57.17.7 problem 10

Internal problem ID [14487]
Book : A First Course in Differential Equations by J. David Logan. Third Edition. Springer-Verlag, NY. 2015.
Section : Chapter 3, Laplace transform. Section 3.4 Impulsive sources. Exercises page 173
Problem number : 10
Date solved : Thursday, October 02, 2025 at 09:37:47 AM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

\begin{align*} x^{\prime \prime }+4 x&=\frac {\operatorname {Heaviside}\left (t -5\right ) \left (t -5\right )}{5}+\left (2-\frac {t}{5}\right ) \operatorname {Heaviside}\left (t -10\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: 0.384 (sec). Leaf size: 49

ode:=diff(diff(x(t),t),t)+4*x(t) = 1/5*(t-5)*Heaviside(t-5)+(2-1/5*t)*Heaviside(t-10); 
ic:=[x(0) = 0, D(x)(0) = 0]; 
dsolve([ode,op(ic)],x(t),method='laplace');
 

\[ x = \frac {\operatorname {Heaviside}\left (t -10\right ) \sin \left (2 t -20\right )}{40}-\frac {\operatorname {Heaviside}\left (t -5\right ) \sin \left (2 t -10\right )}{40}+\frac {\left (-2 t +20\right ) \operatorname {Heaviside}\left (t -10\right )}{40}+\frac {\left (t -5\right ) \operatorname {Heaviside}\left (t -5\right )}{20} \]

✓ Mathematica. Time used: 0.03 (sec). Leaf size: 55

ode=D[x[t],{t,2}]+4*x[t]==1/5*(t-5)*UnitStep[t-5]+(1-1/5*(t-5))*UnitStep[t-10]; 
ic={x[0]==0,Derivative[1][x][0 ]==0}; 
DSolve[{ode,ic},x[t],t,IncludeSingularSolutions->True]
 

\begin{align*} x(t)&\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {1}{40} (2 (t-5)+\sin (10-2 t)) & 5<t\leq 10 \\ \frac {1}{40} (\sin (10-2 t)-\sin (20-2 t)+10) & t>10 \\ \end {array} \\ \end {array} \end{align*}

✓ Sympy. Time used: 2.776 (sec). Leaf size: 97

from sympy import * 
t = symbols("t") 
x = Function("x") 
ode = Eq((t/5 - 2)*Heaviside(t - 10) - (t - 5)*Heaviside(t - 5)/5 + 4*x(t) + 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)
 

\[ x{\left (t \right )} = - \frac {t \theta \left (t - 10\right )}{20} + \frac {t \theta \left (t - 5\right )}{20} + \frac {\sin {\left (t \right )} \cos {\left (t - 20 \right )} \theta \left (t - 10\right )}{20} - \frac {\sin {\left (t \right )} \cos {\left (t - 10 \right )} \theta \left (t - 5\right )}{20} - \frac {\sin {\left (20 \right )} \theta \left (t - 10\right )}{40} + \frac {\theta \left (t - 10\right )}{2} - \frac {\theta \left (t - 5\right )}{4} + \frac {\sin {\left (10 \right )} \theta \left (t - 5\right )}{40} \]