6.10.23 problem 23

Internal problem ID [1827]
Book : Elementary differential equations with boundary value problems. William F. Trench. Brooks/Cole 2001
Section : Chapter 5 linear second order equations. Section 5.7 Variation of Parameters. Page 262
Problem number : 23
Date solved : Tuesday, September 30, 2025 at 05:20:09 AM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

\begin{align*} x^{2} y^{\prime \prime }-2 x \left (x +1\right ) y^{\prime }+\left (x^{2}+2 x +2\right ) y&=x^{3} {\mathrm e}^{x} \end{align*}

✓ Maple. Time used: 0.007 (sec). Leaf size: 20

ode:=x^2*diff(diff(y(x),x),x)-2*x*(1+x)*diff(y(x),x)+(x^2+2*x+2)*y(x) = x^3*exp(x); 
dsolve(ode,y(x), singsol=all);
 

\[ y = \frac {{\mathrm e}^{x} x \left (2 c_1 x +x^{2}+2 c_2 \right )}{2} \]

✓ Mathematica. Time used: 0.522 (sec). Leaf size: 227

ode=x^2*D[y[x],{x,2}]-2*x*D[y[x],x]+(x^2+2*x+2)*y[x]==x^3*Exp[x]; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to e^{i x} x \left (\operatorname {HypergeometricU}(-i,0,-2 i x) \int _1^x-\frac {i e^{(1-i) K[1]} L_i^{-1}(-2 i K[1])}{4 \operatorname {Hypergeometric1F1}(1-i,2,-2 i K[1]) \operatorname {HypergeometricU}(1-i,1,-2 i K[1]) K[1]-2 \operatorname {HypergeometricU}(-i,0,-2 i K[1]) \operatorname {LaguerreL}(-1+i,-2 i K[1])}dK[1]+L_i^{-1}(-2 i x) \int _1^x\frac {i e^{(1-i) K[2]} \operatorname {HypergeometricU}(-i,0,-2 i K[2])}{4 \operatorname {Hypergeometric1F1}(1-i,2,-2 i K[2]) \operatorname {HypergeometricU}(1-i,1,-2 i K[2]) K[2]-2 \operatorname {HypergeometricU}(-i,0,-2 i K[2]) \operatorname {LaguerreL}(-1+i,-2 i K[2])}dK[2]+c_1 \operatorname {HypergeometricU}(-i,0,-2 i x)+c_2 L_i^{-1}(-2 i x)\right ) \end{align*}

✗ Sympy

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-x**3*exp(x) + x**2*Derivative(y(x), (x, 2)) - 2*x*(x + 1)*Derivative(y(x), x) + (x**2 + 2*x + 2)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE Derivative(y(x), x) - (-x**3*exp(x) + x**2*y(x) + x**2*Derivative(y(x), (x, 2)) + 2*x*y(x) + 2*y(x))/(2*x*(x + 1)) cannot be solved by the factorable group method