73.27.16 problem 38.10 (j)

Internal problem ID [15772]
Book : Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section : Chapter 38. Systems of differential equations. A starting point. Additional Exercises. page 786
Problem number : 38.10 (j)
Date solved : Tuesday, January 28, 2025 at 08:06:56 AM
CAS classification : system_of_ODEs

\begin{align*} \frac {d}{d t}x \left (t \right )&=-y \left (t \right )\\ \frac {d}{d t}y \left (t \right )&=4 x \left (t \right )+24 t \end{align*}

With initial conditions

\begin{align*} x \left (0\right ) = 0\\ y \left (0\right ) = 0 \end{align*}

✓ Solution by Maple

Time used: 0.044 (sec). Leaf size: 25

dsolve([diff(x(t),t) = -y(t), diff(y(t),t) = 4*x(t)+24*t, x(0) = 0, y(0) = 0], singsol=all)
 

\begin{align*} x \left (t \right ) &= 3 \sin \left (2 t \right )-6 t \\ y \left (t \right ) &= -6 \cos \left (2 t \right )+6 \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.054 (sec). Leaf size: 182

DSolve[{D[x[t],t]==-y[t],D[y[t],t]==4*x[t]+24*t},{x[0]==0,y[0]==0},{x[t],y[t]},t,IncludeSingularSolutions -> True]
 

\begin{align*} x(t)\to \frac {1}{2} \left (\sin (2 t) \int _1^024 \cos (2 K[2]) K[2]dK[2]-\sin (2 t) \int _1^t24 \cos (2 K[2]) K[2]dK[2]+2 \cos (2 t) \left (\int _1^t12 K[1] \sin (2 K[1])dK[1]-\int _1^012 K[1] \sin (2 K[1])dK[1]\right )\right ) \\ y(t)\to 2 \sin (2 t) \left (\int _1^t12 K[1] \sin (2 K[1])dK[1]-\int _1^012 K[1] \sin (2 K[1])dK[1]\right )-\cos (2 t) \int _1^024 \cos (2 K[2]) K[2]dK[2]+\cos (2 t) \int _1^t24 \cos (2 K[2]) K[2]dK[2] \\ \end{align*}