2.3 Example 3. homogeneous ode where log term is not needed $x^2{y}^{\prime \prime }+3x{y}^{\prime }+4x^{4}y=0$

Solve

$$\begin{array}{}\text{(1)}& x^2{y}^{\prime \prime }+3x{y}^{\prime }+4x^{4}y=0\end{array}$$

Using power series method by expanding around $x=0$. Writing the ode as

$$y^{\prime \prime }\left(x\right)+\frac{3}{x}{y}^{\prime }+4x^{2}y=0$$

Shows that $x=0$ is a singular point. But $\underset{x\to 0}{lim}x\frac{3}{x}=3$. Hence the singularity is removable. This means $x=0$ is a regular singular point. In this case the Frobenius power series will be used instead of the standard power series. Let

$$y\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r}$$

Where $r$ is to be determined. It is the root of the indicial equation. Therefore

$$\begin{array}{rl}{y}^{\prime }\left(x\right)& =\sum _{n=0}^{\mathrm{\infty }}(n+r)a_n{x}^{n+r-1}\\ y^{\prime \prime }\left(x\right)& =\sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)a_n{x}^{n+r-2}\end{array}$$

Substituting the above in (1) gives

$$\begin{array}{rl}{x}^2\sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)a_n{x}^{n+r-2}+3x\sum _{n=0}^{\mathrm{\infty }}(n+r)a_n{x}^{n+r-1}+4x^4\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r}& =0\\ \text{(1A)}& \sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)a_n{x}^{n+r}+\sum _{n=0}^{\mathrm{\infty }}3(n+r)a_n{x}^{n+r}+\sum _{n=0}^{\mathrm{\infty }}4a_n{x}^{n+r+4}& =0\end{array}$$

Here, we need to make all powers on $x$ the same, without making the sums start below zero. This can be done by adjusting the last term above as follows

$$\sum _{n=0}^{\mathrm{\infty }}4a_n{x}^{n+r+4}=\sum _{n=4}^{\mathrm{\infty }}4a_{n-4}{x}^{n+r}$$

And now Eq (1A) becomes

$$\begin{array}{}\text{(1B)}& \sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)a_n{x}^{n+r}+\sum _{n=0}^{\mathrm{\infty }}3(n+r)a_n{x}^{n+r}+\sum _{n=4}^{\mathrm{\infty }}4a_{n-4}{x}^{n+r}=0\end{array}$$

$n=0$ gives the indicial equation

$$\begin{array}{rl}(n+r)(n+r-1)a_n+3(n+r)a_n& =0\\ \left(r\right)(r-1)a_0+3r{a}_0& =0\\ (\left(r\right)(r-1)+3r)a_0& =0\end{array}$$

Since $a_0\ne 0$ then the above becomes

$$\left(r\right)(r-1)+3r=0$$

Hence the roots of the indicial equation are $r_1=0,r_2=-2$. Or $r_1=r_2+N$ where $N=2$. We always take $r_1$ to be the larger of the roots.

When this happens, the solution is given by

$$y\left(x\right)=c_1{y}_1\left(x\right)+c_2{y}_2\left(x\right)$$

Where $y_1\left(x\right)$ is the first solution, which is assumed to be

$$\begin{array}{}\text{(2)}& y_1\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r}\end{array}$$

Where we take $a_0=1$ as it is arbitrary and where $r=r_1=0$. This is the standard Frobenius power series, just like we did to find the indicial equation, the only difference is that now we use $r=r_1$, and hence it is a known value. Once we find $y_1\left(x\right)$, then the second solution is

$$\begin{array}{}\text{(3)}& y_2\left(x\right)=C{y}_1\left(x\right)\ln \left(x\right)+\sum _{n=0}^{\mathrm{\infty }}{b}_n{x}^{n+r}\end{array}$$

We will show below how to find $C$ and $b_n$. First, let us find $y_1\left(x\right)$. From Eq(2)

$$\begin{array}{rl}{y}_1^{\prime }\left(x\right)& =\sum _{n=0}^{\mathrm{\infty }}(n+r)a_n{x}^{n+r-1}\\ y_1^{\prime \prime }\left(x\right)& =\sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)a_n{x}^{n+r-2}\end{array}$$

We need to remember that in the above $r$ is not a symbol any more. It will have the indicial root value, which is $r=r_1=0$ in this case. But we keep $r$ as symbol for now, in order to obtain $a_n\left(r\right)$ as function of $r$ first and use this to find $b_n\left(r\right)$. At the very end we then evaluate everything at $r=r_1=0$. Substituting the above in (1) gives Eq (1B) above (We are following pretty much the same process we did to find the indicial equation here)

$$\begin{array}{}\text{(1B)}& \sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)a_n{x}^{n+r}+\sum _{n=0}^{\mathrm{\infty }}3(n+r)a_n{x}^{n+r}+\sum _{n=4}^{\mathrm{\infty }}4a_{n-4}{x}^{n+r}=0\end{array}$$

Now we are ready to find $a_n$. Now we skip $n=0$ since that was used to obtain the indicial equation, and we know that $a_0=1$ is an arbitrary value to choose.

For $n=1$, Eq (1B) gives

$$\begin{array}{rl}(1+r)(1+r-1)a_1+3(1+r)a_1& =0\\ ((1+r)(1+r-1)+3(1+r))a_1& =0\\ (r^2+4r+3)a_1& =0\end{array}$$

But $r=r_1=0$. The above becomes

$$3a_1=0$$

Hence $a_1=0$.

It is a good idea to use a table to keep record of the $a_n$ values as function of $r$, since this will be used later to find $b_n$.

$n$	$a_n\left(r\right)$	$a_n(r=r_1)$
$0$	$1$	$1$
$1$	$0$	$0$

For $n=2$, Eq (1B) gives

$$\begin{array}{rl}(2+r)(2+r-1)a_2+3(2+r)a_2& =0\\ ((2+r)(2+r-1)+3(2+r))a_2& =0\end{array}$$

But $r=r_1=0$. The above becomes

$$\begin{array}{rl}(\left(2\right)\left(1\right)+3\left(2\right))a_2& =0\\ 8a_2& =0\end{array}$$

Hence $a_2=0$. The table becomes

$n$	$a_n\left(r\right)$	$a_n(r=0)$
$0$	$1$	$1$
$1$	$0$	$0$
$2$	$0$	$0$

For $n=3$, Eq (1B) gives

$$(3+r)(3+r-1)a_3+3(3+r)a_3=0$$

But $r=0$. The above becomes

$$\begin{array}{rl}\left(3\right)\left(2\right)a_3+3\left(3\right)a_3& =0\\ 15a_3& =0\end{array}$$

Hence $a_3=0$ and the table becomes

$n$	$a_n\left(r\right)$	$a_n(r=0)$
$0$	$1$	$1$
$1$	$0$	$0$
$2$	$0$	$0$
$3$	$0$	$0$

For $n\ge 4$ we obtain the recursion equation

$$\begin{array}{rl}(n+r)(n+r-1)a_n+3(n+r)a_n+4a_{n-4}& =0\\ ((n+r)(n+r-1)+3(n+r))a_n+4a_{n-4}& =0\\ \text{(4)}& a_n\left(r\right)& =-\frac{4a_{n-4}\left(r\right)}{(n+r)(n+r-1)+3(n+r)}\end{array}$$

The above is very important, since we will use it to find $b_n\left(r\right)$ later on. For now, we are just finding the $a_n$. Now we find few more $a_n$ terms. From (4) for $n=4$

$$a_4\left(r\right)=-\frac{4a_0\left(r\right)}{(4+r)(4+r-1)+3(4+r)}$$

and $r=r_1=0$ and $a_0=1$, then the above becomes

$$a_4=-\frac{4}{\left(4\right)\left(3\right)+3\left(4\right)}=-\frac{1}{6}$$

The table becomes

$n$	$a_n\left(r\right)$	$a_n(r=0)$
$0$	$1$	$1$
$1$	$0$	$0$
$2$	$0$	$0$
$3$	$0$	$0$
$4$	$-\frac{4}{(4+r)(4+r-1)+3(4+r)}$	$-\frac{1}{6}$

And for $n=5$ from Eq(4)

$$\begin{array}{rl}{a}_5\left(r\right)& =-\frac{4a_1\left(r\right)}{(n+r)(n+r-1)+3(n+r)}\\ & =0\end{array}$$

Since $a_1=0$. Similarly $a_6=0,a_7=0$. For $n=8$

$$a_8\left(r\right)=-\frac{4a_4\left(r\right)}{(8+r)(8+r-1)+3(8+r)}$$

But $a_4\left(r\right)=-\frac{4}{(4+r)(4+r-1)+3(4+r)}$. The above becomes

$$a_8\left(r\right)=\frac{4\frac{4}{(4+r)(4+r-1)+3(4+r)}}{(8+r)(8+r-1)+3(8+r)}=\frac{16}{r^4+28r^3+284r^2+1232r+1920}$$

When $r=r_1=0$ the above becomes

$$a_8\left(r\right)=\frac{1}{120}$$

And so on. The table becomes

$n$	$a_n\left(r\right)$	$a_n(r=0)$
$0$	$1$	$1$
$1$	$0$	$0$
$2$	$0$	$0$
$3$	$0$	$0$
$4$	$-\frac{4}{(4+r)(4+r-1)+3(4+r)}$	$-\frac{1}{6}$
$5$	$0$	$0$
$6$	$0$	$0$
$7$	$0$	$0$
$8$	$\frac{16}{r^4+28r^3+284r^2+1232r+1920}$	$\frac{1}{120}$

Hence $y_1\left(x\right)$ is

$$y_1\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r}$$

But $r=r_1=0$. Therefore

$$\begin{array}{rl}\text{(5)}& y_1\left(x\right)& =\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n& =a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+a_5{x}^5+a_6{x}^6+a_7{x}^7+a_8{x}^8+\cdots \end{array}$$

Using values found for $a_n$ in the above table, then (5) becomes

$$\begin{array}{rl}{y}_1\left(x\right)& =1+a_4{x}^4+a_8{x}^8+\cdots \\ & =1-\frac{1}{6}{x}^4+\frac{1}{120}{x}^8+O\left(x^9\right)\end{array}$$

We are done finding $y_1\left(x\right)$. This was not bad at all. Now comes the hard part. Which is finding $y_2\left(x\right)$. From (3) it is given by

$$\begin{array}{}\text{(3)}& y_2\left(x\right)=C{y}_1\left(x\right)\ln \left(x\right)+\sum _{n=0}^{\mathrm{\infty }}{b}_n{x}^{n+r}\end{array}$$

The first thing to do is to determine if $C$ is zero or not. This is done by finding $\underset{r\to r_2}{lim}{a}_N\left(r\right)$. If this limit exist, then $C=0$, else we need to keep the log term. From the above above we see that $a_N\left(r\right)=a_2\left(r\right)=0$. Recall that $N=2$ since this was the difference between the two roots and $r_2=-2$ (the smaller root). Therefore

$$\underset{r\to r_2}{lim}0=\underset{r\to 0}{lim}0=0$$

Hence the limit exist. Therefore we do not need the log term. This means we can let $C=0$. This is the easy case. Hence (3) becomes

$$\begin{array}{rl}\text{(3A)}& y_2\left(x\right)& =\sum _{n=0}^{\mathrm{\infty }}{b}_n{x}^{n+r}& =x^{-2}\sum _{n=0}^{\mathrm{\infty }}{b}_n{x}^n\end{array}$$

Since $r=r_2=-2$.  Let $b_0=1$. We have to remember now that $b_N=b_2=0$. This is the same we did when the log term was needed in the above example, since $b_N$ is arbitrary, and used to generate $y_1\left(x\right)$. Common practice is to use $b_N=0$. The rest of the $b_n$ are found in similar way, from recursive relation as was done above. Substituting (3A) into $x^2{y}^{\prime \prime }+3x{y}^{\prime }+4x^{4}y=0$ gives Eq. (1B) again, but with $a_n$ replaced by $b_n$

$$\begin{array}{}\text{(1B)}& \sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)b_n{x}^{n+r}+\sum _{n=0}^{\mathrm{\infty }}3(n+r)b_n{x}^{n+r}+\sum _{n=4}^{\mathrm{\infty }}4b_{n-4}{x}^{n+r}=0\end{array}$$

For $n=0$, we skip and let $b_0=1$. For $n=1$ the above gives $b_1=0$. And $b_2=0$ since it is the special term $b_N$. And for $n=3$, we get $b_3=0$. The table for $b_n$ is now

$n$	$b_n\left(r\right)$	$b_n(r=-2)$
$0$	$1$	$1$
$1$	$0$	$0$
$2$	$0$	$0$
$3$	$0$	$0$

For $n\ge 4$, the recursion relation is

$$\begin{array}{rl}(n+r)(n+r-1)b_n+3(n+r)b_n+4b_{n-4}& =0\\ b_n\left(r\right)& =-\frac{4b_{n-4}\left(r\right)}{(n+r)(n+r-1)+3(n+r)}\end{array}$$

For $n=4$

$$\begin{array}{rl}{b}_4\left(r\right)& =-\frac{4b_0\left(r\right)}{(4+r)(4+r-1)+3(4+r)}\\ & =-\frac{4}{(4+r)(4+r-1)+3(4+r)}{b}_0=1\end{array}$$

but $r=-2$. The above becomes

$$b_4=-\frac{4}{(4-2)(4-2-1)+3(4-2)}=-\frac{1}{2}$$

The table becomes

$n$	$b_n\left(r\right)$	$b_n(r=-2)$
$0$	$1$	$1$
$1$	$0$	$0$
$2^{\ast }$	$0$	$0$
$3$	$0$	$0$
$4$	$-\frac{4}{(4+r)(4+r-1)+3(4+r)}$	$-\frac{1}{2}$

We will find that $b_5=b_6=b_7=0$. And for $n=8$

$$b_8\left(r\right)=-\frac{4b_4\left(r\right)}{(8+r)(7+r)+3(8+r)}$$

But $b_4\left(r\right)=-\frac{4}{(4+r)(4+r-1)+3(4+r)}$. Hence

$$b_8\left(r\right)=\frac{4\frac{4}{(4+r)(4+r-1)+3(4+r)}}{(8+r)(7+r)+3(8+r)}=\frac{16}{r^4+28r^3+284r^2+1232r+1920}$$

But $r=-2$.

$$b_8\left(r\right)=\frac{16}{{(-2)}^4+28{(-2)}^3+284{(-2)}^2+1232(-2)+1920}=\frac{1}{24}$$

The table becomes

$n$	$b_n\left(r\right)$	$b_n(r=-2)$
$0$	$1$	$1$
$1$	$0$	$0$
$2^{\ast }$	$0$	$0$
$3$	$0$	$0$
$4$	$-\frac{4}{(4+r)(4+r-1)+3(4+r)}$	$-\frac{1}{2}$
$5$	$0$	$0$
$6$	$0$	$0$
$7$	$0$	$0$
$8$	$\frac{16}{r^4+28r^3+284r^2+1232r+1920}$	$\frac{1}{24}$

And so on. Hence the second solution is

$$\begin{array}{rl}{y}_2\left(x\right)& =\sum _{n=0}^{\mathrm{\infty }}{b}_n{x}^{n+r}\\ & =\sum _{n=0}^{\mathrm{\infty }}{b}_n{x}^{n-2}\\ & =x^{-2}\sum _{n=0}^{\mathrm{\infty }}{b}_n{x}^n\\ & =x^{-2}(b_0+b_{1}x+b_2{x}^2+b_3{x}^3+b_4{x}^4+b_5{x}^5+b_6{x}^6+b_7{x}^7+b_8{x}^8+\cdots )\\ & =x^{-2}(1-\frac{1}{2}{x}^4+\frac{1}{24}{b}_8{x}^8+O\left(x^9\right))\end{array}$$

Therefore the general solution is

$$\begin{array}{rl}y& =C_1{y}_1+C_2{y}_2\\ & =C_1(1-\frac{1}{6}{x}^4+\frac{1}{120}{x}^8+O\left(x^9\right))+C_2\left(x^{-2}(1-\frac{1}{2}{x}^4+\frac{1}{24}{b}_8{x}^8+O\left(x^9\right))\right)\end{array}$$

The following are important items to remember. Always let $b_N=0$ where $N$ is the difference between the roots. When the log term is not needed (as in this problem), $y_2$ is found in very similar way to $y_1$ where $b_0=1$ and the recursion formula is used to find all $b_n$. But when the log term is needed (as in the above problem), it is a little more complicated and need to find $C$ and $b_1$ values by comparing coefficients as was done).

This completes the solution.