64.6.23 problem 23

Internal problem ID [13373]
Book : Differential Equations by Shepley L. Ross. Third edition. John Willey. New Delhi. 2004.
Section : Chapter 2, Miscellaneous Review. Exercises page 60
Problem number : 23
Date solved : Tuesday, January 28, 2025 at 05:33:20 AM
CAS classification : [_linear]

\begin{align*} \left (x +2\right ) y^{\prime }+y&=\left \{\begin {array}{cc} 2 x & 0\le x \le 2 \\ 4 & 2<x \end {array}\right . \end{align*}

With initial conditions

\begin{align*} y \left (0\right )&=4 \end{align*}

✓ Solution by Maple

Time used: 0.157 (sec). Leaf size: 31

dsolve([(x+2)*diff(y(x),x)+y(x)=piecewise(0<=x and x<=2,2*x,x>2,4),y(0) = 4],y(x), singsol=all)
 

\[ y = \frac {\left \{\begin {array}{cc} 8 & x <0 \\ x^{2}+8 & x <2 \\ 4+4 x & 2\le x \end {array}\right .}{x +2} \]

✓ Solution by Mathematica

Time used: 0.077 (sec). Leaf size: 43

DSolve[{(x+2)*D[y[x],x]+y[x]==Piecewise[{{2*x,0<=x<=2},{4,x>2}}],{y[0]==4}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {8}{x+2} & x\leq 0 \\ \frac {4 (x+1)}{x+2} & x>2 \\ \frac {x^2+8}{x+2} & \text {True} \\ \end {array} \\ \end {array} \]