2.2.14 Problem 15

Internal problem ID [9137]
Book : Second order enumerated odes
Section : section 2
Problem number : 15
Date solved : Sunday, March 30, 2025 at 02:23:09 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

Solved as second order Euler type ode

Time used: 0.200 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }-\frac{2y}{x^2}& =x{\mathrm{e}}^{-\sqrt{x}}\end{array}$$

This is second order non-homogeneous ODE. In standard form the ODE is

$$A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=f(x)$$

Where $A=x^2,B=0,C=-2,f(x)=x^3{\mathrm{e}}^{-\sqrt{x}}$. Let the solution be

$$y=y_h+y_p$$

Where $y_h$ is the solution to the homogeneous ODE $A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=0$, and $y_p$ is a particular solution to the non-homogeneous ODE $A{y}^{″}(x)+B{y}^{\prime }(x)+Cy(x)=f(x)$. Solving for $y_h$ from

$$x^2{y}^{\prime \prime }-2y=0$$

This is Euler second order ODE. Let the solution be $y=x^r$, then $y^{\prime }=r{x}^{r-1}$ and $y^{″}=r(r-1)x^{r-2}$. Substituting these back into the given ODE gives

$$x^2(r(r-1))x^{r-2}+0r{x}^{r-1}-2x^r=0$$

Simplifying gives

$$r(r-1)x^r+0x^r-2x^r=0$$

Since $x^r\ne 0$ then dividing throughout by $x^r$ gives

$$r(r-1)+0-2=0$$

Or

$$\begin{array}{}\text{(1)}& r^2-r-2=0\end{array}$$

Equation (1) is the characteristic equation. Its roots determine the form of the general solution. Using the quadratic equation the roots are

$$\begin{array}{rl}{r}_1& =-1\\ r_2& =2\end{array}$$

Since the roots are real and distinct, then the general solution is

$$y=c_1{y}_1+c_2{y}_2$$

Where $y_1=x^{r_1}$ and $y_2=x^{r_2}$. Hence

$$y=\frac{c_1}{x}+c_2{x}^2$$

Next, we find the particular solution to the ODE

$$x^2{y}^{\prime \prime }-2y=x^3{\mathrm{e}}^{-\sqrt{x}}$$

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 $x$ as well. Let

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

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

$$\begin{array}{rl}{y}_1& =\frac{1}{x}\\ y_2& =x^2\end{array}$$

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

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

Where $W(x)$ 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

$$W=\left|\begin{array}{cc}\frac{1}{x}& x^2\\ \frac{d}{dx}\left(\frac{1}{x}\right)& \frac{d}{dx}\left(x^2\right)\end{array}\right|$$

Which gives

$$W=\left|\begin{array}{cc}\frac{1}{x}& x^2\\ -\frac{1}{x^2}& 2x\end{array}\right|$$

Therefore

$$W=\left(\frac{1}{x}\right)\left(2x\right)-\left(x^2\right)(-\frac{1}{x^2})$$

Which simplifies to

$$W=3$$

Which simplifies to

$$W=3$$

Therefore Eq. (2) becomes

$$u_1=-\int \frac{x^5{\mathrm{e}}^{-\sqrt{x}}}{3x^2}dx$$

Which simplifies to

$$u_1=-\int \frac{x^3{\mathrm{e}}^{-\sqrt{x}}}{3}dx$$

Hence

$$u_1=\frac{2x^{7/2}{\mathrm{e}}^{-\sqrt{x}}}{3}+\frac{14x^3{\mathrm{e}}^{-\sqrt{x}}}{3}+28x^{5/2}{\mathrm{e}}^{-\sqrt{x}}+140x^2{\mathrm{e}}^{-\sqrt{x}}+560x^{3/2}{\mathrm{e}}^{-\sqrt{x}}+1680x{\mathrm{e}}^{-\sqrt{x}}+3360\sqrt{x}{\mathrm{e}}^{-\sqrt{x}}+3360{\mathrm{e}}^{-\sqrt{x}}$$

And Eq. (3) becomes

$$u_2=\int \frac{x^2{\mathrm{e}}^{-\sqrt{x}}}{3x^2}dx$$

Which simplifies to

$$u_2=\int \frac{{\mathrm{e}}^{-\sqrt{x}}}{3}dx$$

Hence

$$u_2=-\frac{2\sqrt{x}{\mathrm{e}}^{-\sqrt{x}}}{3}-\frac{2{\mathrm{e}}^{-\sqrt{x}}}{3}$$

Therefore the particular solution, from equation (1) is

$$y_p(x)=\frac{\frac{2x^{7/2}{\mathrm{e}}^{-\sqrt{x}}}{3}+\frac{14x^3{\mathrm{e}}^{-\sqrt{x}}}{3}+28x^{5/2}{\mathrm{e}}^{-\sqrt{x}}+140x^2{\mathrm{e}}^{-\sqrt{x}}+560x^{3/2}{\mathrm{e}}^{-\sqrt{x}}+1680x{\mathrm{e}}^{-\sqrt{x}}+3360\sqrt{x}{\mathrm{e}}^{-\sqrt{x}}+3360{\mathrm{e}}^{-\sqrt{x}}}{x}+x^2(-\frac{2\sqrt{x}{\mathrm{e}}^{-\sqrt{x}}}{3}-\frac{2{\mathrm{e}}^{-\sqrt{x}}}{3})$$

Which simplifies to

$$y_p(x)=\frac{4{\mathrm{e}}^{-\sqrt{x}}(7x^{5/2}+140x^{3/2}+x^3+35x^2+840\sqrt{x}+420x+840)}{x}$$

Therefore the general solution is

$$\begin{array}{rl}y& =y_h+y_p\\ & =\frac{4{\mathrm{e}}^{-\sqrt{x}}(7x^{5/2}+140x^{3/2}+x^3+35x^2+840\sqrt{x}+420x+840)}{x}+\frac{c_1}{x}+c_2{x}^2\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =\frac{4{\mathrm{e}}^{-\sqrt{x}}(7x^{5/2}+140x^{3/2}+x^3+35x^2+840\sqrt{x}+420x+840)}{x}+\frac{c_1}{x}+c_2{x}^2\end{array}$$

Solved as second order ode using Kovacic algorithm

Time used: 0.188 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }-\frac{2y}{x^2}& =x{\mathrm{e}}^{-\sqrt{x}}\end{array}$$

Writing the ode as

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

Comparing (1) and (2) shows that

$$\begin{array}{rl}A& =1\\ \text{(3)}& B& =0C& =-\frac{2}{x^2}\end{array}$$

Applying the Liouville transformation on the dependent variable gives

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

Then (2) becomes

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

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

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

Comparing the above to (5) shows that

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

Therefore eq. (4) becomes

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

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

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

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 2.31: 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 $t=x^2$. There is a pole at $x=0$ 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

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

For the pole at $x=0$ let $b$ be the coefficient of $\frac{1}{x^2}$ 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 $\frac{1}{x^2}$ 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

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

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

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

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

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

The above gives

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

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

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

Let

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

Substituting the above in eq. (1A) gives

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

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

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

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

$$y_1=z_1{e}^{\int -\frac{1}{2}\frac{B}{A}dx}$$

Since $B=0$ then the above reduces to

$$\begin{array}{rl}{y}_1& =z_1\\ & =\frac{1}{x}\end{array}$$

Which simplifies to

$$y_1=\frac{1}{x}$$

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

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

Since $B=0$ then the above becomes

$$\begin{array}{rl}{y}_2& =y_1\int \frac{1}{y_1^2}dx\\ & =\frac{1}{x}\int \frac{1}{\frac{1}{x^2}}dx\\ & =\frac{1}{x}\left(\frac{x^3}{3}\right)\end{array}$$

Therefore the solution is

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

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

$$y^{\prime \prime }-\frac{2y}{x^2}=0$$

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

$$y_h=\frac{c_1}{x}+\frac{c_2{x}^2}{3}$$

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 $x$ as well. Let

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

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

$$\begin{array}{rl}{y}_1& =\frac{1}{x}\\ y_2& =\frac{x^2}{3}\end{array}$$

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

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

Where $W(x)$ 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

$$W=\left|\begin{array}{cc}\frac{1}{x}& \frac{x^2}{3}\\ \frac{d}{dx}\left(\frac{1}{x}\right)& \frac{d}{dx}\left(\frac{x^2}{3}\right)\end{array}\right|$$

Which gives

$$W=\left|\begin{array}{cc}\frac{1}{x}& \frac{x^2}{3}\\ -\frac{1}{x^2}& \frac{2x}{3}\end{array}\right|$$

Therefore

$$W=\left(\frac{1}{x}\right)\left(\frac{2x}{3}\right)-\left(\frac{x^2}{3}\right)(-\frac{1}{x^2})$$

Which simplifies to

$$W=1$$

Which simplifies to

$$W=1$$

Therefore Eq. (2) becomes

$$u_1=-\int \frac{\frac{x^3{\mathrm{e}}^{-\sqrt{x}}}{3}}{1}dx$$

Which simplifies to

$$u_1=-\int \frac{x^3{\mathrm{e}}^{-\sqrt{x}}}{3}dx$$

Hence

$$u_1=\frac{2x^{7/2}{\mathrm{e}}^{-\sqrt{x}}}{3}+\frac{14x^3{\mathrm{e}}^{-\sqrt{x}}}{3}+28x^{5/2}{\mathrm{e}}^{-\sqrt{x}}+140x^2{\mathrm{e}}^{-\sqrt{x}}+560x^{3/2}{\mathrm{e}}^{-\sqrt{x}}+1680x{\mathrm{e}}^{-\sqrt{x}}+3360\sqrt{x}{\mathrm{e}}^{-\sqrt{x}}+3360{\mathrm{e}}^{-\sqrt{x}}$$

And Eq. (3) becomes

$$u_2=\int \frac{{\mathrm{e}}^{-\sqrt{x}}}{1}dx$$

Which simplifies to

$$u_2=\int {\mathrm{e}}^{-\sqrt{x}}dx$$

Hence

$$u_2=-2\sqrt{x}{\mathrm{e}}^{-\sqrt{x}}-2{\mathrm{e}}^{-\sqrt{x}}$$

Therefore the particular solution, from equation (1) is

$$y_p(x)=\frac{\frac{2x^{7/2}{\mathrm{e}}^{-\sqrt{x}}}{3}+\frac{14x^3{\mathrm{e}}^{-\sqrt{x}}}{3}+28x^{5/2}{\mathrm{e}}^{-\sqrt{x}}+140x^2{\mathrm{e}}^{-\sqrt{x}}+560x^{3/2}{\mathrm{e}}^{-\sqrt{x}}+1680x{\mathrm{e}}^{-\sqrt{x}}+3360\sqrt{x}{\mathrm{e}}^{-\sqrt{x}}+3360{\mathrm{e}}^{-\sqrt{x}}}{x}+\frac{x^2(-2\sqrt{x}{\mathrm{e}}^{-\sqrt{x}}-2{\mathrm{e}}^{-\sqrt{x}})}{3}$$

Which simplifies to

$$y_p(x)=\frac{4{\mathrm{e}}^{-\sqrt{x}}(7x^{5/2}+140x^{3/2}+x^3+35x^2+840\sqrt{x}+420x+840)}{x}$$

Therefore the general solution is

$$\begin{array}{rl}y& =y_h+y_p\\ & =(\frac{c_1}{x}+\frac{c_2{x}^2}{3})+\left(\frac{4{\mathrm{e}}^{-\sqrt{x}}(7x^{5/2}+140x^{3/2}+x^3+35x^2+840\sqrt{x}+420x+840)}{x}\right)\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =\frac{4{\mathrm{e}}^{-\sqrt{x}}(7x^{5/2}+140x^{3/2}+x^3+35x^2+840\sqrt{x}+420x+840)}{x}+\frac{c_1}{x}+\frac{c_2{x}^2}{3}\end{array}$$

✓ Maple. Time used: 0.008 (sec). Leaf size: 51

ode:=diff(diff(y(x),x),x)-2/x^2*y(x) = x*exp(-x^(1/2)); 
dsolve(ode,y(x), singsol=all);
 

$$y=\frac{4{\mathrm{e}}^{-\sqrt{x}}(7x^{5/2}+140x^{3/2}+x^3+35x^2+840\sqrt{x}+420x+840)+c_2{x}^3+c_1}{x}$$

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
-> Try solving first the homogeneous part of the ODE 
   checking if the LODE has constant coefficients 
   checking if the LODE is of Euler type 
   <- LODE of Euler type successful 
<- solving first the homogeneous part of the ODE successful
 

✓ Mathematica. Time used: 0.046 (sec). Leaf size: 57

ode=D[y[x],{x,2}]-2/x^2*y[x] == x*Exp[-x^(1/2)]; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to x^2{\int }_1^x\frac{1}{3}{e}^{-\sqrt{K[1]}}dK[1]+c_2{x}^2+\frac{2\mathrm{\Gamma }(8,\sqrt{x})}{3x}+\frac{c_1}{x}$$

✓ Sympy. Time used: 1.005 (sec). Leaf size: 66

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

$$y\left(x\right)=\frac{(C_1{x}^3{e}^{\sqrt{x}}+C_2{e}^{\sqrt{x}}+28x^{\frac{5}{2}}+560x^{\frac{3}{2}}+3360\sqrt{x}+4x^3+140x^2+1680x+3360)e^{-\sqrt{x}}}{x}$$