14.22.11 problem 11

Internal problem ID [2738]
Book : Differential equations and their applications, 4th ed., M. Braun
Section : Section 3.8, Systems of differential equations. The eigenva1ue-eigenvector method. Page 339
Problem number : 11
Date solved : Sunday, March 30, 2025 at 12:16:05 AM
CAS classification : system_of_ODEs

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

With initial conditions

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

✓ Maple. Time used: 0.181 (sec). Leaf size: 23

ode:=[diff(x__1(t),t) = x__1(t)-3*x__2(t)+2*x__3(t), diff(x__2(t),t) = -x__2(t), diff(x__3(t),t) = -x__2(t)-2*x__3(t)]; 
ic:=x__1(0) = -2x__2(0) = 0x__3(0) = 3; 
dsolve([ode,ic]);
 

\begin{align*} x_{1} \left (t \right ) &= -2 \,{\mathrm e}^{-2 t} \\ x_{2} \left (t \right ) &= 0 \\ x_{3} \left (t \right ) &= 3 \,{\mathrm e}^{-2 t} \\ \end{align*}

✓ Mathematica. Time used: 0.005 (sec). Leaf size: 26

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

\begin{align*} \text {x1}(t)\to -2 e^{-2 t} \\ \text {x2}(t)\to 0 \\ \text {x3}(t)\to 3 e^{-2 t} \\ \end{align*}

✓ Sympy. Time used: 0.115 (sec). Leaf size: 48

from sympy import * 
t = symbols("t") 
x__1 = Function("x__1") 
x__2 = Function("x__2") 
x__3 = Function("x__3") 
ode=[Eq(-x__1(t) + 3*x__2(t) - 2*x__3(t) + Derivative(x__1(t), t),0),Eq(x__2(t) + Derivative(x__2(t), t),0),Eq(x__2(t) + 2*x__3(t) + Derivative(x__3(t), t),0)] 
ics = {} 
dsolve(ode,func=[x__1(t),x__2(t),x__3(t)],ics=ics)
 

\[ \left [ x^{1}{\left (t \right )} = - \frac {2 C_{1} e^{- 2 t}}{3} - \frac {5 C_{2} e^{- t}}{2} + C_{3} e^{t}, \ x^{2}{\left (t \right )} = - C_{2} e^{- t}, \ x^{3}{\left (t \right )} = C_{1} e^{- 2 t} + C_{2} e^{- t}\right ] \]