20.22.5 problem Problem 31

Internal problem ID [3960]
Book : Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section : Chapter 10, The Laplace Transform and Some Elementary Applications. Exercises for 10.7. page 704
Problem number : Problem 31
Date solved : Monday, January 27, 2025 at 08:05:02 AM
CAS classification : [[_linear, `class A`]]

\begin{align*} y^{\prime }+3 y&=\left \{\begin {array}{cc} 1 & 0\le t <1 \\ 0 & 1\le t \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: 5.035 (sec). Leaf size: 43

dsolve([diff(y(t),t)+3*y(t)=piecewise(0<=t and t<1,1,t>=1,0),y(0) = 1],y(t), singsol=all)
 

\[ y = \frac {\left (\left \{\begin {array}{cc} 1+2 \,{\mathrm e}^{-3 t} & t <1 \\ 2 \,{\mathrm e}^{-3}+2 & t =1 \\ 2 \,{\mathrm e}^{-3 t}+{\mathrm e}^{-3 t +3} & 1<t \end {array}\right .\right )}{3} \]

✓ Solution by Mathematica

Time used: 0.071 (sec). Leaf size: 47

DSolve[{D[y[t],t]+3*y[t]==Piecewise[{{1,0<=t<1},{0,t >= 1}}],{y[0]==1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} e^{-3 t} & t\leq 0 \\ \frac {1}{3} e^{-3 t} \left (2+e^3\right ) & t>1 \\ \frac {1}{3}+\frac {2 e^{-3 t}}{3} & \text {True} \\ \end {array} \\ \end {array} \]