20.17.15 problem 15

Internal problem ID [3869]
Book : Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section : Chapter 9, First order linear systems. Section 9.5 (Defective coefficient matrix), page 619
Problem number : 15
Date solved : Monday, January 27, 2025 at 08:03:39 AM
CAS classification : system_of_ODEs

\begin{align*} \frac {d}{d t}x_{1} \left (t \right )&=-2 x_{1} \left (t \right )-x_{2} \left (t \right )\\ \frac {d}{d t}x_{2} \left (t \right )&=x_{1} \left (t \right )-4 x_{2} \left (t \right ) \end{align*}

With initial conditions

\begin{align*} x_{1} \left (0\right ) = 0\\ x_{2} \left (0\right ) = -1 \end{align*}

✓ Solution by Maple

Time used: 0.023 (sec). Leaf size: 21

dsolve([diff(x__1(t),t) = -2*x__1(t)-x__2(t), diff(x__2(t),t) = x__1(t)-4*x__2(t), x__1(0) = 0, x__2(0) = -1], singsol=all)
 

\begin{align*} x_{1} \left (t \right ) &= {\mathrm e}^{-3 t} t \\ x_{2} \left (t \right ) &= {\mathrm e}^{-3 t} \left (-1+t \right ) \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.003 (sec). Leaf size: 24

DSolve[{D[x1[t],t]==-2*x1[t]-1*x2[t],D[x2[t],t]==1*x1[t]-4*x2[t]},{x1[0]==0,x2[0]==-1},{x1[t],x2[t]},t,IncludeSingularSolutions -> True]
 

\begin{align*} \text {x1}(t)\to e^{-3 t} t \\ \text {x2}(t)\to e^{-3 t} (t-1) \\ \end{align*}