65.14.1 problem 26.1 (i)

Internal problem ID [13817]
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 (i)
Date solved : Tuesday, January 28, 2025 at 06:04:40 AM
CAS classification : system_of_ODEs

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

With initial conditions

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

✓ Solution by Maple

Time used: 0.056 (sec). Leaf size: 51

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

\begin{align*} x \left (t \right ) &= \frac {5 \,{\mathrm e}^{2 t}}{4}-\frac {29 \,{\mathrm e}^{3 t}}{27}-\frac {t^{2}}{6}-\frac {5 t}{18}-\frac {19}{108} \\ y \left (t \right ) &= \frac {5 \,{\mathrm e}^{2 t}}{2}-\frac {29 \,{\mathrm e}^{3 t}}{27}-\frac {7 t}{9}-\frac {23}{54}-\frac {2 t^{2}}{3} \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.227 (sec). Leaf size: 274

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

\begin{align*} x(t)\to -e^{2 t} \left (\left (2 e^t-1\right ) \int _1^0e^{-3 K[1]} \left (-1+e^{K[1]}\right ) K[1]^2dK[1]+\left (1-2 e^t\right ) \int _1^te^{-3 K[1]} \left (-1+e^{K[1]}\right ) K[1]^2dK[1]-\left (e^t-1\right ) \left (-\int _1^te^{-3 K[2]} \left (-1+2 e^{K[2]}\right ) K[2]^2dK[2]+\int _1^0e^{-3 K[2]} \left (-1+2 e^{K[2]}\right ) K[2]^2dK[2]-1\right )\right ) \\ y(t)\to e^{2 t} \left (-2 \left (e^t-1\right ) \int _1^0e^{-3 K[1]} \left (-1+e^{K[1]}\right ) K[1]^2dK[1]+2 \left (e^t-1\right ) \int _1^te^{-3 K[1]} \left (-1+e^{K[1]}\right ) K[1]^2dK[1]+\left (e^t-2\right ) \left (-\int _1^te^{-3 K[2]} \left (-1+2 e^{K[2]}\right ) K[2]^2dK[2]+\int _1^0e^{-3 K[2]} \left (-1+2 e^{K[2]}\right ) K[2]^2dK[2]-1\right )\right ) \\ \end{align*}