20.20.7 problem 7

Internal problem ID [3897]
Book : Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth edition, 2015
Section : Chapter 9, First order linear systems. Section 9.11 (Chapter review), page 665
Problem number : 7
Date solved : Tuesday, March 04, 2025 at 05:19:01 PM
CAS classification : system_of_ODEs

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

✓ Maple. Time used: 0.033 (sec). Leaf size: 43

ode:=[diff(x__1(t),t) = 3*x__1(t)+4*x__3(t), diff(x__2(t),t) = 2*x__2(t), diff(x__3(t),t) = -4*x__1(t)-5*x__3(t)]; 
dsolve(ode);
 

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

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

ode={D[x1[t],t]==3*x1[t]-0*x2[t]+4*x3[t],D[x2[t],t]==0*x1[t]+2*x2[t]+0*x3[t],D[x3[t],t]==-4*x1[t]+0*x2[t]-5*x3[t]}; 
ic={}; 
DSolve[{ode,ic},{x1[t],x2[t],x3[t]},t,IncludeSingularSolutions->True]
 

\begin{align*} \text {x1}(t)\to e^{-t} (4 c_1 t+4 c_2 t+c_1) \\ \text {x3}(t)\to e^{-t} (c_2-4 (c_1+c_2) t) \\ \text {x2}(t)\to c_3 e^{2 t} \\ \text {x1}(t)\to e^{-t} (4 c_1 t+4 c_2 t+c_1) \\ \text {x3}(t)\to e^{-t} (c_2-4 (c_1+c_2) t) \\ \text {x2}(t)\to 0 \\ \end{align*}

✓ Sympy. Time used: 0.112 (sec). Leaf size: 46

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) - 4*x__3(t) + Derivative(x__1(t), t),0),Eq(-2*x__2(t) + Derivative(x__2(t), t),0),Eq(4*x__1(t) + 5*x__3(t) + Derivative(x__3(t), t),0)] 
ics = {} 
dsolve(ode,func=[x__1(t),x__2(t),x__3(t)],ics=ics)
 

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