begin{array}{}\text{(2)}& z^{\prime \prime }=rz\end{array}$$ Using what is known as the Liouville transformation given by$$\begin{array}{}\text{(3)}& y=z{e}^{\frac{-1}{2}\int adx}\end{array}$$ Where it can be found that $r$ in (2) is given by $$\begin{array}{rl}r& =\frac{1}{4}{a}^2+\frac{1}{2}{a}^{\prime }-b\\ & =\frac{1}{4}{\left(\frac{(2-x^2)}{(x^2-2x)}\right)}^2+\frac{1}{2}\frac{d}{dx}\left(\frac{(2-x^2)}{(x^2-2x)}\right)-\frac{(2x-2)}{(x^2-2x)}\\ \text{(4)}& & =\frac{(x^4-8x^3+24x^2-24x+12)}{4x^2{(x-2)}^2}\end{array}$$

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is $$\begin{array}{}\text{(5)}& z^{\prime \prime }=\frac{(x^4-8x^3+24x^2-24x+12)}{4x^2{(x-2)}^2}z\end{array}$$ Therefore $$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{(x^4-8x^3+24x^2-24x+12)}{4x^2{(x-2)}^2}\end{array}$$

The necessary conditions for case 1 are met.

Step 1 In this we find all ${\left[\sqrt{r}\right]}_c$ and associated ${\alpha }_c^{\pm }$ for each pole. There is one pole at $x=0$ of order 2 and one pole at $x=2$ of order 2. For the pole at $x=0$ since order is $2$ then$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c=0}& =0\\ {\alpha }_{c=0}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4b}\\ {\alpha }_{c=0}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4b}\end{array}$$

Where $b$ is the coefficient of $\frac{1}{{(x-0)}^2}$ in the partial fraction decomposition of $r$ which is $$\begin{array}{}\text{(6)}& \frac{x^4-8x^3+24x^2-24x+12}{4x^2{(x-2)}^2}=\frac{1}{4}-\frac{3}{4}\frac{1}{x}-\frac{1}{4}\frac{1}{(x-2)}+\frac{3}{4}\frac{1}{{(x-2)}^2}+\frac{3}{4}\frac{1}{x^2}\end{array}$$ Hence $b=\frac{3}{4}$. Therefore$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c=0}& =0\\ {\alpha }_{c=0}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4\left(\frac{3}{4}\right)}=\frac{3}{2}\\ {\alpha }_{c=0}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4\left(\frac{3}{4}\right)}=-\frac{1}{2}\end{array}$$

And for the pole at $x=2$ which is order 2, $$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c=2}& =0\\ {\alpha }_{c=2}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4b}\\ {\alpha }_{c=2}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4b}\end{array}$$

Where $b$ is the coefficient of $\frac{1}{{(x-2)}^2}$ in the partial fraction decomposition of $r$ given in Eq (6). Hence $b=\frac{3}{4}$. Therefore the above becomes$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c=2}& =0\\ {\alpha }_{c=2}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4\left(\frac{3}{4}\right)}=\frac{3}{2}\\ {\alpha }_{c=2}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4\left(\frac{3}{4}\right)}=-\frac{1}{2}\end{array}$$

We are done with all the poles. Now we consider $O\left(\mathrm{\infty }\right)$ which is $\mathrm{deg}\left(t\right)-\mathrm{deg}\left(s\right)=4-4=0$. Since $O\left(\mathrm{\infty }\right)=0$, we are in case $2v\le 0$. Hence $v=0$. Then now ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ is the sum of all terms $x^i$ for $0\le i\le v$ in the Laurent series expansion of $\sqrt{r}$ at $\mathrm{\infty }$ which is $$\begin{array}{}\text{(7)}& {\left[\sqrt{r}\right]}_{\mathrm{\infty }}=\frac{1}{2}-\frac{1}{x}+\frac{2}{x^3}+\frac{11}{x^4}+\cdots \end{array}$$ We want only terms for $0\le i\le v$ but $$v=0$$ Therefore only the constant term. Hence$$\begin{array}{}\text{(8)}& {\left[\sqrt{r}\right]}_{\mathrm{\infty }}=\frac{1}{2}\end{array}$$ Which means $$a=\frac{1}{2}$$ As it is the the term that matches ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}=a{x}^v+\cdots$. Hence ${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2=\frac{1}{4}$ and the coefficient of $\frac{1}{x}$ is zero. To find the coefficient of $\frac{1}{x}$ in $r$ long division is done$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{x^4-8x^3+24x^2-24x+12}{4x^4-16x^3+16x^2}\\ & =Q+\frac{R}{4x^4-16x^3+16x^2}\end{array}$$

Where $Q$ is the quotient and $R$ is the remainder. This gives$$r=\frac{1}{4}+\frac{-4x^3+20x^2-24x+12}{4x^4-16x^3+16x^2}$$ Since $v=0$ then the coefficient of $x^{-1}$ in $r$ is found using $\frac{lcoeff\left(R\right)}{lcoeff\left(t\right)}$. This gives $-1$ as seen from above. Hence $b=-1-0=-1$. Therefore$$\begin{array}{rl}{\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)=\frac{1}{2}(\frac{-1}{\frac{1}{2}}-0)=-1\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)=\frac{1}{2}(-\frac{-1}{\frac{1}{2}}-0)=1\end{array}$$

This completes step 1 of the solution. We have found ${\left[\sqrt{r}\right]}_c$ and its associated ${\alpha }_c^{\pm }$ and found ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$. Now we go to step 2 which is to find the $d^{\prime }s$.

step 2 Since we have a pole at $x=c_1=0$ and pole at $x=c_2=1$, and we have one $O\left(\mathrm{\infty }\right)$, each with $\pm$ signs. The following now implements $$d={\alpha }_{\mathrm{\infty }}^{\pm }-\sum _{i=1}^2{\alpha }_{c_i}^{\pm }$$ By trying all possible combinations. There are 8 possible $d$ values. This gives$$\begin{array}{rl}{d}_1& =1-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{+})=1-(\frac{3}{2}+\frac{3}{2})=-2\\ d_2& =1-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{-})=1-(\frac{3}{2}-\frac{1}{2})=0\\ d_3& =1-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{+})=1-(-\frac{1}{2}+\frac{3}{2})=0\\ d_4& =1-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{-})=1-(-\frac{1}{2}-\frac{1}{2})=2\\ d_5& =-1-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{+})=-1-(\frac{3}{2}+\frac{3}{2})=-4\\ d_6& =-1-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{-})=-1-(\frac{3}{2}-\frac{1}{2})=-2\\ d_7& =-1-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{+})=-1-(-\frac{1}{2}+\frac{3}{2})=-2\\ d_8& =-1-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{-})=-1-(-\frac{1}{2}-\frac{1}{2})=0\end{array}$$

Need to complete the solution next.

3.2.8 Example 8

Let $$(x^2+1)y^{\prime \prime }+2x{y}^{\prime }-2y=0$$ Normalizing so that coefficient of $y^{\prime \prime }$ is one gives$$\begin{array}{rl}{y}^{\prime \prime }+\frac{2x}{(x^2+1)}{y}^{\prime }-\frac{2}{(x^2+1)}y& =0\\ \text{(1)}& y^{\prime \prime }+a{y}^{\prime }\left(x\right)+by& =0\end{array}$$

Hence $$\begin{array}{rl}a& =\frac{2x}{(x^2+1)}\\ b& =-\frac{2}{(x^2+1)}\end{array}$$

It is first transformed to the following ode by eliminating the first derivative $$\begin{array}{}\text{(2)}& z^{\prime \prime }=rz\end{array}$$ Using what is known as the Liouville transformation given by$$\begin{array}{}\text{(3)}& y=z{e}^{\frac{-1}{2}\int adx}\end{array}$$ Where it can be found that $r$ in (2) is given by $$\begin{array}{rl}r& =\frac{1}{4}{a}^2+\frac{1}{2}{a}^{\prime }-b\\ & =\frac{1}{4}{\left(\frac{2x}{(x^2+1)}\right)}^2+\frac{1}{2}\frac{d}{dx}\left(\frac{2x}{(x^2+1)}\right)-(-\frac{2}{(x^2+1)})\\ \text{(4)}& & =\frac{2x^2+3}{{(x^2+1)}^2}\end{array}$$

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is $$\begin{array}{}\text{(5)}& z^{\prime \prime }=\frac{2x^2+3}{{(x^2+1)}^2}z\end{array}$$ Therefore $$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{2x^2+3}{{(x^2+1)}^2}\\ \text{(5A)}& & =\frac{2x^2+3}{x^4+2x^2+1}\end{array}$$

The necessary conditions for case 1 are met.

Step 1 In this we find all ${\left[\sqrt{r}\right]}_c$ and associated ${\alpha }_c^{\pm }$ for each pole. There is one pole at $x=-i$ of order 2 and one pole at $x=i$ of order 2. For the pole at $x=-i$ since order is $2$ then$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c_1}& =0\\ {\alpha }_{c_1}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4b}\\ {\alpha }_{c_1}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4b}\end{array}$$

Where $b$ is the coefficient of $\frac{1}{{(x+i)}^2}$ in the partial fraction decomposition of $r$ which is (in Maple this can be found using fullparfrac. $$\begin{array}{}\text{(6)}& \frac{2x^2+3}{{(x^2+1)}^2}=-\frac{1}{4}\frac{1}{{(x-i)}^2}-\frac{1}{4}\frac{1}{{(x+i)}^2}-\frac{5i}{4}\frac{1}{(x-i)}+\frac{5i}{4}\frac{1}{x+i}\end{array}$$ Hence $b=-\frac{1}{4}$. Therefore$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c=0}& =0\\ {\alpha }_{c_1}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4(-\frac{1}{4})}=\frac{1}{2}\\ {\alpha }_{c_1}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4(-\frac{1}{4})}=\frac{1}{2}\end{array}$$

And for the pole at $x=+i$ which is order 2, $$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c_2}& =0\\ {\alpha }_{c_2}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4b}\\ {\alpha }_{c_2}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4b}\end{array}$$

Where $b$ is the coefficient of $\frac{1}{{(x-i)}^2}$ in the partial fraction decomposition of $r$ given in Eq (6). Hence $b=-\frac{1}{4}$. Therefore the above becomes$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c=2}& =0\\ {\alpha }_{c=2}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4(-\frac{1}{4})}=\frac{1}{2}\\ {\alpha }_{c=2}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4(-\frac{1}{4})}=\frac{1}{2}\end{array}$$

We are done with all the poles. Now we consider $O\left(\mathrm{\infty }\right)$ which is $\mathrm{deg}\left(t\right)-\mathrm{deg}\left(s\right)=4-2=2$. Since $O\left(\mathrm{\infty }\right)=2$, then ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}=0$. Now $b$ is the coefficient of $\frac{1}{x^2}$ in $r$ minus coefficient of $\frac{1}{x^2}$ in ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}^2$ which is zero. the coefficient of $\frac{1}{x^2}$ in $r$ is found from $\frac{lcoeff\left(s\right)}{lcoeff\left(t\right)}$ which from Eq (5A) above is $\frac{2}{1}=2$. Hence $b=2-0=2.$$$\begin{array}{rl}{\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4b}=\frac{1}{2}+\frac{1}{2}\sqrt{1+8}=2\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4b}=\frac{1}{2}-\frac{1}{2}\sqrt{1+8}=-1\end{array}$$

This completes step 1 of the solution. We have found ${\left[\sqrt{r}\right]}_c$ and its associated ${\alpha }_c^{\pm }$ and found ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$. Now we go to step 2 which is to find the $d^{\prime }s$.

step 2 Since we have a pole at $x=-i$ and pole at $x=+i$ each of order 2, and we have one $O\left(\mathrm{\infty }\right)$, each with $\pm$ signs. The following now implements $$d={\alpha }_{\mathrm{\infty }}^{\pm }-\sum _{i=1}^2{\alpha }_{c_i}^{\pm }$$ By trying all possible combinations. There are 8 possible $d$ values. This gives$$\begin{array}{rl}{d}_1& =2-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{+})=2-(\frac{1}{2}+\frac{1}{2})=1\\ d_2& =2-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{-})=2-(\frac{1}{2}+\frac{1}{2})=1\\ d_3& =2-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{+})=2-(\frac{1}{2}+\frac{1}{2})=1\\ d_4& =2-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{-})=2-(\frac{1}{2}+\frac{1}{2})=1\\ d_5& =-1-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{+})=-1-(\frac{1}{2}+\frac{1}{2})=-2\\ d_6& =-1-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{-})=-1-(\frac{1}{2}+\frac{1}{2})=-2\\ d_7& =-1-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{+})=-1-(\frac{1}{2}+\frac{1}{2})=-2\\ d_8& =-1-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{-})=-1-(\frac{1}{2}+\frac{1}{2})=-2\end{array}$$

Need to complete the solution next.

3.2.9 Example 9

Let $$(1-x)y^{\prime \prime }+x{y}^{\prime }-y=0$$ Normalizing so that coefficient of $y^{\prime \prime }$ is one gives$$\begin{array}{rl}{y}^{\prime \prime }+\frac{x}{(1-x)}{y}^{\prime }-\frac{1}{(1-x)}y& =0\\ \text{(1)}& y^{\prime \prime }+a{y}^{\prime }\left(x\right)+by& =0\end{array}$$

Hence $$\begin{array}{rl}a& =\frac{x}{(1-x)}\\ b& =-\frac{1}{(1-x)}\end{array}$$

It is first transformed to the following ode by eliminating the first derivative $$\begin{array}{}\text{(2)}& z^{\prime \prime }=rz\end{array}$$ Using what is known as the Liouville transformation given by$$\begin{array}{}\text{(3)}& y=z{e}^{\frac{-1}{2}\int adx}\end{array}$$ Where it can be found that $r$ in (2) is given by $$\begin{array}{rl}r& =\frac{1}{4}a+\frac{1}{2}{a}^{\prime }-b\\ & =\frac{1}{4}{\left(\frac{x}{(1-x)}\right)}^2+\frac{1}{2}\frac{d}{dx}\left(\frac{x}{(1-x)}\right)-(-\frac{1}{(1-x)})\\ \text{(4)}& & =\frac{x^2-4x+6}{4{(x-1)}^2}\end{array}$$

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is $$\begin{array}{}\text{(5)}& z^{\prime \prime }=\frac{x^2-4x+6}{4{(x-1)}^2}z\end{array}$$ Therefore $$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{x^2-4x+6}{4{(x-1)}^2}\\ \text{(5A)}& & =\frac{x^2-4x+6}{4x^2-8x+4}\end{array}$$

The necessary conditions for case 1 are met.

Step 1 In this we find all ${\left[\sqrt{r}\right]}_c$ and associated ${\alpha }_c^{\pm }$ for each pole. There is one pole at $x=1$ of order 2. Hence$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c_1}& =0\\ {\alpha }_{c_1}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4b}\\ {\alpha }_{c_1}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4b}\end{array}$$

Where $b$ is the coefficient of $\frac{1}{{(x-1)}^2}$ in the partial fraction decomposition of $r$ which is (in Maple this can be found using fullparfrac). $$\begin{array}{}\text{(6)}& \frac{x^2-4x+6}{4{(x-1)}^2}=\frac{1}{4}+\frac{3}{4{(x-1)}^2}-\frac{1}{2}\frac{1}{x-1}\end{array}$$ Hence $b=\frac{3}{4}$. Therefore$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c_1}& =0\\ {\alpha }_{c_1}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4\left(\frac{3}{4}\right)}=\frac{3}{2}\\ {\alpha }_{c_1}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4\left(\frac{3}{4}\right)}=-\frac{1}{2}\end{array}$$

We are done with all the poles. Now we consider $O\left(\mathrm{\infty }\right)$ which is $\mathrm{deg}\left(t\right)-\mathrm{deg}\left(s\right)=2-2=0$. Since $O\left(\mathrm{\infty }\right)=0$, we are in case $2v\le 0$. Hence $$v=0$$ Then now ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ is the sum of all terms $x^i$ for $0\le i\le v$ in the Laurent series expansion of $\sqrt{r}$ at $\mathrm{\infty }$ which is $$\begin{array}{}\text{(7)}& {\left[\sqrt{r}\right]}_{\mathrm{\infty }}=\frac{1}{2}-\frac{1}{2x}+\frac{1}{x^3}+\frac{11}{4x^4}+\cdots \end{array}$$ We want only terms for $0\le i\le v$ but $v=0$. Therefore only the constant term. Hence$$\begin{array}{}\text{(8)}& {\left[\sqrt{r}\right]}_{\mathrm{\infty }}=\frac{1}{2}\end{array}$$ Which means $$a=\frac{1}{2}$$ As it is the term that matches ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}=a{x}^v+\cdots$. $$ Now we need to find $b$. This will be the coefficient of $x^{v-1}=\frac{1}{x}$ in $r-{\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2$. But ${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2=\frac{1}{4}$. So coefficient of $\frac{1}{x}$ is zero in ${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2$.  To find the coefficient of $\frac{1}{x}$ in $r$ long division is done$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{x^2-4x+6}{4x^2-8x+4}\\ & =Q+\frac{R}{4x^2-8x+4}\end{array}$$

Where $Q$ is the quotient and $R$ is the remainder. This gives$$r=\frac{1}{4}+\frac{-2x+5}{4x^2-8x+4}$$ Hence the coefficient of $\frac{1}{x}$ is $\frac{lcoeff\left(R\right)}{lcoeff\left(t\right)}=\frac{-2}{4}=-\frac{1}{2}$. Therefore $b=-\frac{1}{2}-0=-\frac{1}{2}$. Hence$$\begin{array}{rl}{\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)=\frac{1}{2}(\frac{-\frac{1}{2}}{\frac{1}{2}}-0)=-\frac{1}{2}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)=\frac{1}{2}(-\frac{-\frac{1}{2}}{\frac{1}{2}}-0)=\frac{1}{2}\end{array}$$

This completes step 1 of the solution. We have found ${\left[\sqrt{r}\right]}_c$ and its associated ${\alpha }_c^{\pm }$ and found ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$. Now we go to step 2 which is to find the $d^{\prime }s$.

step 2 Now $d$ is found using$$d={\alpha }_{\mathrm{\infty }}^{\pm }-\sum _{i=1}^1{\alpha }_{c_i}^{\pm }$$ By trying all possible combinations. There are 4 possible $d$ values. This gives$$\begin{array}{rl}{d}_1& ={\alpha }_{\mathrm{\infty }}^{+}-\left({\alpha }_{c_1}^{+}\right)=-\frac{1}{2}-\frac{3}{2}=-2\\ d_2& ={\alpha }_{\mathrm{\infty }}^{+}-\left({\alpha }_{c_1}^{-}\right)=-\frac{1}{2}-(-\frac{1}{2})=0\\ d_3& ={\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_{c_1}^{+}\right)=\frac{1}{2}-\frac{3}{2}=-1\\ d_4& ={\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_{c_1}^{-}\right)=\frac{1}{2}-(-\frac{1}{2})=1\end{array}$$

Using entry $d=1$ entry above now we find $\omega$ using$$\omega =(\sum _{c}s\left(c\right){\left[\sqrt{r}\right]}_c+\frac{{\alpha }_c^{s\left(c\right)}}{x-c})+s\left(\mathrm{\infty }\right){\left[\sqrt{r}\right]}_{\mathrm{\infty }}$$ Hence$$\begin{array}{rl}\omega & =(0+\frac{{\alpha }_{c_1}^{-}}{x-c_1})+(-)\frac{1}{2}\\ & =\frac{-\frac{1}{2}}{x-1}-\frac{1}{2}\\ & =-\frac{1}{2}\frac{1}{x-1}-\frac{1}{2}\end{array}$$

If this $\omega$ fails to find $p\left(x\right)$, then we will try the entry $d=0$. Which will give$$\omega =-\frac{1}{2}\frac{1}{x-1}+\frac{1}{2}$$ Will finish the solution later.

3.2.10 Example 10

Let $$\begin{array}{rl}\text{(1)}& y^{\prime \prime }-x^2{y}^{\prime }-3xy& =0y^{\prime \prime }+a{y}^{\prime }\left(x\right)+by& =0\end{array}$$

Hence $$\begin{array}{rl}a& =-x^2\\ b& =-3x\end{array}$$

It is first transformed to the following ode by eliminating the first derivative $$\begin{array}{}\text{(2)}& z^{\prime \prime }=rz\end{array}$$ Using what is known as the Liouville transformation given by$$\begin{array}{}\text{(3)}& y=z{e}^{\frac{-1}{2}\int adx}\end{array}$$ Where it can be found that $r$ in (2) is given by $$\begin{array}{rl}r& =\frac{1}{4}{a}^2+\frac{1}{2}{a}^{\prime }-b\\ & =\frac{1}{4}{(-x^2)}^2+\frac{1}{2}\frac{d}{dx}(-x^2)-(-3x)\\ \text{(4)}& & =\frac{1}{4}{x}^4+2x\end{array}$$

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is $$\begin{array}{}\text{(5)}& z^{\prime \prime }=(\frac{1}{4}{x}^4+2x)z\end{array}$$ Therefore $$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{1}{4}{x}^4+2x\end{array}$$

The necessary conditions for case 1 are met.

Step 1 In this we find all ${\left[\sqrt{r}\right]}_c$ and associated ${\alpha }_c^{\pm }$ for each pole. There are no poles. Hence set $\mathrm{\Gamma }$ of poles is empty. Now we consider $O\left(\mathrm{\infty }\right)$ which is $\mathrm{deg}\left(t\right)-\mathrm{deg}\left(s\right)=0-4=-4$. We are in case $2v\le 0$. Hence $-2v=-4$ or $$v=2$$ Then now ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ is the sum of all terms $x^i$ for $0\le i\le v$ in the Laurent series expansion of $\sqrt{r}$ at $\mathrm{\infty }$ which is $$\begin{array}{}\text{(7)}& {\left[\sqrt{r}\right]}_{\mathrm{\infty }}=\frac{x^2}{2}+\frac{2}{x}-\frac{4}{x^4}+\frac{16}{x^7}-\cdots \end{array}$$ We want only terms for $0\le i\le v$ but $v=2$. Therefore need to sum terms $x^0,x^1,x^2$. From the above we see that$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{\mathrm{\infty }}& =\frac{x^2}{2}+0x^1+0x^0\\ & =\frac{x^2}{2}\end{array}$$

Which means $$a=\frac{1}{2}$$ As that is the term which matches ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}=a{x}^2+\cdots$. $$ Now we need to find $b$. This will be the coefficient of $x^{v-1}=x$ in $r$ minus coefficient of $x$ in${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2$. But $${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2=\frac{x^4}{4}$$ Hence the coefficient is zero here. Now we find coefficient of $x$ in $r$. But $r=$ $\frac{1}{4}{x}^4+2x$ hence the coefficient of $x$ is $2$. Therefore$$\begin{array}{rl}b& =2-0\\ & =2\end{array}$$

Hence$$\begin{array}{rl}{\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)=\frac{1}{2}(\frac{2}{\frac{1}{2}}-2)=1\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)=\frac{1}{2}(-\frac{2}{\frac{1}{2}}-2)=-3\end{array}$$

This completes step 1 of the solution. We have found ${\left[\sqrt{r}\right]}_c$ and its associated ${\alpha }_c^{\pm }$ and found ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$. Now we go to step 2 which is to find the $d^{\prime }s$.

step 2 Now $d$ is found using$$d={\alpha }_{\mathrm{\infty }}^{\pm }-\sum _{i=1}^0{\alpha }_{c_i}^{\pm }$$ By trying all possible combinations. There are 4 possible $d$ values. This gives$$\begin{array}{rl}{d}_1& ={\alpha }_{\mathrm{\infty }}^{+}=1\\ d_2& ={\alpha }_{\mathrm{\infty }}^{-}=-3\end{array}$$

Using $d=1$ entry above now we find $\omega$ using$$\omega =(\sum _{c}s\left(c\right){\left[\sqrt{r}\right]}_c+\frac{{\alpha }_c^{s\left(c\right)}}{x-c})+s\left(\mathrm{\infty }\right){\left[\sqrt{r}\right]}_{\mathrm{\infty }}$$ Hence, since there are no poles, only last term above survives giving$$\omega =s\left(\mathrm{\infty }\right){\left[\sqrt{r}\right]}_{\mathrm{\infty }}=(+)\frac{x^2}{2}=\frac{x^2}{2}$$ Will finish the solution next.

3.2.11 Example 11

Let $$\begin{array}{rl}\text{(1)}& y^{\prime \prime }-\frac{2}{5}x{y}^{\prime }+2y& =0y^{\prime \prime }+a{y}^{\prime }\left(x\right)+by& =0\end{array}$$

Hence $$\begin{array}{rl}a& =-\frac{2}{5}x\\ b& =2\end{array}$$

It is first transformed to the following ode by eliminating the first derivative $$\begin{array}{}\text{(2)}& z^{\prime \prime }=rz\end{array}$$ Using what is known as the Liouville transformation given by$$\begin{array}{}\text{(3)}& y=z{e}^{\frac{-1}{2}\int adx}\end{array}$$ Where it can be found that $r$ in (2) is given by $$\begin{array}{rl}r& =\frac{1}{4}{a}^2+\frac{1}{2}{a}^{\prime }-b\\ & =\frac{1}{4}{(-\frac{2}{5}x)}^2+\frac{1}{2}\frac{d}{dx}(-\frac{2}{5}x)-\left(2\right)\\ \text{(4)}& & =\frac{x^2-55}{25}\end{array}$$

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is $$\begin{array}{}\text{(5)}& z^{\prime \prime }=\left(\frac{x^2-55}{25}\right)z\end{array}$$ Therefore $$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{x^2-55}{25}\end{array}$$

The necessary conditions for case 1 are met.

Step 1 In this we find all ${\left[\sqrt{r}\right]}_c$ and associated ${\alpha }_c^{\pm }$ for each pole. There are no poles. Hence set $\mathrm{\Gamma }$ of poles is empty. Now we consider $O\left(\mathrm{\infty }\right)$ which is $\mathrm{deg}\left(t\right)-\mathrm{deg}\left(s\right)=0-2=-2$. We are in case $2v\le 0$. Hence $-2v=-2$ or $$v=1$$ Then now ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ is the sum of all terms $x^i$ for $0\le i\le v$ in the Laurent series expansion of $\sqrt{r}$ at $\mathrm{\infty }$ which is $$\begin{array}{}\text{(7)}& {\left[\sqrt{r}\right]}_{\mathrm{\infty }}=\frac{x}{5}-\frac{11}{2}\frac{1}{x}-\cdots \end{array}$$ We want only terms for $0\le i\le v$ but $v=1$. Therefore need to sum terms $x^0,x^1$. From the above we see that$${\left[\sqrt{r}\right]}_{\mathrm{\infty }}=\frac{x}{5}$$ Which means $$a=\frac{1}{5}$$ As it is the the term that matches ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}=ax$. $$ Now we need to find $b$. This will be the coefficient of $x^{v-1}=x^0$ in $r$ minus the coefficient of $x^0$ in ${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2$. But $$\begin{array}{rl}{\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2& ={\left(\frac{x}{5}\right)}^2\\ & =\frac{x^2}{25}\end{array}$$

Hence the coefficient of $x^0$ is zero. Now we find coefficient of $x^0$ in $r$. Since $r=\frac{x^2}{25}-\frac{55}{25}$ then coefficient of $x^0$ is $\frac{-55}{25}=-\frac{11}{5}$. Hence $b=-\frac{11}{5}-0=-\frac{11}{5}$. Therefore$$\begin{array}{rl}{\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)=\frac{1}{2}(\frac{-\frac{11}{5}}{\frac{1}{5}}-1)=-6\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)=\frac{1}{2}(-\frac{-\frac{11}{5}}{\frac{1}{5}}-1)=5\end{array}$$

This completes step 1 of the solution. We have found ${\left[\sqrt{r}\right]}_c$ and its associated ${\alpha }_c^{\pm }$ and found ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$. Now we go to step 2 which is to find the $d^{\prime }s$.

step 2 Now $d$ is found using$$d={\alpha }_{\mathrm{\infty }}^{\pm }-\sum _{i=1}^0{\alpha }_{c_i}^{\pm }$$ By trying all possible combinations. There are 2 possible $d$ values since no poles. $$\begin{array}{rl}{d}_1& ={\alpha }_{\mathrm{\infty }}^{+}=-6\\ d_2& ={\alpha }_{\mathrm{\infty }}^{-}=5\end{array}$$

Using $d=5$ entry above now we find $\omega$ using$$\omega =(\sum _{c}s\left(c\right){\left[\sqrt{r}\right]}_c+\frac{{\alpha }_c^{s\left(c\right)}}{x-c})+s\left(\mathrm{\infty }\right){\left[\sqrt{r}\right]}_{\mathrm{\infty }}$$ Hence, since there are no poles, only last term above survives giving$$\omega =s\left(\mathrm{\infty }\right){\left[\sqrt{r}\right]}_{\mathrm{\infty }}=(-)\frac{x}{5}=-\frac{x}{5}$$ Will finish the solution next.

3.2.12 Example 12

Let $$\begin{array}{rl}\text{(1)}& y^{\prime \prime }+\frac{x^2-1}{x}{y}^{\prime }+x^{2}y& =0y^{\prime \prime }+a{y}^{\prime }\left(x\right)+by& =0\end{array}$$

Hence $$\begin{array}{rl}a& =\frac{x^2-1}{x}\\ b& =x^2\end{array}$$

It is first transformed to the following ode by eliminating the first derivative $$\begin{array}{}\text{(2)}& z^{\prime \prime }=rz\end{array}$$ Using what is known as the Liouville transformation given by$$\begin{array}{}\text{(3)}& y=z{e}^{\frac{-1}{2}\int adx}\end{array}$$ Where it can be found that $r$ in (2) is given by $$\begin{array}{rl}r& =\frac{1}{4}{a}^2+\frac{1}{2}{a}^{\prime }-b\\ & =\frac{1}{4}{\left(\frac{x^2-1}{x}\right)}^2+\frac{1}{2}\frac{d}{dx}\left(\frac{x^2-1}{x}\right)-x^2\\ \text{(4)}& & =-\frac{3(x^4-1)}{4x^2}\end{array}$$

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is $$\begin{array}{}\text{(5)}& z^{\prime \prime }=-\frac{3(x^4-1)}{4x^2}z\end{array}$$ Therefore $$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{-3(x^4-1)}{4x^2}\end{array}$$

The necessary conditions for case 1 are met.

Step 1 In this we find all ${\left[\sqrt{r}\right]}_c$ and associated ${\alpha }_c^{\pm }$ for each pole. There is one pole at $x=0$ of order 2. Hence

$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c_1}& =0\\ {\alpha }_{c_1}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4b}\\ {\alpha }_{c_1}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4b}\end{array}$$

Where $b$ is the coefficient of $\frac{1}{{(x-0)}^2}$ in the partial fraction decomposition of $r$ which is (in Maple this can be found using fullparfrac). $$\begin{array}{}\text{(6)}& \frac{-3(x^4-1)}{4x^2}=-\frac{3}{4}{x}^2+\frac{3}{4}\frac{1}{x^2}\end{array}$$ Hence $b=\frac{3}{4}$. Therefore$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c_1}& =0\\ {\alpha }_{c_1}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4\left(\frac{3}{4}\right)}=\frac{3}{2}\\ {\alpha }_{c_1}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4\left(\frac{3}{4}\right)}=-\frac{1}{2}\end{array}$$

Now we consider $O\left(\mathrm{\infty }\right)$ which is $\mathrm{deg}\left(t\right)-\mathrm{deg}\left(s\right)=2-4=-2$. We are in case $2v\le 0$. Hence $-2v=-2$ or $$v=1$$ Then now ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ is the sum of all terms $x^i$ for $0\le i\le v$ in the Laurent series expansion of $\sqrt{r}$ at $\mathrm{\infty }$ which is $$\begin{array}{}\text{(7)}& {\left[\sqrt{r}\right]}_{\mathrm{\infty }}=\frac{i\sqrt{3}}{2}x-\frac{i\sqrt{3}}{4}\frac{1}{x^3}+\cdots \end{array}$$ We want only terms for $0\le i\le v$ but $v=1$. Therefore need to sum terms $x^0,x^1$. From the above we see that$${\left[\sqrt{r}\right]}_{\mathrm{\infty }}=\frac{i\sqrt{3}}{2}x$$ Which means $$a=\frac{i\sqrt{3}}{2}$$ As it is the the term that matches ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}=ax$. $$ Now we need to find $b$. This will be the coefficient of $x^{v-1}=x^0$ in $r$ minus the coefficient of $x^0$ in ${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2$. But $$\begin{array}{rl}{\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2& ={\left(\frac{i\sqrt{3}}{2}x\right)}^2\\ & =-\frac{3}{4}{x}^2\end{array}$$

Hence the coefficient is zero here. Now we find coefficient of $x^0$ in $r$. Since $r=-\frac{3}{4}{x}^2+\frac{3}{4}\frac{1}{x^2}$ then coefficient of $x^0$ is zero also. Hence $b=0-0=0$. Therefore$$\begin{array}{rl}{\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)=\frac{1}{2}(0-1)=-\frac{1}{2}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)=\frac{1}{2}(0-1)=-\frac{1}{2}\end{array}$$

This completes step 1 of the solution. We have found ${\left[\sqrt{r}\right]}_c$ and its associated ${\alpha }_c^{\pm }$ and found ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$. Now we go to step 2 which is to find the $d^{\prime }s$.

step 2 Now $d$ is found using$$d={\alpha }_{\mathrm{\infty }}^{\pm }-\sum _{i=1}^0{\alpha }_{c_i}^{\pm }$$ By trying all possible combinations. There are 4 possible $d$ values. This gives$$\begin{array}{rl}{d}_1& ={\alpha }_{\mathrm{\infty }}^{+}-\left({\alpha }_{c_1}^{+}\right)=-\frac{1}{2}-\frac{3}{2}=-1\\ d_2& ={\alpha }_{\mathrm{\infty }}^{+}-\left({\alpha }_{c_1}^{-}\right)=-\frac{1}{2}-(-\frac{1}{2})=0\\ d_3& ={\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_{c_1}^{+}\right)=-\frac{1}{2}-\frac{3}{2}=-2\\ d_4& ={\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_{c_1}^{-}\right)=-\frac{1}{2}-(-\frac{1}{2})=0\end{array}$$

Using first $d=0$ entry above now we find $\omega$ using$$\omega =(\sum _{c}s\left(c\right){\left[\sqrt{r}\right]}_c+\frac{{\alpha }_c^{s\left(c\right)}}{x-c})+s\left(\mathrm{\infty }\right){\left[\sqrt{r}\right]}_{\mathrm{\infty }}$$ Hence$$\omega =(-1)\left(0\right)+\frac{-\frac{1}{2}}{x-0}+(+)\left(\frac{i\sqrt{3}}{2}x\right)=\frac{-1}{2x}+\frac{i\sqrt{3}}{2}x$$ Notice that if have taken the last $d=0$ entry, we will get$$\omega =(-1)\left(0\right)+\frac{-\frac{1}{2}}{x-0}+(-)\left(\frac{i\sqrt{3}}{2}x\right)=\frac{-1}{2x}-\frac{i\sqrt{3}}{2}x$$ In practice, we will try the second one if the first fails. Will finish the solution next.

3.2.13 Example 13

Let $$\begin{array}{rl}\text{(1)}& (1-x)y^{\prime \prime }+x{y}^{\prime }-y& =0y^{\prime \prime }+a{y}^{\prime }\left(x\right)+by& =0\end{array}$$

Hence $$\begin{array}{rl}a& =\frac{x}{1-x}\\ b& =-\frac{1}{1-x}\end{array}$$

It is first transformed to the following ode by eliminating the first derivative $$\begin{array}{}\text{(2)}& z^{\prime \prime }=rz\end{array}$$ Using what is known as the Liouville transformation given by$$\begin{array}{}\text{(3)}& y=z{e}^{\frac{-1}{2}\int adx}\end{array}$$ Where it can be found that $r$ in (2) is given by $$\begin{array}{rl}r& =\frac{1}{4}{a}^2+\frac{1}{2}{a}^{\prime }-b\\ & =\frac{1}{4}{\left(\frac{x}{1-x}\right)}^2+\frac{1}{2}\frac{d}{dx}\left(\frac{x}{1-x}\right)-(-\frac{1}{1-x})\\ \text{(4)}& & =\frac{x^2-4x+6}{4{(x-1)}^2}\end{array}$$

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is $$\begin{array}{}\text{(5)}& z^{\prime \prime }=\frac{x^2-4x+6}{4{(x-1)}^2}z\end{array}$$ Therefore $$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{x^2-4x+6}{4{(x-1)}^2}\end{array}$$

The necessary conditions for case 1 are met.

Step 1 In this we find all ${\left[\sqrt{r}\right]}_c$ and associated ${\alpha }_c^{\pm }$ for each pole. There is one pole at $x=1$ of order 2. Hence

$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c_1}& =0\\ {\alpha }_{c_1}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4b}\\ {\alpha }_{c_1}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4b}\end{array}$$

Where $b$ is the coefficient of $\frac{1}{{(x-1)}^2}$ in the partial fraction decomposition of $r$ which is $$\begin{array}{}\text{(6)}& \frac{x^2-4x+6}{4{(x-1)}^2}=\frac{1}{4}+\frac{3}{4{(x-1)}^2}-\frac{1}{2}\frac{1}{x-1}\end{array}$$ Hence $b=\frac{3}{4}$. Therefore$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c_1}& =0\\ {\alpha }_{c_1}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4\left(\frac{3}{4}\right)}=\frac{3}{2}\\ {\alpha }_{c_1}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4\left(\frac{3}{4}\right)}=-\frac{1}{2}\end{array}$$

Now we consider $O\left(\mathrm{\infty }\right)$ which is $\mathrm{deg}\left(t\right)-\mathrm{deg}\left(s\right)=2-2=0$. We are in case $2v\le 0$. Hence $-2v=0$ or $$v=0$$ Then now ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ is the sum of all terms $x^i$ for $0\le i\le v$ in the Laurent series expansion of $\sqrt{r}$ at $\mathrm{\infty }$ which is $$\begin{array}{}\text{(7)}& {\left[\sqrt{r}\right]}_{\mathrm{\infty }}=\frac{1}{2}-\frac{1}{2x}+\frac{1}{x^3}+\cdots \end{array}$$ But we want only terms for $0\le i\le v$ but $v=0$. Therefore need to sum terms $x^0$. Which is the constant term$$\begin{array}{}\text{(8)}& {\left[\sqrt{r}\right]}_{\mathrm{\infty }}=\frac{1}{2}\end{array}$$ Which means $$a=\frac{1}{2}$$ Now we need to find $b$. Which is given by the coefficient of $\frac{1}{x}$ in $r$ minus coefficient of $\frac{1}{x}$ in ${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2$. But ${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2=\frac{1}{4}$ Hence the coefficient is zero here.  To find the coefficient of $\frac{1}{x}$ in $r$ long division is done (here paper is not clear at all what it means by coefficient of $x^{v-1}$ in $r$ as that depends on the form of $r$ and how it is represented). This method of using long division to find the coefficient works to obtain the correct result. But it is still not clear what the paper actually means by this.$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{x^2-4x+6}{4x^2-8x+4}\\ & =Q+\frac{R}{4x^2-8x+4}\end{array}$$

Where $Q$ is the quotient and $R$ is the remainder. This gives$$r=\frac{1}{4}+\frac{-2x+5}{4x^2-8x+4}$$ For the case of $v=0$ then the coefficient of $x^{-1}$ is  $\frac{lcoeff\left(R\right)}{lcoeff\left(t\right)}=\frac{-2}{4}=-\frac{1}{2}$. Notice that if we just expanded $r$ it will give $\frac{x^2}{4{(x-1)}^2}-\frac{x}{{(x-1)}^2}+\frac{3}{2{(x-1)}^2}$ and we see there is no coefficient of $\frac{1}{x}$ in this representation. So we would have obtain wrong value of $b$ if we just used what the paper said. Now $b=-\frac{1}{2}-0=-\frac{1}{2}$. Therefore $$\begin{array}{rl}{\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)=\frac{1}{2}(\frac{\frac{1}{2}}{\frac{1}{2}}-0)=\frac{1}{2}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)=\frac{1}{2}(-\frac{\frac{1}{2}}{\frac{1}{2}}-0)=-\frac{1}{2}\end{array}$$

This completes step 1 of the solution. We have found ${\left[\sqrt{r}\right]}_c$ and its associated ${\alpha }_c^{\pm }$ and found ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$. Now we go to step 2 which is to find the $d^{\prime }s$.

step 2 Now $d$ is found using$$d={\alpha }_{\mathrm{\infty }}^{\pm }-\sum _{i=1}^1{\alpha }_{c_i}^{\pm }$$ By trying all possible combinations. There are 4 possible $d$ values. This gives$$\begin{array}{rl}{d}_1& ={\alpha }_{\mathrm{\infty }}^{+}-\left({\alpha }_{c_1}^{+}\right)=\frac{1}{2}-\left(\frac{3}{2}\right)=-1\\ d_2& ={\alpha }_{\mathrm{\infty }}^{+}-\left({\alpha }_{c_1}^{-}\right)=\frac{1}{2}-(-\frac{1}{2})=1\\ d_3& ={\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_{c_1}^{+}\right)=-\frac{1}{2}-\left(\frac{3}{2}\right)=-2\\ d_4& ={\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_{c_1}^{-}\right)=-\frac{1}{2}-(-\frac{1}{2})=0\end{array}$$

Using first $d=1$ entry from above we find $\omega$ using$$\omega =(\sum _{c}s\left(c\right){\left[\sqrt{r}\right]}_c+\frac{{\alpha }_c^{s\left(c\right)}}{x-c})+s\left(\mathrm{\infty }\right){\left[\sqrt{r}\right]}_{\mathrm{\infty }}$$ Hence$$\omega =((-1)\left(0\right)+\frac{-\frac{1}{2}}{x-1})+(+)\left(\frac{1}{2}\right)=\frac{-1}{2(x-1)}+\frac{1}{2}$$ And if use the last entry $d=0$ then$$\omega (=(-1)\left(0\right)+\frac{-\frac{1}{2}}{x-1})+(-)\left(\frac{1}{2}x\right)=\frac{-1}{2(x-1)}-\frac{1}{2}$$ In practice, we will try the second one if the first fails. Will finish the solution next.

3.2.14 Example 14

Let $$\begin{array}{rl}\text{(1)}& 3y^{\prime \prime }+x{y}^{\prime }-4y& =0y^{\prime \prime }+a{y}^{\prime }\left(x\right)+by& =0\end{array}$$

Hence $$\begin{array}{rl}a& =\frac{x}{3}\\ b& =-\frac{4}{3}\end{array}$$

It is first transformed to the following ode by eliminating the first derivative $$\begin{array}{}\text{(2)}& z^{\prime \prime }=rz\end{array}$$ Using what is known as the Liouville transformation given by$$\begin{array}{}\text{(3)}& y=z{e}^{\frac{-1}{2}\int adx}\end{array}$$ Where it can be found that $r$ in (2) is given by $$\begin{array}{rl}r& =\frac{1}{4}{a}^2+\frac{1}{2}{a}^{\prime }-b\\ & =\frac{1}{4}{\left(\frac{x}{3}\right)}^2+\frac{1}{2}\frac{d}{dx}\left(\frac{x}{3}\right)-(-\frac{4}{3})\\ \text{(4)}& & =\frac{x^2}{36}+\frac{3}{2}\end{array}$$

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is $$\begin{array}{}\text{(5)}& z^{\prime \prime }=\frac{x^2+54}{36}z\end{array}$$ Therefore $$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{x^2+54}{36}\end{array}$$

The necessary conditions for case 1 are met.

Step 1 In this we find all ${\left[\sqrt{r}\right]}_c$ and associated ${\alpha }_c^{\pm }$ for each pole. There are no poles. Hence $\mathrm{\Gamma }$ is empty.

Now we consider $O\left(\mathrm{\infty }\right)$ which is $\mathrm{deg}\left(t\right)-\mathrm{deg}\left(s\right)=0-2=-2$. We are in case $2v\le 0$. Hence $-2v=-2$ or $$v=1$$ Then now ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ is the sum of all terms $x^i$ for $0\le i\le v$ in the Laurent series expansion of $\sqrt{r}$ at $\mathrm{\infty }$ which is $$\begin{array}{}\text{(7)}& {\left[\sqrt{r}\right]}_{\mathrm{\infty }}=\frac{x}{6}+\frac{9}{2}\frac{1}{x}+\cdots \end{array}$$ But we want only terms for $0\le i\le v$ but $v=1$. Therefore need to sum terms $x^0,x^1$. Therefore$$\begin{array}{}\text{(8)}& {\left[\sqrt{r}\right]}_{\mathrm{\infty }}=\frac{x}{6}\end{array}$$ Which means $$a=\frac{1}{6}$$ Now we need to find $b$. Which is given by the coefficient of $x^{v-1}=x^0$ or the constant term in $r$ minus coefficient of $x^0$ in ${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2$. But ${\left({\left[\sqrt{r}\right]}_{\mathrm{\infty }}\right)}^2=\frac{x^2}{36}$. Hence the coefficient of $x^0$ is zero here. Now we find coefficient of $x^0$ in $r$. Since $r=\frac{x^2}{36}+\frac{54}{36}$ then the coefficient of $x^0$ is $\frac{54}{36}=\frac{3}{2}$. Hence $b=\frac{3}{2}-0=\frac{3}{2}$. Therefore$$\begin{array}{rl}{\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)=\frac{1}{2}(\frac{\frac{3}{2}}{\frac{1}{6}}-1)=4\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)=\frac{1}{2}(-\frac{\frac{3}{2}}{\frac{1}{6}}-1)=-5\end{array}$$

This completes step 1 of the solution. We have found ${\left[\sqrt{r}\right]}_c$ and its associated ${\alpha }_c^{\pm }$ and found ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$. Now we go to step 2 which is to find the $d^{\prime }s$.

step 2 Now $d$ is found using$$d={\alpha }_{\mathrm{\infty }}^{\pm }-\sum _{i=1}^0{\alpha }_{c_i}^{\pm }$$ By trying all possible combinations. There are 2 possible $d$ values (since no poles). This gives$$\begin{array}{rl}{d}_1& ={\alpha }_{\mathrm{\infty }}^{+}=4\\ d_2& ={\alpha }_{\mathrm{\infty }}^{-}=-5\end{array}$$

Using $d=4$ entry from above we find $\omega$ using$$\omega =(\sum _{c}s\left(c\right){\left[\sqrt{r}\right]}_c+\frac{{\alpha }_c^{s\left(c\right)}}{x-c})+s\left(\mathrm{\infty }\right){\left[\sqrt{r}\right]}_{\mathrm{\infty }}$$ Hence$$\omega =\left(0\right)+(+)\left(\frac{x}{6}\right)=\frac{x}{6}$$

3.2.15 Example 15

Let $$\begin{array}{rl}\text{(1)}& (4-x^2)y^{\prime \prime }+x{y}^{\prime }+2y& =0y^{\prime \prime }+a{y}^{\prime }\left(x\right)+by& =0\end{array}$$

Hence $$\begin{array}{rl}a& =\frac{x}{(4-x^2)}\\ b& =\frac{2}{(4-x^2)}\end{array}$$

It is first transformed to the following ode by eliminating the first derivative $$\begin{array}{}\text{(2)}& z^{\prime \prime }=rz\end{array}$$ Using what is known as the Liouville transformation given by$$\begin{array}{}\text{(3)}& y=z{e}^{\frac{-1}{2}\int adx}\end{array}$$ Where it can be found that $r$ in (2) is given by $$\begin{array}{rl}r& =\frac{1}{4}{a}^2+\frac{1}{2}{a}^{\prime }-b\\ & =\frac{1}{4}{\left(\frac{x}{(4-x^2)}\right)}^2+\frac{1}{2}\frac{d}{dx}\left(\frac{x}{(4-x^2)}\right)-\left(\frac{2}{(4-x^2)}\right)\\ \text{(4)}& & =\frac{11x^2-24}{4{(x^2-4)}^2}\end{array}$$

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is $$\begin{array}{}\text{(5)}& z^{\prime \prime }=\frac{11x^2-24}{4{(x^2-4)}^2}z\end{array}$$ Therefore $$\begin{array}{rl}r& =\frac{s}{t}\\ \text{(5A)}& & =\frac{11x^2-24}{4{(x^2-4)}^2}=\frac{11x^2-24}{4x^4-32x^2+64}\end{array}$$

The necessary conditions for case 1 are met.

Step 1 In this we find all ${\left[\sqrt{r}\right]}_c$ and associated ${\alpha }_c^{\pm }$ for each pole. There are two poles at $\pm 2$ each of order 2. For pole at $x=2=c_1$

$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c_1}& =0\\ {\alpha }_{c_1}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4b}\\ {\alpha }_{c_1}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4b}\end{array}$$

Where $b$ is the coefficient of $\frac{1}{{(x-2)}^2}$ in the partial fraction decomposition of $r$ which is $$\begin{array}{}\text{(6)}& \frac{11x^2-24}{4{(x^2-4)}^2}=\frac{5}{16}\frac{1}{{(x+2)}^2}+\frac{5}{16}\frac{1}{{(x-2)}^2}-\frac{17}{32}\frac{1}{(x+2)}+\frac{17}{32}\frac{1}{(x-2)}\end{array}$$ Hence $b=\frac{5}{16}$. Therefore$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c_1}& =0\\ {\alpha }_{c_1}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4\left(\frac{5}{16}\right)}=\frac{5}{4}\\ {\alpha }_{c_1}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4\left(\frac{5}{16}\right)}=-\frac{1}{4}\end{array}$$

And for pole at $x=-2=c_2$$$\begin{array}{rl}{\left[\sqrt{r}\right]}_{c_2}& =0\\ {\alpha }_{c_2}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4\left(\frac{5}{16}\right)}=\frac{5}{4}\\ {\alpha }_{c_2}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4\left(\frac{5}{16}\right)}=-\frac{1}{4}\end{array}$$

Now we consider $O\left(\mathrm{\infty }\right)$ which is $\mathrm{deg}\left(t\right)-\mathrm{deg}\left(s\right)=4-2=2$. Therefore $v=1$. In this case $$\begin{array}{}\text{(7)}& {\left[\sqrt{r}\right]}_{\mathrm{\infty }}=0\end{array}$$ And $${\alpha }_{\mathrm{\infty }}^{\pm }=\frac{1}{2}\pm \frac{1}{2}\sqrt{1+4b}$$ The coefficient of $x^{v-1}=x^0$ is zero in ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}^2$. To find coefficient of $x^0$ in $r=\frac{11x^2-24}{4{(x^2-4)}^2}$ and since $v=0$ then using $b=\frac{lcoeff\left(s\right)}{lcoeff\left(t\right)}$ gives $\frac{11}{4}$. Hence $b=\frac{11}{4}-0=\frac{11}{4}$. Therefore$$\begin{array}{rl}{\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}+\frac{1}{2}\sqrt{1+4\left(\frac{11}{4}\right)}=\frac{1}{2}+\sqrt{3}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}-\frac{1}{2}\sqrt{1+4\left(\frac{11}{4}\right)}=\frac{1}{2}-\sqrt{3}\end{array}$$

This completes step 1 of the solution. We have found ${\left[\sqrt{r}\right]}_c$ and its associated ${\alpha }_c^{\pm }$ and found ${\left[\sqrt{r}\right]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$. Now we go to step 2 which is to find the $d^{\prime }s$.

step 2 Now $d$ is found using$$d={\alpha }_{\mathrm{\infty }}^{\pm }-\sum _{i=1}^0{\alpha }_{c_i}^{\pm }$$ By trying all possible combinations. There are 8 possible $d$ values. These are$$\begin{array}{rl}{d}_1& ={\alpha }_{\mathrm{\infty }}^{+}-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{+})=\frac{1}{2}+\sqrt{3}-(\frac{5}{4}+\frac{5}{4})=\sqrt{3}-2\\ d_2& ={\alpha }_{\mathrm{\infty }}^{+}-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{-})=\frac{1}{2}+\sqrt{3}-(\frac{5}{4}-\frac{1}{4})=\sqrt{3}-\frac{1}{2}\\ d_3& ={\alpha }_{\mathrm{\infty }}^{+}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{+})=\frac{1}{2}+\sqrt{3}-(-\frac{1}{4}+\frac{5}{4})=\sqrt{3}-\frac{1}{2}\\ d_4& ={\alpha }_{\mathrm{\infty }}^{+}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{-})=\frac{1}{2}+\sqrt{3}-(-\frac{1}{4}-\frac{1}{4})=\sqrt{3}+1\\ d_5& ={\alpha }_{\mathrm{\infty }}^{-}-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{+})=\frac{1}{2}-\sqrt{3}-(\frac{5}{4}+\frac{5}{4})=-\sqrt{3}-2\\ d_6& ={\alpha }_{\mathrm{\infty }}^{-}-({\alpha }_{c_1}^{+}+{\alpha }_{c_2}^{-})=\frac{1}{2}-\sqrt{3}-(\frac{5}{4}-\frac{1}{4})=-\sqrt{3}-\frac{1}{2}\\ d_7& ={\alpha }_{\mathrm{\infty }}^{-}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{+})=\frac{1}{2}-\sqrt{3}-(-\frac{1}{4}+\frac{5}{4})=-\sqrt{3}-\frac{1}{2}\\ d_8& ={\alpha }_{\mathrm{\infty }}^{-}-({\alpha }_{c_1}^{-}+{\alpha }_{c_2}^{-})=\frac{1}{2}-\sqrt{3}-(-\frac{1}{4}-\frac{1}{4})=1-\sqrt{3}\end{array}$$

There are no $d\ge 0$ integers. This means case 1 does not apply. We need to try case 2 now.