67.4.26 problem Problem 4(b)

Internal problem ID [13999]
Book : APPLIED DIFFERENTIAL EQUATIONS The Primary Course by Vladimir A. Dobrushkin. CRC Press 2015
Section : Chapter 5.6 Laplace transform. Nonhomogeneous equations. Problems page 368
Problem number : Problem 4(b)
Date solved : Monday, March 31, 2025 at 08:21:22 AM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

\begin{align*} y^{\prime \prime }-3 y^{\prime }+2 y&=\left \{\begin {array}{cc} 0 & 0\le t <1 \\ 1 & 1\le t <2 \\ -1 & 2\le t \end {array}\right . \end{align*}

Using Laplace method With initial conditions

\begin{align*} y \left (0\right )&=3\\ y^{\prime }\left (0\right )&=-1 \end{align*}

✓ Maple. Time used: 0.198 (sec). Leaf size: 121

ode:=diff(diff(y(t),t),t)-3*diff(y(t),t)+2*y(t) = piecewise(0 <= t and t < 1,0,1 <= t and t < 2,1,2 <= t,-1); 
ic:=y(0) = 3, D(y)(0) = -1; 
dsolve([ode,ic],y(t),method='laplace');
 

\[ y = \left \{\begin {array}{cc} 7 \,{\mathrm e}^{t}-4 \,{\mathrm e}^{2 t} & t <1 \\ 7 \,{\mathrm e}-4 \,{\mathrm e}^{2}+\frac {1}{2} & t =1 \\ 7 \,{\mathrm e}^{t}-4 \,{\mathrm e}^{2 t}-{\mathrm e}^{t -1}+\frac {{\mathrm e}^{-2+2 t}}{2}+\frac {1}{2} & t <2 \\ \frac {15 \,{\mathrm e}^{2}}{2}-4 \,{\mathrm e}^{4}-{\mathrm e}-\frac {1}{2} & t =2 \\ 7 \,{\mathrm e}^{t}-4 \,{\mathrm e}^{2 t}+2 \,{\mathrm e}^{t -2}-{\mathrm e}^{t -1}-{\mathrm e}^{-4+2 t}+\frac {{\mathrm e}^{-2+2 t}}{2}-\frac {1}{2} & 2<t \end {array}\right . \]

✓ Mathematica. Time used: 0.04 (sec). Leaf size: 109

ode=D[y[t],{t,2}]-3*D[y[t],t]+2*y[t]==Piecewise[{{0,0<=t<1},{1,1<=t<2},{-1,t>=2}}]; 
ic={y[0]==3,Derivative[1][y][0] ==-1}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

\[ y(t)\to \begin {array}{cc} \{ & \begin {array}{cc} e^t \left (7-4 e^t\right ) & t\leq 1 \\ \frac {1}{2} \left (1-2 e^{t-1}+14 e^t-8 e^{2 t}+e^{2 t-2}\right ) & 1<t\leq 2 \\ \frac {1}{2} \left (-1+4 e^{t-2}-2 e^{t-1}+14 e^t-8 e^{2 t}-2 e^{2 t-4}+e^{2 t-2}\right ) & \text {True} \\ \end {array} \\ \end {array} \]

✗ Sympy

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(-Piecewise((0, (t >= 0) & (t < 1)), (1, (t >= 1) & (t < 2)), (-1, t >= 2)) + 2*y(t) - 3*Derivative(y(t), t) + Derivative(y(t), (t, 2)),0) 
ics = {y(0): 3, Subs(Derivative(y(t), t), t, 0): -1} 
dsolve(ode,func=y(t),ics=ics)