76.25.11 problem 11

Internal problem ID [17824]
Book : Differential equations. An introduction to modern methods and applications. James Brannan, William E. Boyce. Third edition. Wiley 2015
Section : Chapter 6. Systems of First Order Linear Equations. Section 6.3 (Homogeneous Linear Systems with Constant Coefficients). Problems at page 408
Problem number : 11
Date solved : Tuesday, January 28, 2025 at 11:03:27 AM
CAS classification : system_of_ODEs

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

With initial conditions

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

✓ Solution by Maple

Time used: 0.125 (sec). Leaf size: 115

dsolve([diff(x__1(t),t) = x__1(t)+3*x__3(t), diff(x__2(t),t) = -2*x__2(t), diff(x__3(t),t) = 3*x__1(t)-x__3(t), x__1(0) = 2, x__2(0) = -1, x__3(0) = -2], singsol=all)
 

\begin{align*} x_{1} \left (t \right ) &= \left (1-\frac {\sqrt {10}}{5}\right ) {\mathrm e}^{\sqrt {10}\, t}+\left (1+\frac {\sqrt {10}}{5}\right ) {\mathrm e}^{-\sqrt {10}\, t} \\ x_{2} \left (t \right ) &= -{\mathrm e}^{-2 t} \\ x_{3} \left (t \right ) &= \frac {\left (1-\frac {\sqrt {10}}{5}\right ) \sqrt {10}\, {\mathrm e}^{\sqrt {10}\, t}}{3}-\frac {\left (1+\frac {\sqrt {10}}{5}\right ) \sqrt {10}\, {\mathrm e}^{-\sqrt {10}\, t}}{3}-\frac {\left (1-\frac {\sqrt {10}}{5}\right ) {\mathrm e}^{\sqrt {10}\, t}}{3}-\frac {\left (1+\frac {\sqrt {10}}{5}\right ) {\mathrm e}^{-\sqrt {10}\, t}}{3} \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.044 (sec). Leaf size: 32

DSolve[{D[x1[t],t]==-1*x1[t]+0*x2[t]+3*x3[t],D[x2[t],t]==0*x1[t]-2*x2[t]+0*x3[t],D[x3[t],t]==3*x1[t]+0*x2[t]-1*x3[t]},{x1[0]==2,x2[0]==-1,x3[0]==-2},{x1[t],x2[t],x3[t]},t,IncludeSingularSolutions -> True]
 

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