63.15.11 problem 11

Internal problem ID [13115]
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 : 11
Date solved : Wednesday, March 05, 2025 at 09:17:37 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

\begin{align*} x^{\prime \prime }+4 x&=\cos \left (2 t \right ) \operatorname {Heaviside}\left (2 \pi -t \right ) \end{align*}

Using Laplace method With initial conditions

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

✓ Maple. Time used: 10.383 (sec). Leaf size: 26

ode:=diff(diff(x(t),t),t)+4*x(t) = cos(2*t)*Heaviside(2*Pi-t); 
ic:=x(0) = 0, D(x)(0) = 0; 
dsolve([ode,ic],x(t),method='laplace');
 

\[ x \left (t \right ) = -\frac {\sin \left (2 t \right ) \left (-t +\operatorname {Heaviside}\left (t -2 \pi \right ) \left (t -2 \pi \right )\right )}{4} \]

✓ Mathematica. Time used: 86.057 (sec). Leaf size: 129

ode=D[x[t],{t,2}]+4*x[t]==Cos[2*t]*UnitStep[2*Pi-t]; 
ic={x[0]==0,Derivative[1][x][0 ]==0}; 
DSolve[{ode,ic},x[t],t,IncludeSingularSolutions->True]
 

\[ x(t)\to \sin (2 t) \left (\int _1^t\frac {1}{2} \cos ^2(2 K[2]) \theta (2 \pi -K[2])dK[2]-\int _1^0\frac {1}{2} \cos ^2(2 K[2]) \theta (2 \pi -K[2])dK[2]\right )-\cos (2 t) \int _1^0-\frac {1}{4} \sin (4 K[1]) \theta (2 \pi -K[1])dK[1]+\cos (2 t) \int _1^t-\frac {1}{4} \sin (4 K[1]) \theta (2 \pi -K[1])dK[1] \]

✓ Sympy. Time used: 1.114 (sec). Leaf size: 34

from sympy import * 
t = symbols("t") 
x = Function("x") 
ode = Eq(4*x(t) - cos(2*t)*Heaviside(-t + 2*pi) + Derivative(x(t), (t, 2)),0) 
ics = {x(0): 0, Subs(Derivative(x(t), t), t, 0): 0} 
dsolve(ode,func=x(t),ics=ics)
 

\[ x{\left (t \right )} = \left (\frac {t \theta \left (- t + 2 \pi \right )}{4} - \frac {\pi \theta \left (- t + 2 \pi \right )}{2} + \frac {\pi }{2}\right ) \sin {\left (2 t \right )} \]