20.16.19 problem 19

Internal problem ID [3852]
Book : Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section : Chapter 9, First order linear systems. Section 9.4 (Nondefective coefficient matrix), page 607
Problem number : 19
Date solved : Monday, January 27, 2025 at 08:03:26 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 )+3 x_{3} \left (t \right )\\ \frac {d}{d t}x_{2} \left (t \right )&=3 x_{1} \left (t \right )+x_{2} \left (t \right )\\ \frac {d}{d t}x_{3} \left (t \right )&=2 x_{1} \left (t \right )-x_{2} \left (t \right )+3 x_{3} \left (t \right ) \end{align*}

With initial conditions

\begin{align*} x_{1} \left (0\right ) = -4\\ x_{2} \left (0\right ) = 4\\ x_{3} \left (0\right ) = 4 \end{align*}

✓ Solution by Maple

Time used: 0.014 (sec). Leaf size: 46

dsolve([diff(x__1(t),t) = 2*x__1(t)-x__2(t)+3*x__3(t), diff(x__2(t),t) = 3*x__1(t)+x__2(t), diff(x__3(t),t) = 2*x__1(t)-x__2(t)+3*x__3(t), x__1(0) = -4, x__2(0) = 4, x__3(0) = 4], singsol=all)
 

\begin{align*} x_{1} \left (t \right ) &= {\mathrm e}^{4 t}-2 \,{\mathrm e}^{2 t}-3 \\ x_{2} \left (t \right ) &= 9+{\mathrm e}^{4 t}-6 \,{\mathrm e}^{2 t} \\ x_{3} \left (t \right ) &= {\mathrm e}^{4 t}-2 \,{\mathrm e}^{2 t}+5 \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.006 (sec). Leaf size: 48

DSolve[{D[x1[t],t]==2*x1[t]-x2[t]+3*x3[t],D[x2[t],t]==3*x1[t]+x2[t],D[x3[t],t]==2*x1[t]-x2[t]+3*x3[t]},{x1[0]==-4,x2[0]==4,x3[0]==4},{x1[t],x2[t],x3[t]},t,IncludeSingularSolutions -> True]
 

\begin{align*} \text {x1}(t)\to -2 e^{2 t}+e^{4 t}-3 \\ \text {x2}(t)\to \left (e^{2 t}-3\right )^2 \\ \text {x3}(t)\to -2 e^{2 t}+e^{4 t}+5 \\ \end{align*}