71.15.1 problem 4 (a)

Internal problem ID [14543]
Book : Ordinary Differential Equations by Charles E. Roberts, Jr. CRC Press. 2010
Section : Chapter 5. The Laplace Transform Method. Exercises 5.4, page 265
Problem number : 4 (a)
Date solved : Tuesday, January 28, 2025 at 06:43:10 AM
CAS classification : [[_linear, `class A`]]

\begin{align*} y^{\prime }+2 y&=\left \{\begin {array}{cc} 2 & 0\le x <1 \\ 1 & 1\le x \end {array}\right . \end{align*}

Using Laplace method With initial conditions

\begin{align*} y \left (0\right )&=1 \end{align*}

✓ Solution by Maple

Time used: 13.502 (sec). Leaf size: 22

dsolve([diff(y(x),x)+2*y(x)=piecewise(0<=x and x<1,2,1<=x,1),y(0) = 1],y(x), singsol=all)
 

\[ y = \left \{\begin {array}{cc} 1 & x <1 \\ \frac {1}{2}+\frac {{\mathrm e}^{-2 x +2}}{2} & 1\le x \end {array}\right . \]

✓ Solution by Mathematica

Time used: 0.066 (sec). Leaf size: 37

DSolve[{D[y[x],x]+2*y[x]==Piecewise[{ {2,0<=x<1},{1,1<=x}}],{y[0]==1}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \begin {array}{cc} \{ & \begin {array}{cc} e^{-2 x} & x\leq 0 \\ 1 & 0<x\leq 1 \\ \frac {1}{2} \left (1+e^{2-2 x}\right ) & \text {True} \\ \end {array} \\ \end {array} \]