12.10.35 problem 35

Internal problem ID [1839]
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 : 35
Date solved : Monday, January 27, 2025 at 05:37:00 AM
CAS classification : [[_2nd_order, _exact, _linear, _nonhomogeneous]]

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

With initial conditions

\begin{align*} y \left (0\right )&=0\\ y^{\prime }\left (0\right )&=0 \end{align*}

✓ Solution by Maple

Time used: 0.010 (sec). Leaf size: 20

dsolve([(x+1)*(2*x+3)*diff(y(x),x$2)+2*(x+2)*diff(y(x),x)-2*y(x)=(2*x+3)^2,y(0) = 0, D(y)(0) = 0],y(x), singsol=all)
 

\[ y = \frac {x^{2} \left (4 x +9\right )}{6 x +6} \]

✓ Solution by Mathematica

Time used: 2.103 (sec). Leaf size: 22

DSolve[{(x+1)*(2*x+3)*D[y[x],{x,2}]+2*(x+2)*D[y[x],x]-2*y[x]==(2*x+3)^2,{y[0]==0,Derivative[1][y][0] ==0}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {x^2 (4 x+9)}{6 (x+1)} \]