72.10.12 problem 11 (b)

Internal problem ID [14823]
Book : DIFFERENTIAL EQUATIONS by Paul Blanchard, Robert L. Devaney, Glen R. Hall. 4th edition. Brooks/Cole. Boston, USA. 2012
Section : Chapter 3. Linear Systems. Exercises section 3.2. page 277
Problem number : 11 (b)
Date solved : Tuesday, January 28, 2025 at 07:17:14 AM
CAS classification : system_of_ODEs

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

With initial conditions

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

✓ Solution by Maple

Time used: 0.008 (sec). Leaf size: 33

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

\begin{align*} x \left (t \right ) &= \frac {2 \,{\mathrm e}^{-3 t}}{5}-\frac {2 \,{\mathrm e}^{2 t}}{5} \\ y &= \frac {{\mathrm e}^{-3 t}}{5}+\frac {4 \,{\mathrm e}^{2 t}}{5} \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.004 (sec). Leaf size: 42

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

\begin{align*} x(t)\to -\frac {2}{5} e^{-3 t} \left (e^{5 t}-1\right ) \\ y(t)\to \frac {1}{5} e^{-3 t} \left (4 e^{5 t}+1\right ) \\ \end{align*}