72.16.16 problem 16

Internal problem ID [14833]
Book : DIFFERENTIAL EQUATIONS by Paul Blanchard, Robert L. Devaney, Glen R. Hall. 4th edition. Brooks/Cole. Boston, USA. 2012
Section : Chapter 4. Forcing and Resonance. Section 4.1 page 399
Problem number : 16
Date solved : Thursday, March 13, 2025 at 04:20:44 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

\begin{align*} y^{\prime \prime }+4 y^{\prime }+20 y&={\mathrm e}^{-\frac {t}{2}} \end{align*}

With initial conditions

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

✓ Maple. Time used: 0.029 (sec). Leaf size: 30

ode:=diff(diff(y(t),t),t)+4*diff(y(t),t)+20*y(t) = exp(-1/2*t); 
ic:=y(0) = 0, D(y)(0) = 0; 
dsolve([ode,ic],y(t), singsol=all);
 

\[ y = \frac {4 \,{\mathrm e}^{-\frac {t}{2}}}{73}+\frac {\left (-3 \sin \left (4 t \right )-8 \cos \left (4 t \right )\right ) {\mathrm e}^{-2 t}}{146} \]

✓ Mathematica. Time used: 0.168 (sec). Leaf size: 127

ode=D[y[t],{t,2}]+4*D[y[t],t]+20*y[t]==Exp[-t/2]; 
ic={y[0]==0,Derivative[1][y][0] ==0}; 
DSolve[{ode,ic},y[t],t,IncludeSingularSolutions->True]
 

\[ y(t)\to e^{-2 t} \left (-\sin (4 t) \int _1^0\frac {1}{4} e^{\frac {3 K[1]}{2}} \cos (4 K[1])dK[1]+\sin (4 t) \int _1^t\frac {1}{4} e^{\frac {3 K[1]}{2}} \cos (4 K[1])dK[1]+\cos (4 t) \left (\int _1^t-\frac {1}{4} e^{\frac {3 K[2]}{2}} \sin (4 K[2])dK[2]-\int _1^0-\frac {1}{4} e^{\frac {3 K[2]}{2}} \sin (4 K[2])dK[2]\right )\right ) \]

✓ Sympy. Time used: 0.260 (sec). Leaf size: 34

from sympy import * 
t = symbols("t") 
y = Function("y") 
ode = Eq(20*y(t) + 4*Derivative(y(t), t) + Derivative(y(t), (t, 2)) - exp(-t/2),0) 
ics = {y(0): 0, Subs(Derivative(y(t), t), t, 0): 0} 
dsolve(ode,func=y(t),ics=ics)
 

\[ y{\left (t \right )} = \left (- \frac {3 \sin {\left (4 t \right )}}{146} - \frac {4 \cos {\left (4 t \right )}}{73}\right ) e^{- 2 t} + \frac {4 e^{- \frac {t}{2}}}{73} \]