14.16.8 problem 24

Internal problem ID [2678]
Book : Differential equations and their applications, 4th ed., M. Braun
Section : Chapter 2. Second order differential equations. Section 2.9, The method of Laplace transform. Excercises page 232
Problem number : 24
Date solved : Monday, January 27, 2025 at 06:06:17 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

\begin{align*} y^{\prime \prime }-3 y^{\prime }+2 y&={\mathrm e}^{-t} \end{align*}

Using Laplace method With initial conditions

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

✓ Solution by Maple

Time used: 0.896 (sec). Leaf size: 47

dsolve([diff(y(t),t$2)-3*diff(y(t),t)+2*y(t)=exp(-t),y(t__0) = 1, D(y)(t__0) = 0],y(t), singsol=all)
 

\[ y = 2 \,{\mathrm e}^{t -t_{0}}-{\mathrm e}^{2 t -2 t_{0}}+\frac {{\mathrm e}^{-t}}{6}-\frac {{\mathrm e}^{-2 t_{0} +t}}{2}+\frac {{\mathrm e}^{-3 t_{0} +2 t}}{3} \]

✓ Solution by Mathematica

Time used: 0.027 (sec). Leaf size: 38

DSolve[{D[y[t],{t,2}]-3*D[y[t],t]+2*y[t]==Exp[-t],{y[t0]==0,Derivative[1][y][t0] ==0}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {1}{6} e^{-t-3 \text {t0}} \left (e^t-e^{\text {t0}}\right )^2 \left (2 e^t+e^{\text {t0}}\right ) \]