72.16.40 problem 42

Internal problem ID [14936]
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 : 42
Date solved : Tuesday, January 28, 2025 at 07:21:15 AM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

\begin{align*} y^{\prime \prime }+4 y&=6+t^{2}+{\mathrm e}^{t} \end{align*}

With initial conditions

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

✓ Solution by Maple

Time used: 0.020 (sec). Leaf size: 27

dsolve([diff(y(t),t$2)+4*y(t)=6+t^2+exp(t),y(0) = 0, D(y)(0) = 0],y(t), singsol=all)
 

\[ y = -\frac {\sin \left (2 t \right )}{10}-\frac {63 \cos \left (2 t \right )}{40}+\frac {11}{8}+\frac {t^{2}}{4}+\frac {{\mathrm e}^{t}}{5} \]

✓ Solution by Mathematica

Time used: 0.099 (sec). Leaf size: 127

DSolve[{D[y[t],{t,2}]+4*y[t]==6+t^2+Exp[t],{y[0]==0,Derivative[1][y][0] ==0}},y[t],t,IncludeSingularSolutions -> True]
 

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