6.23.2 problem section 10.6, problem 2

Internal problem ID [2287]
Book : Elementary differential equations with boundary value problems. William F. Trench. Brooks/Cole 2001
Section : Chapter 10 Linear system of Differential equations. Section 10.6, constant coefficient homogeneous system III. Page 566
Problem number : section 10.6, problem 2
Date solved : Tuesday, September 30, 2025 at 05:25:50 AM
CAS classification : system_of_ODEs

\begin{align*} \frac {d}{d t}y_{1} \left (t \right )&=-11 y_{1} \left (t \right )+4 y_{2} \left (t \right )\\ \frac {d}{d t}y_{2} \left (t \right )&=-26 y_{1} \left (t \right )+9 y_{2} \left (t \right ) \end{align*}
Maple. Time used: 0.136 (sec). Leaf size: 59
ode:=[diff(y__1(t),t) = -11*y__1(t)+4*y__2(t), diff(y__2(t),t) = -26*y__1(t)+9*y__2(t)]; 
dsolve(ode);
 
\begin{align*} y_{1} \left (t \right ) &= {\mathrm e}^{-t} \left (\sin \left (2 t \right ) c_1 +\cos \left (2 t \right ) c_2 \right ) \\ y_{2} \left (t \right ) &= \frac {{\mathrm e}^{-t} \left (5 \sin \left (2 t \right ) c_1 -\sin \left (2 t \right ) c_2 +\cos \left (2 t \right ) c_1 +5 \cos \left (2 t \right ) c_2 \right )}{2} \\ \end{align*}
Mathematica. Time used: 0.004 (sec). Leaf size: 64
ode={D[ y1[t],t]==-11*y1[t]+4*y2[t],D[ y2[t],t]==-26*y1[t]+9*y2[t]}; 
ic={}; 
DSolve[{ode,ic},{y1[t],y2[t]},t,IncludeSingularSolutions->True]
 
\begin{align*} \text {y1}(t)&\to e^{-t} (c_1 \cos (2 t)+(2 c_2-5 c_1) \sin (2 t))\\ \text {y2}(t)&\to e^{-t} (c_2 \cos (2 t)+(5 c_2-13 c_1) \sin (2 t)) \end{align*}
Sympy. Time used: 0.065 (sec). Leaf size: 58
from sympy import * 
t = symbols("t") 
y__1 = Function("y__1") 
y__2 = Function("y__2") 
ode=[Eq(11*y__1(t) - 4*y__2(t) + Derivative(y__1(t), t),0),Eq(26*y__1(t) - 9*y__2(t) + Derivative(y__2(t), t),0)] 
ics = {} 
dsolve(ode,func=[y__1(t),y__2(t)],ics=ics)
 
\[ \left [ y^{1}{\left (t \right )} = \left (\frac {C_{1}}{13} + \frac {5 C_{2}}{13}\right ) e^{- t} \cos {\left (2 t \right )} - \left (\frac {5 C_{1}}{13} - \frac {C_{2}}{13}\right ) e^{- t} \sin {\left (2 t \right )}, \ y^{2}{\left (t \right )} = - C_{1} e^{- t} \sin {\left (2 t \right )} + C_{2} e^{- t} \cos {\left (2 t \right )}\right ] \]