72.21.1 problem 1

Internal problem ID [14900]
Book : DIFFERENTIAL EQUATIONS by Paul Blanchard, Robert L. Devaney, Glen R. Hall. 4th edition. Brooks/Cole. Boston, USA. 2012
Section : Chapter 6. Laplace transform. Section 6.6. page 624
Problem number : 1
Date solved : Monday, March 31, 2025 at 01:01:56 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

\begin{align*} y^{\prime \prime }+2 y^{\prime }+2 y&={\mathrm e}^{-2 t} \sin \left (4 t \right ) \end{align*}

Using Laplace method With initial conditions

\begin{align*} y \left (0\right )&=2\\ y^{\prime }\left (0\right )&=-2 \end{align*}

✓ Maple. Time used: 0.171 (sec). Leaf size: 37

ode:=diff(diff(y(t),t),t)+2*diff(y(t),t)+2*y(t) = exp(-2*t)*sin(4*t); 
ic:=y(0) = 2, D(y)(0) = -2; 
dsolve([ode,ic],y(t),method='laplace');
 

\[ y = \frac {{\mathrm e}^{-2 t} \left (-7 \sin \left (4 t \right )+4 \cos \left (4 t \right )\right )}{130}+\frac {128 \,{\mathrm e}^{-t} \left (\cos \left (t \right )+\frac {\sin \left (t \right )}{8}\right )}{65} \]

✓ Mathematica. Time used: 0.131 (sec). Leaf size: 116

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

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

✓ Sympy. Time used: 0.398 (sec). Leaf size: 37

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

\[ y{\left (t \right )} = \left (\frac {\left (- 7 \sin {\left (4 t \right )} + 4 \cos {\left (4 t \right )}\right ) e^{- t}}{130} + \frac {16 \sin {\left (t \right )}}{65} + \frac {128 \cos {\left (t \right )}}{65}\right ) e^{- t} \]