2.2.39 Problem 38

Internal problem ID [8842]
Book : Own collection of miscellaneous problems
Section : section 2.0
Problem number : 38
Date solved : Friday, April 25, 2025 at 05:13:31 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

Solved as second order Bessel ode

Time used: 0.631 (sec)

Solve

$$\begin{array}{rl}{y}^{\prime \prime }-2x^{2}y-x^4+1& =0\end{array}$$

Writing the ode as

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

Let the solution be

$$\begin{array}{rl}y& =y_h+y_p\end{array}$$

Where $y_h$ is the solution to the homogeneous ODE and $y_p$ is a particular solution to the non-homogeneous ODE. Bessel ode has the form

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

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

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

With the standard solution

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

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

$$\begin{array}{rl}\alpha & =\frac{1}{2}\\ \beta & =\frac{i\sqrt{2}}{2}\\ n& =\frac{1}{4}\\ \gamma & =2\end{array}$$

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

$$\begin{array}{r}y=c_1\sqrt{x}\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})+c_2\sqrt{x}\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})\end{array}$$

Therefore the homogeneous solution $y_h$ is

$$y_h=c_1\sqrt{x}\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})+c_2\sqrt{x}\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})$$

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

$$\begin{array}{rl}{y}_1& =\sqrt{x}\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})\\ y_2& =\sqrt{x}\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})\end{array}$$

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

$$W=\left|\begin{array}{cc}\sqrt{x}\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})& \sqrt{x}\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})\\ \frac{d}{dx}\left(\sqrt{x}\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})\right)& \frac{d}{dx}\left(\sqrt{x}\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})\right)\end{array}\right|$$

Which gives

$$W=\left|\begin{array}{cc}\sqrt{x}\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})& \sqrt{x}\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})\\ \frac{\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})}{2\sqrt{x}}+i{x}^{3/2}(-\mathrm{BesselJ}(\frac{5}{4},\frac{i\sqrt{2}{x}^2}{2})-\frac{i\sqrt{2}\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})}{4x^2})\sqrt{2}& \frac{\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})}{2\sqrt{x}}+i{x}^{3/2}(-\mathrm{BesselY}(\frac{5}{4},\frac{i\sqrt{2}{x}^2}{2})-\frac{i\sqrt{2}\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})}{4x^2})\sqrt{2}\end{array}\right|$$

Therefore

$$W=\left(\sqrt{x}\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})\right)(\frac{\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})}{2\sqrt{x}}+i{x}^{3/2}(-\mathrm{BesselY}(\frac{5}{4},\frac{i\sqrt{2}{x}^2}{2})-\frac{i\sqrt{2}\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})}{4x^2})\sqrt{2})-\left(\sqrt{x}\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})\right)(\frac{\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})}{2\sqrt{x}}+i{x}^{3/2}(-\mathrm{BesselJ}(\frac{5}{4},\frac{i\sqrt{2}{x}^2}{2})-\frac{i\sqrt{2}\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})}{4x^2})\sqrt{2})$$

Which simplifies to

$$W=-i{x}^2\sqrt{2}(\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})\mathrm{BesselY}(\frac{5}{4},\frac{i\sqrt{2}{x}^2}{2})-\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})\mathrm{BesselJ}(\frac{5}{4},\frac{i\sqrt{2}{x}^2}{2}))$$

Which simplifies to

$$W=\frac{4}{\pi }$$

Therefore Eq. (2) becomes

$$u_1=-\int \frac{x^{5/2}\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})(x^4-1)}{\frac{4x^2}{\pi }}dx$$

Which simplifies to

$$u_1=-\int \frac{\sqrt{x}\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})(x^4-1)\pi }{4}dx$$

Hence

$$u_1=-{\int }_0^x\frac{\sqrt{\alpha }\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{\alpha }^2}{2})({\alpha }^4-1)\pi }{4}d\alpha$$

And Eq. (3) becomes

$$u_2=\int \frac{x^{5/2}\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})(x^4-1)}{\frac{4x^2}{\pi }}dx$$

Which simplifies to

$$u_2=\int \frac{\sqrt{x}\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})(x^4-1)\pi }{4}dx$$

Hence

$$u_2={\int }_0^x\frac{\sqrt{\alpha }\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{\alpha }^2}{2})({\alpha }^4-1)\pi }{4}d\alpha$$

Therefore the particular solution, from equation (1) is

$$y_p(x)=-{\int }_0^x\frac{\sqrt{\alpha }\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{\alpha }^2}{2})({\alpha }^4-1)\pi }{4}d\alpha \sqrt{x}\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})+\sqrt{x}\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2}){\int }_0^x\frac{\sqrt{\alpha }\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{\alpha }^2}{2})({\alpha }^4-1)\pi }{4}d\alpha$$

Which simplifies to

$$y_p(x)=\frac{\sqrt{x}\pi (-\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2}){\int }_0^x\sqrt{\alpha }\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{\alpha }^2}{2})({\alpha }^4-1)d\alpha +\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2}){\int }_0^x\sqrt{\alpha }\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{\alpha }^2}{2})({\alpha }^4-1)d\alpha )}{4}$$

Therefore the general solution is

$$\begin{array}{rl}y& =y_h+y_p\\ & =(c_1\sqrt{x}\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})+c_2\sqrt{x}\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2}))+\left(\frac{\sqrt{x}\pi (-\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2}){\int }_0^x\sqrt{\alpha }\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{\alpha }^2}{2})({\alpha }^4-1)d\alpha +\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2}){\int }_0^x\sqrt{\alpha }\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{\alpha }^2}{2})({\alpha }^4-1)d\alpha )}{4}\right)\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =c_1\sqrt{x}\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})+c_2\sqrt{x}\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2})+\frac{\sqrt{x}\pi (-\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2}){\int }_0^x\sqrt{\alpha }\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{\alpha }^2}{2})({\alpha }^4-1)d\alpha +\mathrm{BesselY}(\frac{1}{4},\frac{i\sqrt{2}{x}^2}{2}){\int }_0^x\sqrt{\alpha }\mathrm{BesselJ}(\frac{1}{4},\frac{i\sqrt{2}{\alpha }^2}{2})({\alpha }^4-1)d\alpha )}{4}\end{array}$$

✓ Maple. Time used: 0.004 (sec). Leaf size: 40

ode:=diff(diff(y(x),x),x)-2*x^2*y(x)-x^4+1 = 0; 
dsolve(ode,y(x), singsol=all);
 

$$y=\sqrt{x}\mathrm{BesselI}(\frac{1}{4},\frac{\sqrt{2}{x}^2}{2})c_2+\sqrt{x}\mathrm{BesselK}(\frac{1}{4},\frac{\sqrt{2}{x}^2}{2})c_1-\frac{x^2}{2}$$

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 
   trying a symmetry of the form [xi=0, eta=F(x)] 
   checking if the LODE is missing y 
   -> Trying a Liouvillian solution using Kovacics algorithm 
   <- No Liouvillian solutions exists 
   -> Trying a solution in terms of special functions: 
      -> Bessel 
      <- Bessel successful 
   <- special function solution successful 
<- solving first the homogeneous part of the ODE successful
 

✓ Mathematica. Time used: 4.079 (sec). Leaf size: 288

ode=D[y[x],{x,2}]-2*x^2*y[x]-x^4+1==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

$$y(x)\to \mathrm{ParabolicCylinderD}(-\frac{1}{2},2^{3/4}x)({\int }_1^x\frac{(K[1{]}^4-1)\mathrm{ParabolicCylinderD}(-\frac{1}{2},i{2}^{3/4}K[1])}{i{2}^{3/4}\mathrm{HermiteH}(-\frac{1}{2},\sqrt[4]{2}K[1])\mathrm{HermiteH}(\frac{1}{2},i\sqrt[4]{2}K[1])+\mathrm{ParabolicCylinderD}(-\frac{1}{2},i{2}^{3/4}K[1])(2\sqrt{2}K[1]\mathrm{ParabolicCylinderD}(-\frac{1}{2},2^{3/4}K[1])-2^{3/4}\mathrm{ParabolicCylinderD}(\frac{1}{2},2^{3/4}K[1]))}dK[1]+c_1)+\mathrm{ParabolicCylinderD}(-\frac{1}{2},i{2}^{3/4}x)({\int }_1^x\frac{i(K[2{]}^4-1)\mathrm{ParabolicCylinderD}(-\frac{1}{2},2^{3/4}K[2])}{2^{3/4}\mathrm{HermiteH}(-\frac{1}{2},\sqrt[4]{2}K[2])\mathrm{HermiteH}(\frac{1}{2},i\sqrt[4]{2}K[2])+i\mathrm{ParabolicCylinderD}(-\frac{1}{2},i{2}^{3/4}K[2])(2^{3/4}\mathrm{ParabolicCylinderD}(\frac{1}{2},2^{3/4}K[2])-2\sqrt{2}K[2]\mathrm{ParabolicCylinderD}(-\frac{1}{2},2^{3/4}K[2]))}dK[2]+c_2)$$

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-x**4 - 2*x**2*y(x) + Derivative(y(x), (x, 2)) + 1,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : solve: Cannot solve -x**4 - 2*x**2*y(x) + Derivative(y(x), (x, 2)) + 1