63.15.5 problem 6(e)

Internal problem ID [13109]
Book : A First Course in Differential Equations by J. David Logan. Third Edition. Springer-Verlag, NY. 2015.
Section : Chapter 3, Laplace transform. Section 3.2.1 Initial value problems. Exercises page 156
Problem number : 6(e)
Date solved : Wednesday, March 05, 2025 at 09:17:28 PM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

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

Using Laplace method With initial conditions

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

✓ Maple. Time used: 9.253 (sec). Leaf size: 24

ode:=diff(diff(x(t),t),t)-2*diff(x(t),t)+2*x(t) = exp(-t); 
ic:=x(0) = 0, D(x)(0) = 1; 
dsolve([ode,ic],x(t),method='laplace');
 

\[ x \left (t \right ) = \frac {{\mathrm e}^{-t}}{5}+\frac {\left (-\cos \left (t \right )+7 \sin \left (t \right )\right ) {\mathrm e}^{t}}{5} \]

✓ Mathematica. Time used: 0.069 (sec). Leaf size: 96

ode=D[x[t],{t,2}]-2*D[x[t],t]+2*x[t]==Exp[-t]; 
ic={x[0]==0,Derivative[1][x][0 ]==1}; 
DSolve[{ode,ic},x[t],t,IncludeSingularSolutions->True]
 

\[ x(t)\to e^t \left (\sin (t) \left (-\int _1^0e^{-2 K[1]} \cos (K[1])dK[1]\right )+\sin (t) \int _1^te^{-2 K[1]} \cos (K[1])dK[1]-\cos (t) \int _1^0-e^{-2 K[2]} \sin (K[2])dK[2]+\cos (t) \int _1^t-e^{-2 K[2]} \sin (K[2])dK[2]+\sin (t)\right ) \]

✓ Sympy. Time used: 0.236 (sec). Leaf size: 24

from sympy import * 
t = symbols("t") 
x = Function("x") 
ode = Eq(2*x(t) - 2*Derivative(x(t), t) + Derivative(x(t), (t, 2)) - exp(-t),0) 
ics = {x(0): 0, Subs(Derivative(x(t), t), t, 0): 1} 
dsolve(ode,func=x(t),ics=ics)
 

\[ x{\left (t \right )} = \left (\frac {7 \sin {\left (t \right )}}{5} - \frac {\cos {\left (t \right )}}{5}\right ) e^{t} + \frac {e^{- t}}{5} \]