7.25.15 problem 15

Internal problem ID [635]
Book : Elementary Differential Equations. By C. Henry Edwards, David E. Penney and David Calvis. 6th edition. 2008
Section : Chapter 5. Linear systems of differential equations. Section 5.4 (The eigenvalue method for homogeneous systems). Problems at page 378
Problem number : 15
Date solved : Saturday, March 29, 2025 at 05:01:13 PM
CAS classification : system_of_ODEs

\begin{align*} \frac {d}{d t}x_{1} \left (t \right )&=7 x_{1} \left (t \right )-5 x_{2} \left (t \right )\\ \frac {d}{d t}x_{2} \left (t \right )&=4 x_{1} \left (t \right )+3 x_{2} \left (t \right ) \end{align*}

✓ Maple. Time used: 0.144 (sec). Leaf size: 58

ode:=[diff(x__1(t),t) = 7*x__1(t)-5*x__2(t), diff(x__2(t),t) = 4*x__1(t)+3*x__2(t)]; 
dsolve(ode);
 

\begin{align*} x_{1} \left (t \right ) &= {\mathrm e}^{5 t} \left (\sin \left (4 t \right ) c_1 +\cos \left (4 t \right ) c_2 \right ) \\ x_{2} \left (t \right ) &= \frac {2 \,{\mathrm e}^{5 t} \left (\sin \left (4 t \right ) c_1 +2 \sin \left (4 t \right ) c_2 -2 \cos \left (4 t \right ) c_1 +\cos \left (4 t \right ) c_2 \right )}{5} \\ \end{align*}

✓ Mathematica. Time used: 0.005 (sec). Leaf size: 72

ode={D[x1[t],t]==7*x1[t]-5*x2[t],D[x2[t],t]==4*x1[t]+3*x2[t]}; 
ic={}; 
DSolve[{ode,ic},{x1[t],x2[t]},t,IncludeSingularSolutions->True]
 

\begin{align*} \text {x1}(t)\to \frac {1}{4} e^{5 t} (4 c_1 \cos (4 t)+(2 c_1-5 c_2) \sin (4 t)) \\ \text {x2}(t)\to \frac {1}{2} e^{5 t} (2 c_2 \cos (4 t)+(2 c_1-c_2) \sin (4 t)) \\ \end{align*}

✓ Sympy. Time used: 0.134 (sec). Leaf size: 58

from sympy import * 
t = symbols("t") 
x__1 = Function("x__1") 
x__2 = Function("x__2") 
ode=[Eq(-7*x__1(t) + 5*x__2(t) + Derivative(x__1(t), t),0),Eq(-4*x__1(t) - 3*x__2(t) + Derivative(x__2(t), t),0)] 
ics = {} 
dsolve(ode,func=[x__1(t),x__2(t)],ics=ics)
 

\[ \left [ x^{1}{\left (t \right )} = \left (\frac {C_{1}}{2} - C_{2}\right ) e^{5 t} \cos {\left (4 t \right )} - \left (C_{1} + \frac {C_{2}}{2}\right ) e^{5 t} \sin {\left (4 t \right )}, \ x^{2}{\left (t \right )} = C_{1} e^{5 t} \cos {\left (4 t \right )} - C_{2} e^{5 t} \sin {\left (4 t \right )}\right ] \]