20.17.10 problem 10

Internal problem ID [3864]
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 : 10
Date solved : Tuesday, March 04, 2025 at 05:18:16 PM
CAS classification : system_of_ODEs

\begin{align*} \frac {d}{d t}x_{1} \left (t \right )&=3 x_{1} \left (t \right )+x_{2} \left (t \right )\\ \frac {d}{d t}x_{2} \left (t \right )&=-x_{1} \left (t \right )+5 x_{2} \left (t \right )\\ \frac {d}{d t}x_{3} \left (t \right )&=4 x_{3} \left (t \right ) \end{align*}

✓ Maple. Time used: 0.030 (sec). Leaf size: 37

ode:=[diff(x__1(t),t) = 3*x__1(t)+x__2(t), diff(x__2(t),t) = -x__1(t)+5*x__2(t), diff(x__3(t),t) = 4*x__3(t)]; 
dsolve(ode);
 

\begin{align*} x_{1} \left (t \right ) &= {\mathrm e}^{4 t} \left (c_{2} t +c_{1} \right ) \\ x_{2} \left (t \right ) &= {\mathrm e}^{4 t} \left (c_{2} t +c_{1} +c_{2} \right ) \\ x_{3} \left (t \right ) &= c_3 \,{\mathrm e}^{4 t} \\ \end{align*}

✓ Mathematica. Time used: 0.023 (sec). Leaf size: 102

ode={D[x1[t],t]==3*x1[t]+1*x2[t]+0*x3[t],D[x2[t],t]==-1*x1[t]+5*x2[t]+0*x3[t],D[x3[t],t]==0*x1[t]+0*x2[t]+4*x3[t]}; 
ic={}; 
DSolve[{ode,ic},{x1[t],x2[t],x3[t]},t,IncludeSingularSolutions->True]
 

\begin{align*} \text {x1}(t)\to e^{4 t} (c_1 (-t)+c_2 t+c_1) \\ \text {x2}(t)\to e^{4 t} ((c_2-c_1) t+c_2) \\ \text {x3}(t)\to c_3 e^{4 t} \\ \text {x1}(t)\to e^{4 t} (c_1 (-t)+c_2 t+c_1) \\ \text {x2}(t)\to e^{4 t} ((c_2-c_1) t+c_2) \\ \text {x3}(t)\to 0 \\ \end{align*}

✓ Sympy. Time used: 0.117 (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(-3*x__1(t) - x__2(t) + Derivative(x__1(t), t),0),Eq(x__1(t) - 5*x__2(t) + Derivative(x__2(t), t),0),Eq(-4*x__3(t) + Derivative(x__3(t), t),0)] 
ics = {} 
dsolve(ode,func=[x__1(t),x__2(t),x__3(t)],ics=ics)
 

\[ \left [ x^{1}{\left (t \right )} = - C_{2} t e^{4 t} - \left (C_{1} - C_{2}\right ) e^{4 t}, \ x^{2}{\left (t \right )} = - C_{1} e^{4 t} - C_{2} t e^{4 t}, \ x^{3}{\left (t \right )} = C_{3} e^{4 t}\right ] \]