14.23.13 problem Problem 13

Internal problem ID [3985]
Book : Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section : Chapter 10, The Laplace Transform and Some Elementary Applications. Exercises for 10.8. page 710
Problem number : Problem 13
Date solved : Tuesday, September 30, 2025 at 07:00:01 AM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

\begin{align*} y^{\prime \prime }+2 y^{\prime }+5 y&=4 \sin \left (t \right )+\delta \left (t -\frac {\pi }{6}\right ) \end{align*}

Using Laplace method With initial conditions

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

✓ Maple. Time used: 0.289 (sec). Leaf size: 69

ode:=diff(diff(y(t),t),t)+2*diff(y(t),t)+5*y(t) = 4*sin(t)+Dirac(t-1/6*Pi); 
ic:=[y(0) = 0, D(y)(0) = 1]; 
dsolve([ode,op(ic)],y(t),method='laplace');
 

\[ y = -\frac {\operatorname {Heaviside}\left (t -\frac {\pi }{6}\right ) \left (\cos \left (t \right )^{2} \sqrt {3}-\cos \left (t \right ) \sin \left (t \right )-\frac {\sqrt {3}}{2}\right ) {\mathrm e}^{-t +\frac {\pi }{6}}}{2}+\frac {\left (4 \cos \left (t \right )^{2}+3 \cos \left (t \right ) \sin \left (t \right )-2\right ) {\mathrm e}^{-t}}{5}-\frac {2 \cos \left (t \right )}{5}+\frac {4 \sin \left (t \right )}{5} \]

✓ Mathematica. Time used: 0.4 (sec). Leaf size: 75

ode=D[y[t],{t,2}]+2*D[y[t],t]+5*y[t]==4*Sin[t]+DiracDelta[t-Pi/6]; 
ic={y[0]==0,Derivative[1][y][0] ==1}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

\begin{align*} y(t)&\to \frac {1}{20} e^{-t} \left (-5 e^{\pi /6} \theta (6 t-\pi ) \left (\sqrt {3} \cos (2 t)-\sin (2 t)\right )+16 e^t \sin (t)+6 \sin (2 t)-8 e^t \cos (t)+8 \cos (2 t)\right ) \end{align*}

✓ Sympy. Time used: 56.768 (sec). Leaf size: 138

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(-Dirac(t - pi/6) + 5*y(t) - 4*sin(t) + 2*Derivative(y(t), t) + Derivative(y(t), (t, 2)),0) 
ics = {y(0): 0, Subs(Derivative(y(t), t), t, 0): 1} 
dsolve(ode,func=y(t),ics=ics)
 

\[ y{\left (t \right )} = \left (\left (- \frac {\int \left (\operatorname {Dirac}{\left (t - \frac {\pi }{6} \right )} + 4 \sin {\left (t \right )}\right ) e^{t} \sin {\left (2 t \right )}\, dt}{2} + \frac {\int \limits ^{0} \operatorname {Dirac}{\left (t - \frac {\pi }{6} \right )} e^{t} \sin {\left (2 t \right )}\, dt}{2} + \frac {\int \limits ^{0} 4 e^{t} \sin {\left (t \right )} \sin {\left (2 t \right )}\, dt}{2}\right ) \cos {\left (2 t \right )} + \left (\frac {\int \left (\operatorname {Dirac}{\left (t - \frac {\pi }{6} \right )} + 4 \sin {\left (t \right )}\right ) e^{t} \cos {\left (2 t \right )}\, dt}{2} - \frac {\int \limits ^{0} \operatorname {Dirac}{\left (t - \frac {\pi }{6} \right )} e^{t} \cos {\left (2 t \right )}\, dt}{2} - \frac {\int \limits ^{0} 4 e^{t} \sin {\left (t \right )} \cos {\left (2 t \right )}\, dt}{2} + \frac {1}{2}\right ) \sin {\left (2 t \right )}\right ) e^{- t} \]