63.15.8 problem 6(h)

Internal problem ID [13112]
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(h)
Date solved : Wednesday, March 05, 2025 at 09:17:35 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

\begin{align*} x^{\prime \prime }+9 x&=\sin \left (3 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: 7.551 (sec). Leaf size: 18

ode:=diff(diff(x(t),t),t)+9*x(t) = sin(3*t); 
ic:=x(0) = 0, D(x)(0) = 0; 
dsolve([ode,ic],x(t),method='laplace');
 

\[ x \left (t \right ) = \frac {\sin \left (3 t \right )}{18}-\frac {\cos \left (3 t \right ) t}{6} \]

✓ Mathematica. Time used: 0.061 (sec). Leaf size: 93

ode=D[x[t],{t,2}]+9*x[t]==Sin[3*t]; 
ic={x[0]==0,Derivative[1][x][0 ]==0}; 
DSolve[{ode,ic},x[t],t,IncludeSingularSolutions->True]
 

\[ x(t)\to \sin (3 t) \left (\int _1^t\frac {1}{6} \sin (6 K[2])dK[2]-\int _1^0\frac {1}{6} \sin (6 K[2])dK[2]\right )-\cos (3 t) \int _1^0-\frac {1}{3} \sin ^2(3 K[1])dK[1]+\cos (3 t) \int _1^t-\frac {1}{3} \sin ^2(3 K[1])dK[1] \]

✓ Sympy. Time used: 0.135 (sec). Leaf size: 17

from sympy import * 
t = symbols("t") 
x = Function("x") 
ode = Eq(9*x(t) - sin(3*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)
 

\[ x{\left (t \right )} = - \frac {t \cos {\left (3 t \right )}}{6} + \frac {\sin {\left (3 t \right )}}{18} \]