65.14.2 problem 26.1 (ii)

Internal problem ID [13818]
Book : AN INTRODUCTION TO ORDINARY DIFFERENTIAL EQUATIONS by JAMES C. ROBINSON. Cambridge University Press 2004
Section : Chapter 26, Explicit solutions of coupled linear systems. Exercises page 257
Problem number : 26.1 (ii)
Date solved : Tuesday, January 28, 2025 at 06:04:41 AM
CAS classification : system_of_ODEs

\begin{align*} \frac {d}{d t}x \left (t \right )&=x \left (t \right )-4 y \left (t \right )+\cos \left (2 t \right )\\ \frac {d}{d t}y \left (t \right )&=x \left (t \right )+y \left (t \right ) \end{align*}

With initial conditions

\begin{align*} x \left (0\right ) = 1\\ y \left (0\right ) = 1 \end{align*}

✓ Solution by Maple

Time used: 0.299 (sec). Leaf size: 65

dsolve([diff(x(t),t) = x(t)-4*y(t)+cos(2*t), diff(y(t),t) = x(t)+y(t), x(0) = 1, y(0) = 1], singsol=all)
 

\begin{align*} x \left (t \right ) &= \frac {26 \cos \left (2 t \right ) {\mathrm e}^{t}}{17}-\frac {32 \sin \left (2 t \right ) {\mathrm e}^{t}}{17}+\frac {2 \sin \left (2 t \right )}{17}-\frac {9 \cos \left (2 t \right )}{17} \\ y \left (t \right ) &= \frac {13 \sin \left (2 t \right ) {\mathrm e}^{t}}{17}+\frac {16 \cos \left (2 t \right ) {\mathrm e}^{t}}{17}+\frac {\cos \left (2 t \right )}{17}-\frac {4 \sin \left (2 t \right )}{17} \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.085 (sec). Leaf size: 259

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

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