72.21.6 problem 6

Internal problem ID [14976]
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 : 6
Date solved : Tuesday, January 28, 2025 at 07:26:07 AM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

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

Using Laplace method With initial conditions

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

✓ Solution by Maple

Time used: 8.059 (sec). Leaf size: 18

dsolve([diff(y(t),t$2)+4*y(t)=sin(2*t),y(0) = 0, D(y)(0) = 0],y(t), singsol=all)
 

\[ y = \frac {\sin \left (2 t \right )}{8}-\frac {\cos \left (2 t \right ) t}{4} \]

✓ Solution by Mathematica

Time used: 0.067 (sec). Leaf size: 93

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

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