2.1.47 Problem 47

Internal problem ID [8758]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 47
Date solved : Friday, April 25, 2025 at 05:03:19 PM
CAS classification : [[_2nd_order, _missing_y]]

Solved as second order linear exact ode

Time used: 0.162 (sec)

Solve

$$\begin{array}{rl}t{y}^{\prime \prime }+4y^{\prime }& =t^2\end{array}$$

An ode of the form

$$\begin{array}{rl}p\left(t\right)y^{\prime \prime }+q\left(t\right)y^{\prime }+r\left(t\right)y& =s\left(t\right)\end{array}$$

is exact if

$$\begin{array}{}\text{(1)}& p^{″}(t)-q^{\prime }(t)+r(t)& =0\end{array}$$

For the given ode we have

$$\begin{array}{rl}p(x)& =t\\ q(x)& =4\\ r(x)& =0\\ s(x)& =t^2\end{array}$$

Hence

$$\begin{array}{rl}{p}^{″}(x)& =0\\ q^{\prime }(x)& =0\end{array}$$

Therefore (1) becomes

$$\begin{array}{rl}0-\left(0\right)+\left(0\right)& =0\end{array}$$

Hence the ode is exact. Since we now know the ode is exact, it can be written as

$$\begin{array}{rl}{(p\left(t\right)y^{\prime }+(q\left(t\right)-p^{\prime }\left(t\right))y)}^{\prime }& =s(x)\end{array}$$

Integrating gives

$$\begin{array}{rl}p\left(t\right)y^{\prime }+(q\left(t\right)-p^{\prime }\left(t\right))y& =\int s\left(t\right)dt\end{array}$$

Substituting the above values for $p,q,r,s$ gives

$$\begin{array}{rl}{y}^{\prime }t+3y& =\int t^{2}dt\end{array}$$

We now have a first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }t+3y=\frac{t^3}{3}+c_1\end{array}$$

In canonical form a linear first order is

$$\begin{array}{rl}{y}^{\prime }+q(t)y& =p(t)\end{array}$$

Comparing the above to the given ode shows that

$$\begin{array}{rl}q(t)& =\frac{3}{t}\\ p(t)& =\frac{t^3+3c_1}{3t}\end{array}$$

The integrating factor $\mu$ is

$$\begin{array}{rl}\mu & =e^{\int qdt}\\ & ={\mathrm{e}}^{\int \frac{3}{t}dt}\\ & =t^3\end{array}$$

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}t}\left(\mu y\right)& =\mu p\\ \frac{\mathrm{d}}{\mathrm{d}t}\left(\mu y\right)& =\left(\mu \right)\left(\frac{t^3+3c_1}{3t}\right)\\ \frac{\mathrm{d}}{\mathrm{d}t}\left(y{t}^3\right)& =\left(t^3\right)\left(\frac{t^3+3c_1}{3t}\right)\\ \mathrm{d}\left(y{t}^3\right)& =\left(\frac{(t^3+3c_1)t^2}{3}\right)\mathrm{d}t\end{array}$$

Integrating gives

$$\begin{array}{rl}y{t}^3& =\int \frac{(t^3+3c_1)t^2}{3}dt\\ & =\frac{{(t^3+3c_1)}^2}{18}+c_2\end{array}$$

Dividing throughout by the integrating factor $t^3$ gives the final solution

$$y=\frac{\frac{{(t^3+3c_1)}^2}{18}+c_2}{t^3}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =\frac{\frac{{(t^3+3c_1)}^2}{18}+c_2}{t^3}\end{array}$$

Solved as second order missing y ode

Time used: 0.124 (sec)

Solve

$$\begin{array}{rl}t{y}^{\prime \prime }+4y^{\prime }& =t^2\end{array}$$

This is second order ode with missing dependent variable $y$. Let

$$\begin{array}{rl}u(t)& =y^{\prime }\end{array}$$

Then

$$\begin{array}{rl}{u}^{\prime }(t)& =y^{\prime \prime }\end{array}$$

Hence the ode becomes

$$\begin{array}{r}t{u}^{\prime }\left(t\right)+4u\left(t\right)-t^2=0\end{array}$$

Which is now solved for $u(t)$ as first order ode.

In canonical form a linear first order is

$$\begin{array}{rl}{u}^{\prime }+q(t)u& =p(t)\end{array}$$

Comparing the above to the given ode shows that

$$\begin{array}{rl}q(t)& =\frac{4}{t}\\ p(t)& =t\end{array}$$

The integrating factor $\mu$ is

$$\begin{array}{rl}\mu & =e^{\int qdt}\\ & ={\mathrm{e}}^{\int \frac{4}{t}dt}\\ & =t^4\end{array}$$

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}t}\left(\mu u\right)& =\mu p\\ \frac{\mathrm{d}}{\mathrm{d}t}\left(\mu u\right)& =\left(\mu \right)\left(t\right)\\ \frac{\mathrm{d}}{\mathrm{d}t}\left(u{t}^4\right)& =\left(t^4\right)\left(t\right)\\ \mathrm{d}\left(u{t}^4\right)& =t^5\mathrm{d}t\end{array}$$

Integrating gives

$$\begin{array}{rl}u{t}^4& =\int t^{5}dt\\ & =\frac{t^6}{6}+c_1\end{array}$$

Dividing throughout by the integrating factor $t^4$ gives the final solution

$$u=\frac{t^6+6c_1}{6t^4}$$

In summary, these are the solution found for $y\left(t\right)$

$$\begin{array}{rl}u& =\frac{t^6+6c_1}{6t^4}\end{array}$$

For solution $u=\frac{t^6+6c_1}{6t^4}$, since $u=y^{\prime }\left(t\right)$ then we now have a new first order ode to solve which is

$$\begin{array}{r}{y}^{\prime }\left(t\right)=\frac{t^6+6c_1}{6t^4}\end{array}$$

Since the ode has the form $y^{\prime }=f(t)$, then we only need to integrate $f(t)$.

$$\begin{array}{rl}\int dy& =\int \frac{t^6+6c_1}{6t^4}dt\\ y& =\frac{t^3}{18}-\frac{c_1}{3t^3}+c_2\end{array}$$

In summary, these are the solution found for $(y)$

$$\begin{array}{rl}y& =\frac{t^3}{18}-\frac{c_1}{3t^3}+c_2\end{array}$$

Summary of solutions found

$$\begin{array}{rl}y& =\frac{t^3}{18}-\frac{c_1}{3t^3}+c_2\end{array}$$

Solved as second order integrable as is ode

Time used: 0.167 (sec)

Solve

$$\begin{array}{rl}t{y}^{\prime \prime }+4y^{\prime }& =t^2\end{array}$$

Integrating both sides of the ODE w.r.t $t$ gives

$$\begin{array}{rl}\int (t{y}^{\prime \prime }+4y^{\prime })dt& =\int t^{2}dt\\ y^{\prime }t+3y=\frac{t^3}{3}+c_1\end{array}$$

Which is now solved for $y$. In canonical form a linear first order is

$$\begin{array}{rl}{y}^{\prime }+q(t)y& =p(t)\end{array}$$

Comparing the above to the given ode shows that

$$\begin{array}{rl}q(t)& =\frac{3}{t}\\ p(t)& =\frac{t^3+3c_1}{3t}\end{array}$$

The integrating factor $\mu$ is

$$\begin{array}{rl}\mu & =e^{\int qdt}\\ & ={\mathrm{e}}^{\int \frac{3}{t}dt}\\ & =t^3\end{array}$$

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}t}\left(\mu y\right)& =\mu p\\ \frac{\mathrm{d}}{\mathrm{d}t}\left(\mu y\right)& =\left(\mu \right)\left(\frac{t^3+3c_1}{3t}\right)\\ \frac{\mathrm{d}}{\mathrm{d}t}\left(y{t}^3\right)& =\left(t^3\right)\left(\frac{t^3+3c_1}{3t}\right)\\ \mathrm{d}\left(y{t}^3\right)& =\left(\frac{(t^3+3c_1)t^2}{3}\right)\mathrm{d}t\end{array}$$

Integrating gives

$$\begin{array}{rl}y{t}^3& =\int \frac{(t^3+3c_1)t^2}{3}dt\\ & =\frac{{(t^3+3c_1)}^2}{18}+c_2\end{array}$$

Dividing throughout by the integrating factor $t^3$ gives the final solution

$$y=\frac{\frac{{(t^3+3c_1)}^2}{18}+c_2}{t^3}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =\frac{\frac{{(t^3+3c_1)}^2}{18}+c_2}{t^3}\end{array}$$

Solved as second order integrable as is ode (ABC method)

Time used: 0.064 (sec)

Solve

$$\begin{array}{rl}t{y}^{\prime \prime }+4y^{\prime }& =t^2\end{array}$$

Writing the ode as

$$t{y}^{\prime \prime }+4y^{\prime }=t^2$$

Integrating both sides of the ODE w.r.t $t$ gives

$$\begin{array}{rl}\int (t{y}^{\prime \prime }+4y^{\prime })dt& =\int t^{2}dt\\ y^{\prime }t+3y=\frac{t^3}{3}+c_1\end{array}$$

Which is now solved for $y$. In canonical form a linear first order is

$$\begin{array}{rl}{y}^{\prime }+q(t)y& =p(t)\end{array}$$

Comparing the above to the given ode shows that

$$\begin{array}{rl}q(t)& =\frac{3}{t}\\ p(t)& =\frac{t^3+3c_1}{3t}\end{array}$$

The integrating factor $\mu$ is

$$\begin{array}{rl}\mu & =e^{\int qdt}\\ & ={\mathrm{e}}^{\int \frac{3}{t}dt}\\ & =t^3\end{array}$$

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}t}\left(\mu y\right)& =\mu p\\ \frac{\mathrm{d}}{\mathrm{d}t}\left(\mu y\right)& =\left(\mu \right)\left(\frac{t^3+3c_1}{3t}\right)\\ \frac{\mathrm{d}}{\mathrm{d}t}\left(y{t}^3\right)& =\left(t^3\right)\left(\frac{t^3+3c_1}{3t}\right)\\ \mathrm{d}\left(y{t}^3\right)& =\left(\frac{(t^3+3c_1)t^2}{3}\right)\mathrm{d}t\end{array}$$

Integrating gives

$$\begin{array}{rl}y{t}^3& =\int \frac{(t^3+3c_1)t^2}{3}dt\\ & =\frac{{(t^3+3c_1)}^2}{18}+c_2\end{array}$$

Dividing throughout by the integrating factor $t^3$ gives the final solution

$$y=\frac{\frac{{(t^3+3c_1)}^2}{18}+c_2}{t^3}$$

In canonical form a linear first order is

$$\begin{array}{rl}{y}^{\prime }+q(t)y& =p(t)\end{array}$$

Comparing the above to the given ode shows that

$$\begin{array}{rl}q(t)& =\frac{3}{t}\\ p(t)& =\frac{t^3+3c_1}{3t}\end{array}$$

The integrating factor $\mu$ is

$$\begin{array}{rl}\mu & =e^{\int qdt}\\ & ={\mathrm{e}}^{\int \frac{3}{t}dt}\\ & =t^3\end{array}$$

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}t}\left(\mu y\right)& =\mu p\\ \frac{\mathrm{d}}{\mathrm{d}t}\left(\mu y\right)& =\left(\mu \right)\left(\frac{t^3+3c_1}{3t}\right)\\ \frac{\mathrm{d}}{\mathrm{d}t}\left(y{t}^3\right)& =\left(t^3\right)\left(\frac{t^3+3c_1}{3t}\right)\\ \mathrm{d}\left(y{t}^3\right)& =\left(\frac{(t^3+3c_1)t^2}{3}\right)\mathrm{d}t\end{array}$$

Integrating gives

$$\begin{array}{rl}y{t}^3& =\int \frac{(t^3+3c_1)t^2}{3}dt\\ & =\frac{{(t^3+3c_1)}^2}{18}+c_2\end{array}$$

Dividing throughout by the integrating factor $t^3$ gives the final solution

$$y=\frac{\frac{{(t^3+3c_1)}^2}{18}+c_2}{t^3}$$

Will add steps showing solving for IC soon.

Solved as second order ode using non constant coeff transformation on B method

Time used: 0.151 (sec)

Solve

$$\begin{array}{rl}t{y}^{\prime \prime }+4y^{\prime }& =t^2\end{array}$$

Given an ode of the form

$$\begin{array}{rl}A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =F(t)\end{array}$$

This method reduces the order ode the ODE by one by applying the transformation

$$\begin{array}{rl}y& =Bv\end{array}$$

This results in

$$\begin{array}{rl}{y}^{\prime }& =B^{\prime }v+v^{\prime }B\\ y^{″}& =B^{″}v+B^{\prime }{v}^{\prime }+v^{″}B+v^{\prime }{B}^{\prime }\\ & =v^{″}B+2v^{\prime }+B^{\prime }+B^{″}v\end{array}$$

And now the original ode becomes

$$\begin{array}{rl}A(v^{″}B+2v^{\prime }{B}^{\prime }+B^{″}v)+B(B^{\prime }v+v^{\prime }B)+CBv& =0\\ \text{(1)}& AB{v}^{″}+(2A{B}^{\prime }+B^2)v^{\prime }+(A{B}^{″}+B{B}^{\prime }+CB)v& =0\end{array}$$

If the term $A{B}^{″}+B{B}^{\prime }+CB$ is zero, then this method works and can be used to solve

$$AB{v}^{″}+(2A{B}^{\prime }+B^2)v^{\prime }=0$$

By Using $u=v^{\prime }$ which reduces the order of the above ode to one. The new ode is

$$AB{u}^{\prime }+(2A{B}^{\prime }+B^2)u=0$$

The above ode is first order ode which is solved for $u$. Now a new ode $v^{\prime }=u$ is solved for $v$ as first order ode. Then the final solution is obtain from $y=Bv$.

This method works only if the term $A{B}^{″}+B{B}^{\prime }+CB$ is zero. The given ODE shows that

$$\begin{array}{rl}A& =t\\ B& =4\\ C& =0\\ F& =t^2\end{array}$$

The above shows that for this ode

$$\begin{array}{rl}A{B}^{″}+B{B}^{\prime }+CB& =\left(t\right)\left(0\right)+\left(4\right)\left(0\right)+\left(0\right)\left(4\right)\\ & =0\end{array}$$

Hence the ode in $v$ given in (1) now simplifies to

$$\begin{array}{rl}4t{v}^{″}+\left(16\right)v^{\prime }& =0\end{array}$$

Now by applying $v^{\prime }=u$ the above becomes

$$\begin{array}{r}4t{u}^{\prime }\left(t\right)+16u\left(t\right)=0\end{array}$$

Which is now solved for $u$. In canonical form a linear first order is

$$\begin{array}{rl}{u}^{\prime }+q(t)u& =p(t)\end{array}$$

Comparing the above to the given ode shows that

$$\begin{array}{rl}q(t)& =\frac{4}{t}\\ p(t)& =0\end{array}$$

The integrating factor $\mu$ is

$$\begin{array}{rl}\mu & =e^{\int qdt}\\ & ={\mathrm{e}}^{\int \frac{4}{t}dt}\\ & =t^4\end{array}$$

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}t}\mu u& =0\\ \frac{\mathrm{d}}{\mathrm{d}t}\left(u{t}^4\right)& =0\end{array}$$

Integrating gives

$$\begin{array}{rl}u{t}^4& =\int 0dt+c_1\\ & =c_1\end{array}$$

Dividing throughout by the integrating factor $t^4$ gives the final solution

$$u=\frac{c_1}{t^4}$$

The ode for $v$ now becomes

$$v^{\prime }\left(t\right)=\frac{c_1}{t^4}$$

Which is now solved for $v$. Since the ode has the form $v^{\prime }=f(t)$, then we only need to integrate $f(t)$.

$$\begin{array}{rl}\int dv& =\int \frac{c_1}{t^4}dt\\ v& =-\frac{c_1}{3t^3}+c_2\end{array}$$

Replacing $v$ above by $\frac{y\left(t\right)}{4}$, then the homogeneous solution is

$$\begin{array}{rl}{y}_h(t)& =Bv\\ & =-\frac{4(-3c_2{t}^3+c_1)}{3t^3}\end{array}$$

And now the particular solution $y_p(t)$ will be found. The particular solution $y_p$ can be found using either the method of undetermined coefficients, or the method of variation of parameters. The method of variation of parameters will be used as it is more general and can be used when the coefficients of the ODE depend on $t$ as well. Let

$$\begin{array}{}\text{(1)}& y_p(t)=u_1{y}_1+u_2{y}_2\end{array}$$

Where $u_1,u_2$ to be determined, and $y_1,y_2$ are the two basis solutions (the two linearly independent solutions of the homogeneous ODE) found earlier when solving the homogeneous ODE as

$$\begin{array}{rl}{y}_1& =4\\ y_2& =\frac{1}{t^3}\end{array}$$

In the Variation of parameters $u_1,u_2$ are found using

$$\begin{array}{}\text{(2)}& u_1& =-\int \frac{y_{2}f(t)}{aW(t)}\text{(3)}& u_2& =\int \frac{y_{1}f(t)}{aW(t)}\end{array}$$

Where $W(t)$ is the Wronskian and $a$ is the coefficient in front of $y^{″}$ in the given ODE. The Wronskian is given by $W=\left|\begin{array}{cc}{y}_1& y_2\\ y_1^{\prime }& y_2^{\prime }\end{array}\right|$. Hence

$$W=\left|\begin{array}{cc}4& \frac{1}{t^3}\\ \frac{d}{dt}\left(4\right)& \frac{d}{dt}\left(\frac{1}{t^3}\right)\end{array}\right|$$

Which gives

$$W=\left|\begin{array}{cc}4& \frac{1}{t^3}\\ 0& -\frac{3}{t^4}\end{array}\right|$$

Therefore

$$W=\left(4\right)(-\frac{3}{t^4})-\left(\frac{1}{t^3}\right)\left(0\right)$$

Which simplifies to

$$W=-\frac{12}{t^4}$$

Which simplifies to

$$W=-\frac{12}{t^4}$$

Therefore Eq. (2) becomes

$$u_1=-\int \frac{\frac{1}{t}}{-\frac{12}{t^3}}dt$$

Which simplifies to

$$u_1=-\int -\frac{t^2}{12}dt$$

Hence

$$u_1=\frac{t^3}{36}$$

And Eq. (3) becomes

$$u_2=\int \frac{4t^2}{-\frac{12}{t^3}}dt$$

Which simplifies to

$$u_2=\int -\frac{t^5}{3}dt$$

Hence

$$u_2=-\frac{t^6}{18}$$

Therefore the particular solution, from equation (1) is

$$y_p(t)=\frac{t^3}{18}$$

Hence the complete solution is

$$\begin{array}{rl}y(t)& =y_h+y_p\\ & =(-\frac{4(-3c_2{t}^3+c_1)}{3t^3})+\left(\frac{t^3}{18}\right)\\ & =\frac{t^6+72c_2{t}^3-24c_1}{18t^3}\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y\left(t\right)& =\frac{t^6+72c_2{t}^3-24c_1}{18t^3}\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.245 (sec)

Solve

$$\begin{array}{rl}t{y}^{\prime \prime }+4y^{\prime }& =t^2\end{array}$$

Writing the ode as

$$\begin{array}{}\text{(1)}& t{y}^{\prime \prime }+4y^{\prime }& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =t\\ \text{(3)}& B& =4C& =0\end{array}$$

Applying the Liouville transformation on the dependent variable gives

$$\begin{array}{rl}z(t)& =y{e}^{\int \frac{B}{2A}dt}\end{array}$$

Then (2) becomes

$$\begin{array}{}\text{(4)}& z^{″}(t)=rz(t)\end{array}$$

Where $r$ is given by

$$\begin{array}{rl}\text{(5)}& r& =\frac{s}{t}& =\frac{2A{B}^{\prime }-2B{A}^{\prime }+B^2-4AC}{4A^2}\end{array}$$

Substituting the values of $A,B,C$ from (3) in the above and simplifying gives

$$\begin{array}{}\text{(6)}& r& =\frac{2}{t^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =2\\ t& =t^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(t)& =\left(\frac{2}{t^2}\right)z(t)\end{array}$$

Equation (7) is now solved. After finding $z(t)$ then $y$ is found using the inverse transformation

$$\begin{array}{rl}y& =z\left(t\right)e^{-\int \frac{B}{2A}dt}\end{array}$$

The first step is to determine the case of Kovacic algorithm this ode belongs to. There are 3 cases depending on the order of poles of $r$ and the order of $r$ at $\mathrm{\infty }$. The following table summarizes these cases.

Case	Allowed pole order for $r$	Allowed value for $\mathcal{O}(\mathrm{\infty })$
1	$\{0,1,2,4,6,8,\cdots \}$	$\{\cdots ,-6,-4,-2,0,2,3,4,5,6,\cdots \}$
2	Need to have at least one pole that is either order $2$ or odd order greater than $2$. Any other pole order is allowed as long as the above condition is satisfied. Hence the following set of pole orders are all allowed. $\{1,2\}$,$\{1,3\}$,$\{2\}$,$\{3\}$,$\{3,4\}$,$\{1,2,5\}$.	no condition
3	$\{1,2\}$	$\{2,3,4,5,6,7,\cdots \}$

Table 2.13: Necessary conditions for each Kovacic case

The order of $r$ at $\mathrm{\infty }$ is the degree of $t$ minus the degree of $s$. Therefore

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =2-0\\ & =2\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=t^2$. There is a pole at $t=0$ of order $2$. Since there is no odd order pole larger than $2$ and the order at $\mathrm{\infty }$ is $2$ then the necessary conditions for case one are met. Since there is a pole of order $2$ then necessary conditions for case two are met. Since pole order is not larger than $2$ and the order at $\mathrm{\infty }$ is $2$ then the necessary conditions for case three are met. Therefore

$$\begin{array}{rl}L& =[1,2,4,6,12]\end{array}$$

Attempting to find a solution using case $n=1$.

Looking at poles of order 2. The partial fractions decomposition of $r$ is

$$r=\frac{2}{t^2}$$

For the pole at $t=0$ let $b$ be the coefficient of $\frac{1}{t^2}$ in the partial fractions decomposition of $r$ given above. Therefore $b=2$. Hence

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

Since the order of $r$ at $\mathrm{\infty }$ is 2 then $[\sqrt{r}{]}_{\mathrm{\infty }}=0$. Let $b$ be the coefficient of $\frac{1}{t^2}$ in the Laurent series expansion of $r$ at $\mathrm{\infty }$. which can be found by dividing the leading coefficient of $s$ by the leading coefficient of $t$ from

$$\begin{array}{rlrl}r& =\frac{s}{t}& & =\frac{2}{t^2}\end{array}$$

Since the $\text{gcd}(s,t)=1$. This gives $b=2$. Hence

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

The following table summarizes the findings so far for poles and for the order of $r$ at $\mathrm{\infty }$ where $r$ is

$$r=\frac{2}{t^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$0$	$2$	$0$	$2$	$-1$

Order of $r$ at $\mathrm{\infty }$	$[\sqrt{r}{]}_{\mathrm{\infty }}$	${\alpha }_{\mathrm{\infty }}^{+}$	${\alpha }_{\mathrm{\infty }}^{-}$
$2$	$0$	$2$	$-1$

Now that the all $[\sqrt{r}{]}_c$ and its associated ${\alpha }_c^{\pm }$ have been determined for all the poles in the set $\mathrm{\Gamma }$ and $[\sqrt{r}{]}_{\mathrm{\infty }}$ and its associated ${\alpha }_{\mathrm{\infty }}^{\pm }$ have also been found, the next step is to determine possible non negative integer $d$ from these using

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{s(\mathrm{\infty })}-\sum _{c\in \mathrm{\Gamma }}{\alpha }_c^{s(c)}\end{array}$$

Where $s(c)$ is either $+$ or $-$ and $s(\mathrm{\infty })$ is the sign of ${\alpha }_{\mathrm{\infty }}^{\pm }$. This is done by trial over all set of families $s=(s(c){)}_{c\in \mathrm{\Gamma }\cup \mathrm{\infty }}$ until such $d$ is found to work in finding candidate $\omega$. Trying ${\alpha }_{\mathrm{\infty }}^{-}=-1$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_{c_1}^{-}\right)\\ & =-1-(-1)\\ & =0\end{array}$$

Since $d$ an integer and $d\ge 0$ then it can be used to find $\omega$ using

$$\begin{array}{rl}\omega & =\sum _{c\in \mathrm{\Gamma }}(s(c)[\sqrt{r}{]}_c+\frac{{\alpha }_c^{s(c)}}{t-c})+s(\mathrm{\infty })[\sqrt{r}{]}_{\mathrm{\infty }}\end{array}$$

The above gives

$$\begin{array}{rl}\omega & =((-)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{-}}{t-c_1})+(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =-\frac{1}{t}+(-)\left(0\right)\\ & =-\frac{1}{t}\\ & =-\frac{1}{t}\end{array}$$

Now that $\omega$ is determined, the next step is find a corresponding minimal polynomial $p(t)$ of degree $d=0$ to solve the ode. The polynomial $p(t)$ needs to satisfy the equation

$$\begin{array}{}\text{(1A)}& p^{″}+2\omega p^{\prime }+({\omega }^{\prime }+{\omega }^2-r)p=0\end{array}$$

Let

$$\begin{array}{}\text{(2A)}& p(t)& =1\end{array}$$

Substituting the above in eq. (1A) gives

$$\begin{array}{rl}\left(0\right)+2(-\frac{1}{t})\left(0\right)+(\left(\frac{1}{t^2}\right)+{(-\frac{1}{t})}^2-\left(\frac{2}{t^2}\right))& =0\\ 0=0\end{array}$$

The equation is satisfied since both sides are zero. Therefore the first solution to the ode $z^{″}=rz$ is

$$\begin{array}{rl}{z}_1(t)& =p{e}^{\int \omega dt}\\ & ={\mathrm{e}}^{\int -\frac{1}{t}dt}\\ & =\frac{1}{t}\end{array}$$

The first solution to the original ode in $y$ is found from

$$\begin{array}{rl}{y}_1& =z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dt}\\ & =z_1{e}^{-\int \frac{1}{2}\frac{4}{t}dt}\\ & =z_1{e}^{-2\ln \left(t\right)}\\ & =z_1\left(\frac{1}{t^2}\right)\end{array}$$

Which simplifies to

$$y_1=\frac{1}{t^3}$$

The second solution $y_2$ to the original ode is found using reduction of order

$$y_2=y_1\int \frac{e^{\int -\frac{B}{A}dt}}{y_1^2}dt$$

Substituting gives

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{4}{t}dt}}{{\left(y_1\right)}^2}dt\\ & =y_1\int \frac{e^{-4\ln \left(t\right)}}{{\left(y_1\right)}^2}dt\\ & =y_1\left(\frac{t^3}{3}\right)\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left(\frac{1}{t^3}\right)+c_2\left(\frac{1}{t^3}\left(\frac{t^3}{3}\right)\right)\end{array}$$

This is second order nonhomogeneous ODE. Let the solution be

$$y=y_h+y_p$$

Where $y_h$ is the solution to the homogeneous ODE $A{y}^{″}(t)+B{y}^{\prime }(t)+Cy(t)=0$, and $y_p$ is a particular solution to the nonhomogeneous ODE $A{y}^{″}(t)+B{y}^{\prime }(t)+Cy(t)=f(t)$. $y_h$ is the solution to

$$t{y}^{\prime \prime }+4y^{\prime }=0$$

The homogeneous solution is found using the Kovacic algorithm which results in

$$y_h=\frac{c_1}{t^3}+\frac{c_2}{3}$$

The particular solution $y_p$ can be found using either the method of undetermined coefficients, or the method of variation of parameters. The method of variation of parameters will be used as it is more general and can be used when the coefficients of the ODE depend on $t$ as well. Let

$$\begin{array}{}\text{(1)}& y_p(t)=u_1{y}_1+u_2{y}_2\end{array}$$

Where $u_1,u_2$ to be determined, and $y_1,y_2$ are the two basis solutions (the two linearly independent solutions of the homogeneous ODE) found earlier when solving the homogeneous ODE as

$$\begin{array}{rl}{y}_1& =\frac{1}{t^3}\\ y_2& =\frac{1}{3}\end{array}$$

In the Variation of parameters $u_1,u_2$ are found using

$$\begin{array}{}\text{(2)}& u_1& =-\int \frac{y_{2}f(t)}{aW(t)}\text{(3)}& u_2& =\int \frac{y_{1}f(t)}{aW(t)}\end{array}$$

Where $W(t)$ is the Wronskian and $a$ is the coefficient in front of $y^{″}$ in the given ODE. The Wronskian is given by $W=\left|\begin{array}{cc}{y}_1& y_2\\ y_1^{\prime }& y_2^{\prime }\end{array}\right|$. Hence

$$W=\left|\begin{array}{cc}\frac{1}{t^3}& \frac{1}{3}\\ \frac{d}{dt}\left(\frac{1}{t^3}\right)& \frac{d}{dt}\left(\frac{1}{3}\right)\end{array}\right|$$

Which gives

$$W=\left|\begin{array}{cc}\frac{1}{t^3}& \frac{1}{3}\\ -\frac{3}{t^4}& 0\end{array}\right|$$

Therefore

$$W=\left(\frac{1}{t^3}\right)\left(0\right)-\left(\frac{1}{3}\right)(-\frac{3}{t^4})$$

Which simplifies to

$$W=\frac{1}{t^4}$$

Which simplifies to

$$W=\frac{1}{t^4}$$

Therefore Eq. (2) becomes

$$u_1=-\int \frac{\frac{t^2}{3}}{\frac{1}{t^3}}dt$$

Which simplifies to

$$u_1=-\int \frac{t^5}{3}dt$$

Hence

$$u_1=-\frac{t^6}{18}$$

And Eq. (3) becomes

$$u_2=\int \frac{\frac{1}{t}}{\frac{1}{t^3}}dt$$

Which simplifies to

$$u_2=\int t^{2}dt$$

Hence

$$u_2=\frac{t^3}{3}$$

Therefore the particular solution, from equation (1) is

$$y_p(t)=\frac{t^3}{18}$$

Therefore the general solution is

$$\begin{array}{rl}y& =y_h+y_p\\ & =(\frac{c_1}{t^3}+\frac{c_2}{3})+\left(\frac{t^3}{18}\right)\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =\frac{c_1}{t^3}+\frac{c_2}{3}+\frac{t^3}{18}\end{array}$$

✓ Maple. Time used: 0.001 (sec). Leaf size: 17

ode:=diff(diff(y(t),t),t)*t+4*diff(y(t),t) = t^2; 
dsolve(ode,y(t), singsol=all);
 

$$y=\frac{t^3}{18}-\frac{c_1}{3t^3}+c_2$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
-> Calling odsolve with the ODE, diff(_b(_a),_a) = -(-_a^2+4*_b(_a))/_a, _b(_a) 
   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   <- 1st order linear successful 
<- high order exact linear fully integrable successful
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & t\left(\frac{d}{dt}\frac{d}{dt}y\left(t\right)\right)+4\frac{d}{dt}y\left(t\right)=t^2\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d}{dt}\frac{d}{dt}y\left(t\right)\\ \bullet & & \text{Make substitution}u=\frac{d}{dt}y\left(t\right)\text{to reduce order of ODE}\\ & & t\left(\frac{d}{dt}u\left(t\right)\right)+4u\left(t\right)=t^2\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dt}u\left(t\right)=\frac{-4u\left(t\right)+t^2}{t}\\ \bullet & & \text{Collect w.r.t.}u\left(t\right)\text{and simplify}\\ & & \frac{d}{dt}u\left(t\right)=-\frac{4u\left(t\right)}{t}+t\\ \bullet & & \text{Group terms with}u\left(t\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE}\\ & & \frac{d}{dt}u\left(t\right)+\frac{4u\left(t\right)}{t}=t\\ \bullet & & \text{The ODE is linear; multiply by an integrating factor}\mu \left(t\right)\\ & & \mu \left(t\right)(\frac{d}{dt}u\left(t\right)+\frac{4u\left(t\right)}{t})=\mu \left(t\right)t\\ \bullet & & \text{Assume the lhs of the ODE is the total derivative}\frac{d}{dt}\left(u\left(t\right)\mu \left(t\right)\right)\\ & & \mu \left(t\right)(\frac{d}{dt}u\left(t\right)+\frac{4u\left(t\right)}{t})=\left(\frac{d}{dt}u\left(t\right)\right)\mu \left(t\right)+u\left(t\right)\left(\frac{d}{dt}\mu \left(t\right)\right)\\ \bullet & & \text{Isolate}\frac{d}{dt}\mu \left(t\right)\\ & & \frac{d}{dt}\mu \left(t\right)=\frac{4\mu \left(t\right)}{t}\\ \bullet & & \text{Solve to find the integrating factor}\\ & & \mu \left(t\right)=t^4\\ \bullet & & \text{Integrate both sides with respect to}t\\ & & \int \left(\frac{d}{dt}\left(u\left(t\right)\mu \left(t\right)\right)\right)dt=\int \mu \left(t\right)tdt+\mathit{C}\mathit{1}\\ \bullet & & \text{Evaluate the integral on the lhs}\\ & & u\left(t\right)\mu \left(t\right)=\int \mu \left(t\right)tdt+\mathit{C}\mathit{1}\\ \bullet & & \text{Solve for}u\left(t\right)\\ & & u\left(t\right)=\frac{\int \mu \left(t\right)tdt+\mathit{C}\mathit{1}}{\mu \left(t\right)}\\ \bullet & & \text{Substitute}\mu \left(t\right)=t^4\\ & & u\left(t\right)=\frac{\int t^{5}dt+\mathit{C}\mathit{1}}{t^4}\\ \bullet & & \text{Evaluate the integrals on the rhs}\\ & & u\left(t\right)=\frac{\frac{t^6}{6}+\mathit{C}\mathit{1}}{t^4}\\ \bullet & & \text{Simplify}\\ & & u\left(t\right)=\frac{t^6+6\mathit{C}\mathit{1}}{6t^4}\\ \bullet & & \text{Solve 1st ODE for}u\left(t\right)\\ & & u\left(t\right)=\frac{t^6+6\mathit{C}\mathit{1}}{6t^4}\\ \bullet & & \text{Make substitution}u=\frac{d}{dt}y\left(t\right)\\ & & \frac{d}{dt}y\left(t\right)=\frac{t^6+6\mathit{C}\mathit{1}}{6t^4}\\ \bullet & & \text{Integrate both sides to solve for}y\left(t\right)\\ & & \int \left(\frac{d}{dt}y\left(t\right)\right)dt=\int \frac{t^6+6\mathit{C}\mathit{1}}{6t^4}dt+\mathit{C}\mathit{2}\\ \bullet & & \text{Compute integrals}\\ & & y\left(t\right)=\frac{t^3}{18}-\frac{\mathit{C}\mathit{1}}{3t^3}+\mathit{C}\mathit{2}\end{array}$$

✓ Mathematica. Time used: 0.034 (sec). Leaf size: 24

ode=t*D[y[t],{t,2}]+4*D[y[t],t]==t^2; 
ic={}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

$$y(t)\to \frac{t^3}{18}-\frac{c_1}{3t^3}+c_2$$

✓ Sympy. Time used: 0.208 (sec). Leaf size: 14

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(-t**2 + t*Derivative(y(t), (t, 2)) + 4*Derivative(y(t), t),0) 
ics = {} 
dsolve(ode,func=y(t),ics=ics)
 

$$y\left(t\right)=C_1+\frac{C_2}{t^3}+\frac{t^3}{18}$$