2.1.53 Problem 53

Internal problem ID [8765]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 53
Date solved : Wednesday, March 05, 2025 at 06:46:10 AM
CAS classification : [[_Emden, _Fowler], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]

Solve

$$\begin{array}{rl}t{y}^{\prime \prime }-y^{\prime }+4t^{3}y& =0\end{array}$$

Solved as second order ode using change of variable on x method 2

Time used: 0.109 (sec)

In normal form the ode

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

Becomes

$$\begin{array}{}\text{(2)}& y^{\prime \prime }+p\left(t\right)y^{\prime }+q\left(t\right)y& =0\end{array}$$

Where

$$\begin{array}{rl}p\left(t\right)& =-\frac{1}{t}\\ q\left(t\right)& =4t^2\end{array}$$

Applying change of variables $\tau =g\left(t\right)$ to (2) gives

$$\begin{array}{}\text{(3)}& \frac{d^2}{d{\tau }^2}y\left(\tau \right)+p_1\left(\frac{d}{d\tau }y\left(\tau \right)\right)+q_{1}y\left(\tau \right)& =0\end{array}$$

Where $\tau$ is the new independent variable, and

$$\begin{array}{}\text{(4)}& p_1\left(\tau \right)& =\frac{{\tau }^{\prime \prime }\left(t\right)+p\left(t\right){\tau }^{\prime }\left(t\right)}{{{\tau }^{\prime }\left(t\right)}^2}\text{(5)}& q_1\left(\tau \right)& =\frac{q\left(t\right)}{{{\tau }^{\prime }\left(t\right)}^2}\end{array}$$

Let $p_1=0$. Eq (4) simplifies to

$$\begin{array}{rl}{\tau }^{\prime \prime }\left(t\right)+p\left(t\right){\tau }^{\prime }\left(t\right)& =0\end{array}$$

This ode is solved resulting in

$$\begin{array}{rl}\tau & =\int {\mathrm{e}}^{-\int p\left(t\right)dt}dt\\ & =\int {\mathrm{e}}^{-\int -\frac{1}{t}dt}dt\\ & =\int e^{\ln \left(t\right)}dt\\ & =\int tdt\\ \text{(6)}& & =\frac{t^2}{2}\end{array}$$

Using (6) to evaluate $q_1$ from (5) gives

$$\begin{array}{rl}{q}_1\left(\tau \right)& =\frac{q\left(t\right)}{{{\tau }^{\prime }\left(t\right)}^2}\\ & =\frac{4t^2}{t^2}\\ \text{(7)}& & =4\end{array}$$

Substituting the above in (3) and noting that now $p_1=0$ results in

$$\begin{array}{rl}\frac{d^2}{d{\tau }^2}y\left(\tau \right)+q_{1}y\left(\tau \right)& =0\\ \frac{d^2}{d{\tau }^2}y\left(\tau \right)+4y\left(\tau \right)& =0\end{array}$$

The above ode is now solved for $y\left(\tau \right)$.This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

$$A{y}^{″}(\tau )+B{y}^{\prime }(\tau )+Cy(\tau )=0$$

Where in the above $A=1,B=0,C=4$. Let the solution be $y\left(\tau \right)=e^{\lambda \tau }$. Substituting this into the ODE gives

$$\begin{array}{}\text{(1)}& {\lambda }^2{\mathrm{e}}^{\tau \lambda }+4{\mathrm{e}}^{\tau \lambda }=0\end{array}$$

Since exponential function is never zero, then dividing Eq(2) throughout by $e^{\lambda \tau }$ gives

$$\begin{array}{}\text{(2)}& {\lambda }^2+4=0\end{array}$$

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula

$${\lambda }_{1,2}=\frac{-B}{2A}\pm \frac{1}{2A}\sqrt{B^2-4AC}$$

Substituting $A=1,B=0,C=4$ into the above gives

$$\begin{array}{rl}{\lambda }_{1,2}& =\frac{0}{(2)\left(1\right)}\pm \frac{1}{(2)\left(1\right)}\sqrt{0^2-(4)\left(1\right)\left(4\right)}\\ & =\pm 2i\end{array}$$

Hence

$$\begin{array}{rl}{\lambda }_1& =+2i\\ {\lambda }_2& =-2i\end{array}$$

Which simplifies to

$$\begin{array}{rl}{\lambda }_1& =2i\\ {\lambda }_2& =-2i\end{array}$$

Since roots are complex conjugate of each others, then let the roots be

$${\lambda }_{1,2}=\alpha \pm i\beta$$

Where $\alpha =0$ and $\beta =2$. Therefore the final solution, when using Euler relation, can be written as

$$y\left(\tau \right)=e^{\alpha \tau }(c_1\cos (\beta \tau )+c_2\sin (\beta \tau ))$$

Which becomes

$$y\left(\tau \right)=e^0(c_1\cos \left(2\tau \right)+c_2\sin \left(2\tau \right))$$

Or

$$y\left(\tau \right)=c_1\cos \left(2\tau \right)+c_2\sin \left(2\tau \right)$$

Will add steps showing solving for IC soon.

The above solution is now transformed back to $y$ using (6) which results in

$$y=c_1\cos \left(t^2\right)+c_2\sin \left(t^2\right)$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_1\cos \left(t^2\right)+c_2\sin \left(t^2\right)\end{array}$$

Solved as second order ode using change of variable on x method 1

Time used: 0.168 (sec)

In normal form the ode

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

Becomes

$$\begin{array}{}\text{(2)}& y^{\prime \prime }+p\left(t\right)y^{\prime }+q\left(t\right)y& =0\end{array}$$

Where

$$\begin{array}{rl}p\left(t\right)& =-\frac{1}{t}\\ q\left(t\right)& =4t^2\end{array}$$

Applying change of variables $\tau =g\left(t\right)$ to (2) results

$$\begin{array}{}\text{(3)}& \frac{d^2}{d{\tau }^2}y\left(\tau \right)+p_1\left(\frac{d}{d\tau }y\left(\tau \right)\right)+q_{1}y\left(\tau \right)& =0\end{array}$$

Where $\tau$ is the new independent variable, and

$$\begin{array}{}\text{(4)}& p_1\left(\tau \right)& =\frac{{\tau }^{\prime \prime }\left(t\right)+p\left(t\right){\tau }^{\prime }\left(t\right)}{{{\tau }^{\prime }\left(t\right)}^2}\text{(5)}& q_1\left(\tau \right)& =\frac{q\left(t\right)}{{{\tau }^{\prime }\left(t\right)}^2}\end{array}$$

Let $q_1=c^2$ where $c$ is some constant. Therefore from (5)

$$\begin{array}{rl}{\tau }^{\prime }& =\frac{1}{c}\sqrt{q}\\ \text{(6)}& & =\frac{2\sqrt{t^2}}{c}{\tau }^{″}& =\frac{2t}{c\sqrt{t^2}}\end{array}$$

Substituting the above into (4) results in

$$\begin{array}{rl}{p}_1\left(\tau \right)& =\frac{{\tau }^{\prime \prime }\left(t\right)+p\left(t\right){\tau }^{\prime }\left(t\right)}{{{\tau }^{\prime }\left(t\right)}^2}\\ & =\frac{\frac{2t}{c\sqrt{t^2}}-\frac{1}{t}\frac{2\sqrt{t^2}}{c}}{{\left(\frac{2\sqrt{t^2}}{c}\right)}^2}\\ & =0\end{array}$$

Therefore ode (3) now becomes

$$\begin{array}{rl}y{\left(\tau \right)}^{″}+p_{1}y{\left(\tau \right)}^{\prime }+q_{1}y\left(\tau \right)& =0\\ \text{(7)}& \frac{d^2}{d{\tau }^2}y\left(\tau \right)+c^{2}y\left(\tau \right)& =0\end{array}$$

The above ode is now solved for $y\left(\tau \right)$. Since the ode is now constant coefficients, it can be easily solved to give

$$\begin{array}{rl}y\left(\tau \right)& =c_1\cos \left(c\tau \right)+c_2\sin \left(c\tau \right)\end{array}$$

Now from (6)

$$\begin{array}{rl}\tau & =\int \frac{1}{c}\sqrt{q}dt\\ & =\frac{\int 2\sqrt{t^2}dt}{c}\\ & =\frac{t\sqrt{t^2}}{c}\end{array}$$

Substituting the above into the solution obtained gives

$$y=c_1\cos \left(t\sqrt{t^2}\right)+c_2\sin \left(t\sqrt{t^2}\right)$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_1\cos \left(t\sqrt{t^2}\right)+c_2\sin \left(t\sqrt{t^2}\right)\end{array}$$

Solved as second order Bessel ode

Time used: 0.056 (sec)

Writing the ode as

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

Bessel ode has the form

$$\begin{array}{}\text{(2)}& t^2{y}^{\prime \prime }+y^{\prime }t+(-n^2+t^2)y=0\end{array}$$

The generalized form of Bessel ode is given by Bowman (1958) as the following

$$\begin{array}{}\text{(3)}& t^2{y}^{\prime \prime }+(1-2\alpha )t{y}^{\prime }+({\beta }^2{\gamma }^2{t}^{2\gamma }-n^2{\gamma }^2+{\alpha }^2)y=0\end{array}$$

With the standard solution

$$\begin{array}{}\text{(4)}& y& =t^{\alpha }(c_1\mathrm{BesselJ}(n,\beta t^{\gamma })+c_2\mathrm{BesselY}(n,\beta t^{\gamma }))\end{array}$$

Comparing (3) to (1) and solving for $\alpha ,\beta ,n,\gamma$ gives

$$\begin{array}{rl}\alpha & =1\\ \beta & =1\\ n& =\frac{1}{2}\\ \gamma & =2\end{array}$$

Substituting all the above into (4) gives the solution as

$$\begin{array}{r}y=\frac{c_{1}t\sqrt{2}\sin \left(t^2\right)}{\sqrt{\pi }\sqrt{t^2}}-\frac{c_{2}t\sqrt{2}\cos \left(t^2\right)}{\sqrt{\pi }\sqrt{t^2}}\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =\frac{c_{1}t\sqrt{2}\sin \left(t^2\right)}{\sqrt{\pi }\sqrt{t^2}}-\frac{c_{2}t\sqrt{2}\cos \left(t^2\right)}{\sqrt{\pi }\sqrt{t^2}}\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.306 (sec)

Writing the ode as

$$\begin{array}{}\text{(1)}& t{y}^{\prime \prime }-y^{\prime }+4t^{3}y& =0\text{(2)}& A{y}^{\prime \prime }+B{y}^{\prime }+Cy& =0\end{array}$$

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =t\\ \text{(3)}& B& =-1C& =4t^3\end{array}$$

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

$$\begin{array}{}\text{(6)}& r& =\frac{-16t^4+3}{4t^2}\end{array}$$

Comparing the above to (5) shows that

$$\begin{array}{rl}s& =-16t^4+3\\ t& =4t^2\end{array}$$

Therefore eq. (4) becomes

$$\begin{array}{}\text{(7)}& z^{″}(t)& =\left(\frac{-16t^4+3}{4t^2}\right)z(t)\end{array}$$

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 2.10: Necessary conditions for each Kovacic case

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

$$\begin{array}{rl}O\left(\mathrm{\infty }\right)& =\text{deg}(t)-\text{deg}(s)\\ & =2-4\\ & =-2\end{array}$$

The poles of $r$ in eq. (7) and the order of each pole are determined by solving for the roots of $t=4t^2$. 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. Therefore

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

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

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

$$r=-4t^2+\frac{3}{4t^2}$$

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 $b=\frac{3}{4}$. Hence

$$\begin{array}{rl}[\sqrt{r}{]}_c& =0\\ {\alpha }_c^{+}& =\frac{1}{2}+\sqrt{1+4b}& & =\frac{3}{2}\\ {\alpha }_c^{-}& =\frac{1}{2}-\sqrt{1+4b}& & =-\frac{1}{2}\end{array}$$

Since the order of $r$ at $\mathrm{\infty }$ is $O_r(\mathrm{\infty })=-2$ then

$$\begin{array}{rlrlrl}v& =\frac{-O_r(\mathrm{\infty })}{2}& & =\frac{2}{2}& & =1\end{array}$$

$[\sqrt{r}{]}_{\mathrm{\infty }}$ is the sum of terms involving $t^i$ for $0\le i\le v$ in the Laurent series for $\sqrt{r}$ at $\mathrm{\infty }$. Therefore

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^v{a}_i{t}^i\\ \text{(8)}& & =\sum _{i=0}^1{a}_i{t}^i\end{array}$$

Let $a$ be the coefficient of $t^v=t^1$ in the above sum. The Laurent series of $\sqrt{r}$ at $\mathrm{\infty }$ is

$$\begin{array}{}\text{(9)}& \sqrt{r}\approx 2it-\frac{3i}{16t^3}-\frac{9i}{1024t^7}-\frac{27i}{32768t^{11}}-\frac{405i}{4194304t^{15}}-\frac{1701i}{134217728t^{19}}-\frac{15309i}{8589934592t^{23}}-\frac{72171i}{274877906944t^{27}}+\dots \end{array}$$

Comparing Eq. (9) with Eq. (8) shows that

$$a=2i$$

From Eq. (9) the sum up to $v=1$ gives

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =\sum _{i=0}^1{a}_i{t}^i\\ \text{(10)}& & =2it\end{array}$$

Now we need to find $b$, where $b$ be the coefficient of $t^{v-1}=t^0=1$ in $r$ minus the coefficient of same term but in ${([\sqrt{r}{]}_{\mathrm{\infty }})}^2$ where $[\sqrt{r}{]}_{\mathrm{\infty }}$ was found above in Eq (10). Hence

$${([\sqrt{r}{]}_{\mathrm{\infty }})}^2=-4t^2$$

This shows that the coefficient of $1$ in the above is $0$. Now we need to find the coefficient of $1$ in $r$. How this is done depends on if $v=0$ or not. Since $v=1$ which is not zero, then starting $r=\frac{s}{t}$, we do long division and write this in the form

$$r=Q+\frac{R}{t}$$

Where $Q$ is the quotient and $R$ is the remainder. Then the coefficient of $1$ in $r$ will be the coefficient this term in the quotient. Doing long division gives

$$\begin{array}{rl}r& =\frac{s}{t}\\ & =\frac{-16t^4+3}{4t^2}\\ & =Q+\frac{R}{4t^2}\\ & =(-4t^2)+\left(\frac{3}{4t^2}\right)\\ & =-4t^2+\frac{3}{4t^2}\end{array}$$

We see that the coefficient of the term $t$ in the quotient is $0$. Now $b$ can be found.

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

Hence

$$\begin{array}{rl}[\sqrt{r}{]}_{\mathrm{\infty }}& =2it\\ {\alpha }_{\mathrm{\infty }}^{+}& =\frac{1}{2}(\frac{b}{a}-v)& & =\frac{1}{2}(\frac{0}{2i}-1)& & =-\frac{1}{2}\\ {\alpha }_{\mathrm{\infty }}^{-}& =\frac{1}{2}(-\frac{b}{a}-v)& & =\frac{1}{2}(-\frac{0}{2i}-1)& & =-\frac{1}{2}\end{array}$$

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

$$r=\frac{-16t^4+3}{4t^2}$$

pole $c$ location	pole order	$[\sqrt{r}{]}_c$	${\alpha }_c^{+}$	${\alpha }_c^{-}$
$0$	$2$	$0$	$\frac{3}{2}$	$-\frac{1}{2}$

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

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 ${\alpha }_{\mathrm{\infty }}^{-}=-\frac{1}{2}$ then

$$\begin{array}{rl}d& ={\alpha }_{\mathrm{\infty }}^{-}-\left({\alpha }_{c_1}^{-}\right)\\ & =-\frac{1}{2}-(-\frac{1}{2})\\ & =0\end{array}$$

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

$$\begin{array}{rl}\omega & =((-)[\sqrt{r}{]}_{c_1}+\frac{{\alpha }_{c_1}^{-}}{t-c_1})+(-)[\sqrt{r}{]}_{\mathrm{\infty }}\\ & =-\frac{1}{2t}+(-)\left(2it\right)\\ & =-\frac{1}{2t}-2it\\ & =-\frac{1}{2t}-2it\end{array}$$

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

$$\begin{array}{rl}\left(0\right)+2(-\frac{1}{2t}-2it)\left(0\right)+((\frac{1}{2t^2}-2i)+{(-\frac{1}{2t}-2it)}^2-\left(\frac{-16t^4+3}{4t^2}\right))& =0\\ 0=0\end{array}$$

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

$$\begin{array}{rl}{z}_1(t)& =p{e}^{\int \omega dt}\\ & ={\mathrm{e}}^{\int (-\frac{1}{2t}-2it)dt}\\ & =\frac{{\mathrm{e}}^{-i{t}^2}}{\sqrt{t}}\end{array}$$

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

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

Which simplifies to

$$y_1={\mathrm{e}}^{-i{t}^2}$$

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

$$\begin{array}{rl}{y}_2& =y_1\int \frac{e^{\int -\frac{-1}{t}dt}}{{\left(y_1\right)}^2}dt\\ & =y_1\int \frac{e^{\ln \left(t\right)}}{{\left(y_1\right)}^2}dt\\ & =y_1(-\frac{i{\mathrm{e}}^{2i{t}^2}}{4})\end{array}$$

Therefore the solution is

$$\begin{array}{rl}y& =c_1{y}_1+c_2{y}_2\\ & =c_1\left({\mathrm{e}}^{-i{t}^2}\right)+c_2\left({\mathrm{e}}^{-i{t}^2}(-\frac{i{\mathrm{e}}^{2i{t}^2}}{4})\right)\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_1{\mathrm{e}}^{-i{t}^2}-\frac{i{c}_2{\mathrm{e}}^{i{t}^2}}{4}\end{array}$$

Solved as second order ode adjoint method

Time used: 0.411 (sec)

In normal form the ode

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

Becomes

$$\begin{array}{}\text{(2)}& y^{\prime \prime }+p\left(t\right)y^{\prime }+q\left(t\right)y& =r\left(t\right)\end{array}$$

Where

$$\begin{array}{rl}p\left(t\right)& =-\frac{1}{t}\\ q\left(t\right)& =4t^2\\ r\left(t\right)& =0\end{array}$$

The Lagrange adjoint ode is given by

$$\begin{array}{rl}{\xi }^{{}^{″}}-(\xi p{)}^{\prime }+\xi q& =0\\ {\xi }^{{}^{″}}-{(-\frac{\xi \left(t\right)}{t})}^{\prime }+\left(4t^2\xi \left(t\right)\right)& =0\\ {\xi }^{\prime \prime }\left(t\right)+\frac{{\xi }^{\prime }\left(t\right)}{t}+\frac{(4t^4-1)\xi \left(t\right)}{t^2}& =0\end{array}$$

Which is solved for $\xi (t)$. Writing the ode as

$$\begin{array}{}\text{(1)}& {\xi }^{\prime \prime }{t}^2+{\xi }^{\prime }t+(4t^4-1)\xi =0\end{array}$$

Bessel ode has the form

$$\begin{array}{}\text{(2)}& {\xi }^{\prime \prime }{t}^2+{\xi }^{\prime }t+(-n^2+t^2)\xi =0\end{array}$$

The generalized form of Bessel ode is given by Bowman (1958) as the following

$$\begin{array}{}\text{(3)}& {\xi }^{\prime \prime }{t}^2+(1-2\alpha )t{\xi }^{\prime }+({\beta }^2{\gamma }^2{t}^{2\gamma }-n^2{\gamma }^2+{\alpha }^2)\xi =0\end{array}$$

With the standard solution

$$\begin{array}{}\text{(4)}& \xi & =t^{\alpha }(c_1\mathrm{BesselJ}(n,\beta t^{\gamma })+c_2\mathrm{BesselY}(n,\beta t^{\gamma }))\end{array}$$

Comparing (3) to (1) and solving for $\alpha ,\beta ,n,\gamma$ gives

$$\begin{array}{rl}\alpha & =0\\ \beta & =1\\ n& =-\frac{1}{2}\\ \gamma & =2\end{array}$$

Substituting all the above into (4) gives the solution as

$$\begin{array}{r}\xi =\frac{c_1\sqrt{2}\cos \left(t^2\right)}{\sqrt{\pi }\sqrt{t^2}}+\frac{c_2\sqrt{2}\sin \left(t^2\right)}{\sqrt{\pi }\sqrt{t^2}}\end{array}$$

Will add steps showing solving for IC soon.

The original ode now reduces to first order ode

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

Or

$$\begin{array}{rl}{y}^{\prime }+y(-\frac{1}{t}-\frac{-\frac{c_1\sqrt{2}\cos \left(t^2\right)t}{\sqrt{\pi }{\left(t^2\right)}^{3/2}}-\frac{2c_{1}t\sqrt{2}\sin \left(t^2\right)}{\sqrt{\pi }\sqrt{t^2}}-\frac{c_2\sqrt{2}\sin \left(t^2\right)t}{\sqrt{\pi }{\left(t^2\right)}^{3/2}}+\frac{2c_{2}t\sqrt{2}\cos \left(t^2\right)}{\sqrt{\pi }\sqrt{t^2}}}{\frac{c_1\sqrt{2}\cos \left(t^2\right)}{\sqrt{\pi }\sqrt{t^2}}+\frac{c_2\sqrt{2}\sin \left(t^2\right)}{\sqrt{\pi }\sqrt{t^2}}})& =0\end{array}$$

Which is now a first order ode. This 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

$$\begin{array}{rl}q(t)& =-\frac{2t(c_2\cos \left(t^2\right)-c_1\sin \left(t^2\right))}{c_1\cos \left(t^2\right)+c_2\sin \left(t^2\right)}\\ p(t)& =0\end{array}$$

The integrating factor $\mu$ is

$$\begin{array}{rl}\mu & =e^{\int qdt}\\ & ={\mathrm{e}}^{\int -\frac{2t(c_2\cos \left(t^2\right)-c_1\sin \left(t^2\right))}{c_1\cos \left(t^2\right)+c_2\sin \left(t^2\right)}dt}\\ & =\frac{1}{c_1\cos \left(t^2\right)+c_2\sin \left(t^2\right)}\end{array}$$

The ode becomes

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}t}\mu y& =0\\ \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{y}{c_1\cos \left(t^2\right)+c_2\sin \left(t^2\right)}\right)& =0\end{array}$$

Integrating gives

$$\begin{array}{rl}\frac{y}{c_1\cos \left(t^2\right)+c_2\sin \left(t^2\right)}& =\int 0dt+c_3\\ & =c_3\end{array}$$

Dividing throughout by the integrating factor $\frac{1}{c_1\cos \left(t^2\right)+c_2\sin \left(t^2\right)}$ gives the final solution

$$y=(c_1\cos \left(t^2\right)+c_2\sin \left(t^2\right))c_3$$

Hence, the solution found using Lagrange adjoint equation method is

$$\begin{array}{rl}y& =(c_1\cos \left(t^2\right)+c_2\sin \left(t^2\right))c_3\end{array}$$

The constants can be merged to give

$$y=c_1\cos \left(t^2\right)+c_2\sin \left(t^2\right)$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_1\cos \left(t^2\right)+c_2\sin \left(t^2\right)\end{array}$$

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

ode:=t*diff(diff(y(t),t),t)-diff(y(t),t)+4*t^3*y(t) = 0; 
dsolve(ode,y(t), singsol=all);
 

$$y=c_1\sin \left(t^2\right)+c_2\cos \left(t^2\right)$$

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
trying a symmetry of the form [xi=0, eta=F(x)] 
<- linear_1 successful`
 

Maple step by step

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & t\left(\frac{d^2}{d{t}^2}y\left(t\right)\right)-\frac{d}{dt}y\left(t\right)+4t^{3}y\left(t\right)=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d^2}{d{t}^2}y\left(t\right)\\ \bullet & & \text{Isolate 2nd derivative}\\ & & \frac{d^2}{d{t}^2}y\left(t\right)=-4t^{2}y\left(t\right)+\frac{\frac{d}{dt}y\left(t\right)}{t}\\ \bullet & & \text{Group terms with}y\left(t\right)\text{on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\\ & & \frac{d^2}{d{t}^2}y\left(t\right)-\frac{\frac{d}{dt}y\left(t\right)}{t}+4t^{2}y\left(t\right)=0\\ ◻& & \text{Check to see if}{t}_0=0\text{is a regular singular point}\\ & \circ & \text{Define functions}\\ & & [P_2\left(t\right)=-\frac{1}{t},P_3\left(t\right)=4t^2]\\ & \circ & t\cdot P_2\left(t\right)\text{is analytic at}t=0\\ & & (t\cdot P_2\left(t\right))|{}_{t=0}=-1\\ & \circ & t^2\cdot P_3\left(t\right)\text{is analytic at}t=0\\ & & (t^2\cdot P_3\left(t\right))|{}_{t=0}=0\\ & \circ & t=0\text{is a regular singular point}\\ & & \text{Check to see if}{t}_0=0\text{is a regular singular point}\\ & & t_0=0\\ \bullet & & \text{Multiply by denominators}\\ & & t\left(\frac{d^2}{d{t}^2}y\left(t\right)\right)-\frac{d}{dt}y\left(t\right)+4t^{3}y\left(t\right)=0\\ \bullet & & \text{Assume series solution for}y\left(t\right)\\ & & y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^{k+r}\\ ◻& & \text{Rewrite ODE with series expansions}\\ & \circ & \text{Convert}{t}^3\cdot y\left(t\right)\text{to series expansion}\\ & & t^3\cdot y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^{k+r+3}\\ & \circ & \text{Shift index using}k->k-3\\ & & t^3\cdot y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=3}}{a}_{k-3}{t}^{k+r}\\ & \circ & \text{Convert}\frac{d}{dt}y\left(t\right)\text{to series expansion}\\ & & \frac{d}{dt}y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)t^{k+r-1}\\ & \circ & \text{Shift index using}k->k+1\\ & & \frac{d}{dt}y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1}}{a}_{k+1}(k+1+r)t^{k+r}\\ & \circ & \text{Convert}t\cdot \left(\frac{d^2}{d{t}^2}y\left(t\right)\right)\text{to series expansion}\\ & & t\cdot \left(\frac{d^2}{d{t}^2}y\left(t\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k(k+r)(k+r-1)t^{k+r-1}\\ & \circ & \text{Shift index using}k->k+1\\ & & t\cdot \left(\frac{d^2}{d{t}^2}y\left(t\right)\right)=\stackrel{\mathrm{\infty }}{\sum _{k=-1}}{a}_{k+1}(k+1+r)(k+r)t^{k+r}\\ & & \text{Rewrite ODE with series expansions}\\ & & a_{0}r(-2+r)t^{-1+r}+a_1(1+r)(-1+r)t^r+a_2(2+r)r{t}^{1+r}+a_3(3+r)(1+r)t^{2+r}+(\stackrel{\mathrm{\infty }}{\sum _{k=3}}(a_{k+1}(k+1+r)(k+r-1)+4a_{k-3})t^{k+r})=0\\ \bullet & & a_0\text{cannot be 0 by assumption, giving the indicial equation}\\ & & r(-2+r)=0\\ \bullet & & \text{Values of r that satisfy the indicial equation}\\ & & r\in \{0,2\}\\ \bullet & & \text{The coefficients of each power of}t\text{must be 0}\\ & & [a_1(1+r)(-1+r)=0,a_2(2+r)r=0,a_3(3+r)(1+r)=0]\\ \bullet & & \text{Solve for the dependent coefficient(s)}\\ & & \{a_1=0,a_2=0,a_3=0\}\\ \bullet & & \text{Each term in the series must be 0, giving the recursion relation}\\ & & a_{k+1}(k+1+r)(k+r-1)+4a_{k-3}=0\\ \bullet & & \text{Shift index using}k->k+3\\ & & a_{k+4}(k+4+r)(k+2+r)+4a_k=0\\ \bullet & & \text{Recursion relation that defines series solution to ODE}\\ & & a_{k+4}=-\frac{4a_k}{(k+4+r)(k+2+r)}\\ \bullet & & \text{Recursion relation for}r=0\\ & & a_{k+4}=-\frac{4a_k}{(k+4)(k+2)}\\ \bullet & & \text{Solution for}r=0\\ & & [y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^k,a_{k+4}=-\frac{4a_k}{(k+4)(k+2)},a_1=0,a_2=0,a_3=0]\\ \bullet & & \text{Recursion relation for}r=2\\ & & a_{k+4}=-\frac{4a_k}{(k+6)(k+4)}\\ \bullet & & \text{Solution for}r=2\\ & & [y\left(t\right)=\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^{k+2},a_{k+4}=-\frac{4a_k}{(k+6)(k+4)},a_1=0,a_2=0,a_3=0]\\ \bullet & & \text{Combine solutions and rename parameters}\\ & & [y\left(t\right)=(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{a}_k{t}^k)+(\stackrel{\mathrm{\infty }}{\sum _{k=0}}{b}_k{t}^{k+2}),a_{4+k}=-\frac{4a_k}{(4+k)(k+2)},a_1=0,a_2=0,a_3=0,b_{4+k}=-\frac{4b_k}{(k+6)(4+k)},b_1=0,b_2=0,b_3=0]\end{array}$$

✓ Mathematica. Time used: 0.02 (sec). Leaf size: 20

ode=t*D[y[t],{t,2}]-D[y[t],t]+4*t^3*y[t]==0; 
ic={}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

$$y(t)\to c_1\cos \left(t^2\right)+c_2\sin \left(t^2\right)$$

✓ Sympy. Time used: 0.164 (sec). Leaf size: 20

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(4*t**3*y(t) + t*Derivative(y(t), (t, 2)) - Derivative(y(t), t),0) 
ics = {} 
dsolve(ode,func=y(t),ics=ics)
 

$$y\left(t\right)=t(C_1{J}_{\frac{1}{2}}\left(t^2\right)+C_2{Y}_{\frac{1}{2}}\left(t^2\right))$$