12.22.20 problem section 10.5, problem 20
Internal
problem
ID
[2273]
Book
:
Elementary
differential
equations
with
boundary
value
problems.
William
F.
Trench.
Brooks/Cole
2001
Section
:
Chapter
10
Linear
system
of
Differential
equations.
Section
10.5,
constant
coefficient
homogeneous
system
II.
Page
555
Problem
number
:
section
10.5,
problem
20
Date
solved
:
Tuesday, March 04, 2025 at 01:52:50 PM
CAS
classification
:
system_of_ODEs
\begin{align*} \frac {d}{d t}y_{1} \left (t \right )&=-7 y_{1} \left (t \right )-4 y_{2} \left (t \right )+4 y_{3} \left (t \right )\\ \frac {d}{d t}y_{2} \left (t \right )&=y_{1} \left (t \right )+y_{3} \left (t \right )\\ \frac {d}{d t}y_{3} \left (t \right )&=-9 y_{1} \left (t \right )-5 y_{2} \left (t \right )+6 y_{3} \left (t \right ) \end{align*}
With initial conditions
\begin{align*} y_{1} \left (0\right ) = -6\\ y_{2} \left (0\right ) = 9\\ y_{3} \left (0\right ) = -1 \end{align*}
✓ Maple. Time used: 0.052 (sec). Leaf size: 79
ode:=[diff(y__1(t),t) = -7*y__1(t)-4*y__2(t)+4*y__3(t), diff(y__2(t),t) = y__1(t)+y__3(t), diff(y__3(t),t) = -9*y__1(t)-5*y__2(t)+6*y__3(t)];
ic:=y__1(0) = -6y__2(0) = 9y__3(0) = -1;
dsolve([ode,ic]);
\begin{align*}
y_{1} \left (t \right ) &= -\frac {4 \,{\mathrm e}^{-t} \left (\frac {13 \sin \left (2 t \right )}{2}+\frac {39 \cos \left (2 t \right )}{2}\right )}{13} \\
y_{2} \left (t \right ) &= \frac {9 \,{\mathrm e}^{t}}{2}-\frac {7 \,{\mathrm e}^{-t} \sin \left (2 t \right )}{2}+\frac {9 \,{\mathrm e}^{-t} \cos \left (2 t \right )}{2} \\
y_{3} \left (t \right ) &= \frac {9 \,{\mathrm e}^{t}}{2}-\frac {7 \,{\mathrm e}^{-t} \sin \left (2 t \right )}{2}-\frac {11 \,{\mathrm e}^{-t} \cos \left (2 t \right )}{2} \\
\end{align*}
✓ Mathematica. Time used: 0.008 (sec). Leaf size: 50
ode={D[ y1[t],t]==-7*y1[t]-4*y2[t]+4*y3[t],D[ y2[t],t]==-1*y1[t]-0*y2[t]+1*y3[t],D[ y3[t],t]==-9*y1[t]-5*y2[t]+6*y3[t]};
ic={y1[0]==-6,y2[0]==9,y3[0]==-1};
DSolve[{ode,ic},{y1[t],y2[t],y3[t]},t,IncludeSingularSolutions->True]
\begin{align*}
\text {y1}(t)\to -2 e^{-3 t}-4 e^t \\
\text {y2}(t)\to e^t (9-4 t) \\
\text {y3}(t)\to e^t (1-4 t)-2 e^{-3 t} \\
\end{align*}
✓ Sympy. Time used: 0.174 (sec). Leaf size: 112
from sympy import *
t = symbols("t")
y__1 = Function("y__1")
y__2 = Function("y__2")
y__3 = Function("y__3")
ode=[Eq(7*y__1(t) + 4*y__2(t) - 4*y__3(t) + Derivative(y__1(t), t),0),Eq(-y__1(t) - y__3(t) + Derivative(y__2(t), t),0),Eq(9*y__1(t) + 5*y__2(t) - 6*y__3(t) + Derivative(y__3(t), t),0)]
ics = {}
dsolve(ode,func=[y__1(t),y__2(t),y__3(t)],ics=ics)
\[
\left [ y^{1}{\left (t \right )} = - \left (\frac {4 C_{1}}{17} - \frac {16 C_{2}}{17}\right ) e^{- t} \cos {\left (2 t \right )} - \left (\frac {16 C_{1}}{17} + \frac {4 C_{2}}{17}\right ) e^{- t} \sin {\left (2 t \right )}, \ y^{2}{\left (t \right )} = C_{3} e^{t} + \left (\frac {5 C_{1}}{17} + \frac {14 C_{2}}{17}\right ) e^{- t} \sin {\left (2 t \right )} + \left (\frac {14 C_{1}}{17} - \frac {5 C_{2}}{17}\right ) e^{- t} \cos {\left (2 t \right )}, \ y^{3}{\left (t \right )} = - C_{1} e^{- t} \sin {\left (2 t \right )} + C_{2} e^{- t} \cos {\left (2 t \right )} + C_{3} e^{t}\right ]
\]