90.27.4 problem 4

Internal problem ID [25414]
Book : Ordinary Differential Equations. By William Adkins and Mark G Davidson. Springer. NY. 2010. ISBN 978-1-4614-3617-1
Section : Chapter 6. Discontinuous Functions and the Laplace Transform. Exercises at page 425
Problem number : 4
Date solved : Friday, October 03, 2025 at 12:01:17 AM
CAS classification : [[_linear, `class A`]]

\begin{align*} y^{\prime }+2 y&=\left \{\begin {array}{cc} t & 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 )&=0 \\ \end{align*}

✓ Maple. Time used: 0.075 (sec). Leaf size: 40

ode:=diff(y(t),t)+2*y(t) = piecewise(t < 1 and 0 <= t,t,1 <= t,0); 
ic:=[y(0) = 0]; 
dsolve([ode,op(ic)],y(t),method='laplace');
 

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

✓ Mathematica. Time used: 0.048 (sec). Leaf size: 47

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

\begin{align*} y(t)&\to \begin {array}{cc} \{ & \begin {array}{cc} 0 & t\leq 0 \\ \frac {1}{4} \left (2 t+e^{-2 t}-1\right ) & 0<t\leq 1 \\ \frac {1}{4} e^{-2 t} \left (1+e^2\right ) & \text {True} \\ \end {array} \\ \end {array} \end{align*}

✗ Sympy

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(-Piecewise((t, (t >= 0) & (t < 1)), (0, t >= 1)) + 2*y(t) + Derivative(y(t), t),0) 
ics = {y(0): 0} 
dsolve(ode,func=y(t),ics=ics)