9.4.37 problem problem 48
Internal
problem
ID
[1001]
Book
:
Differential
equations
and
linear
algebra,
4th
ed.,
Edwards
and
Penney
Section
:
Section
7.3,
The
eigenvalue
method
for
linear
systems.
Page
395
Problem
number
:
problem
48
Date
solved
:
Tuesday, March 04, 2025 at 12:07:11 PM
CAS
classification
:
system_of_ODEs
\begin{align*} \frac {d}{d t}x_{1} \left (t \right )&=47 x_{1} \left (t \right )-8 x_{2} \left (t \right )+5 x_{3} \left (t \right )-5 x_{4} \left (t \right )\\ \frac {d}{d t}x_{2} \left (t \right )&=-10 x_{1} \left (t \right )+32 x_{2} \left (t \right )+18 x_{3} \left (t \right )-2 x_{4} \left (t \right )\\ \frac {d}{d t}x_{3} \left (t \right )&=139 x_{1} \left (t \right )-40 x_{2} \left (t \right )-167 x_{3} \left (t \right )-121 x_{4} \left (t \right )\\ \frac {d}{d t}x_{4} \left (t \right )&=-232 x_{1} \left (t \right )+64 x_{2} \left (t \right )+360 x_{3} \left (t \right )+248 x_{4} \left (t \right ) \end{align*}
✓ Maple. Time used: 0.050 (sec). Leaf size: 124
ode:=[diff(x__1(t),t) = 47*x__1(t)-8*x__2(t)+5*x__3(t)-5*x__4(t), diff(x__2(t),t) = -10*x__1(t)+32*x__2(t)+18*x__3(t)-2*x__4(t), diff(x__3(t),t) = 139*x__1(t)-40*x__2(t)-167*x__3(t)-121*x__4(t), diff(x__4(t),t) = -232*x__1(t)+64*x__2(t)+360*x__3(t)+248*x__4(t)];
dsolve(ode);
\begin{align*}
x_{1} \left (t \right ) &= c_1 \,{\mathrm e}^{16 t}+c_2 \,{\mathrm e}^{48 t}+c_3 \,{\mathrm e}^{64 t}+c_4 \,{\mathrm e}^{32 t} \\
x_{2} \left (t \right ) &= 2 c_1 \,{\mathrm e}^{16 t}-\frac {c_2 \,{\mathrm e}^{48 t}}{3}+c_3 \,{\mathrm e}^{64 t}+\frac {5 c_4 \,{\mathrm e}^{32 t}}{2} \\
x_{3} \left (t \right ) &= -c_1 \,{\mathrm e}^{16 t}+\frac {c_2 \,{\mathrm e}^{48 t}}{3}+2 c_3 \,{\mathrm e}^{64 t}+\frac {c_4 \,{\mathrm e}^{32 t}}{2} \\
x_{4} \left (t \right ) &= 2 c_1 \,{\mathrm e}^{16 t}+\frac {2 c_2 \,{\mathrm e}^{48 t}}{3}-3 c_3 \,{\mathrm e}^{64 t}-\frac {c_4 \,{\mathrm e}^{32 t}}{2} \\
\end{align*}
✓ Mathematica. Time used: 0.007 (sec). Leaf size: 448
ode={D[ x1[t],t]==47*x1[t]-8*x2[t]+5*x3[t]-5*x4[t],D[ x2[t],t]==-10*x1[t]+32*x2[t]+18*x3[t]-2*x4[t],D[ x3[t],t]==139*x1[t]-40*x2[t]-167*x3[t]-121*x4[t],D[ x4[t],t]==-232*x1[t]+64*x2[t]+360*x3[t]+248*x4[t]};
ic={};
DSolve[{ode,ic},{x1[t],x2[t],x3[t],x4[t]},t,IncludeSingularSolutions->True]
\begin{align*}
\text {x1}(t)\to \frac {1}{16} e^{16 t} \left (c_1 \left (-38 e^{16 t}-6 e^{32 t}+27 e^{48 t}+33\right )-\left (e^{16 t}-1\right ) \left (8 c_2 \left (e^{16 t}+e^{32 t}-1\right )+c_3 \left (9 e^{16 t}+39 e^{32 t}-53\right )+c_4 \left (7 e^{16 t}+25 e^{32 t}-27\right )\right )\right ) \\
\text {x2}(t)\to \frac {1}{16} e^{16 t} \left (c_1 \left (-95 e^{16 t}+2 e^{32 t}+27 e^{48 t}+66\right )-8 c_2 \left (-5 e^{16 t}+e^{48 t}+2\right )-\left (e^{16 t}-1\right ) \left (c_3 \left (49 e^{16 t}+39 e^{32 t}-106\right )+c_4 \left (31 e^{16 t}+25 e^{32 t}-54\right )\right )\right ) \\
\text {x3}(t)\to \frac {1}{16} e^{16 t} \left (c_1 \left (-19 e^{16 t}-2 e^{32 t}+54 e^{48 t}-33\right )+8 c_2 \left (e^{16 t}-2 e^{48 t}+1\right )+31 c_3 e^{16 t}+10 c_3 e^{32 t}-78 c_3 e^{48 t}+17 c_4 e^{16 t}+6 c_4 e^{32 t}-50 c_4 e^{48 t}+53 c_3+27 c_4\right ) \\
\text {x4}(t)\to -\frac {1}{16} e^{16 t} \left (c_1 \left (-19 e^{16 t}+4 e^{32 t}+81 e^{48 t}-66\right )+8 c_2 \left (e^{16 t}-3 e^{48 t}+2\right )+31 c_3 e^{16 t}-20 c_3 e^{32 t}-117 c_3 e^{48 t}+17 c_4 e^{16 t}-12 c_4 e^{32 t}-75 c_4 e^{48 t}+106 c_3+54 c_4\right ) \\
\end{align*}
✓ Sympy. Time used: 0.242 (sec). Leaf size: 136
from sympy import *
t = symbols("t")
x__1 = Function("x__1")
x__2 = Function("x__2")
x__3 = Function("x__3")
x__4 = Function("x__4")
ode=[Eq(-47*x__1(t) + 8*x__2(t) - 5*x__3(t) + 5*x__4(t) + Derivative(x__1(t), t),0),Eq(10*x__1(t) - 32*x__2(t) - 18*x__3(t) + 2*x__4(t) + Derivative(x__2(t), t),0),Eq(-139*x__1(t) + 40*x__2(t) + 167*x__3(t) + 121*x__4(t) + Derivative(x__3(t), t),0),Eq(232*x__1(t) - 64*x__2(t) - 360*x__3(t) - 248*x__4(t) + Derivative(x__4(t), t),0)]
ics = {}
dsolve(ode,func=[x__1(t),x__2(t),x__3(t),x__4(t)],ics=ics)
\[
\left [ x^{1}{\left (t \right )} = \frac {C_{1} e^{16 t}}{2} - 2 C_{2} e^{32 t} + \frac {3 C_{3} e^{48 t}}{2} - \frac {C_{4} e^{64 t}}{3}, \ x^{2}{\left (t \right )} = C_{1} e^{16 t} - 5 C_{2} e^{32 t} - \frac {C_{3} e^{48 t}}{2} - \frac {C_{4} e^{64 t}}{3}, \ x^{3}{\left (t \right )} = - \frac {C_{1} e^{16 t}}{2} - C_{2} e^{32 t} + \frac {C_{3} e^{48 t}}{2} - \frac {2 C_{4} e^{64 t}}{3}, \ x^{4}{\left (t \right )} = C_{1} e^{16 t} + C_{2} e^{32 t} + C_{3} e^{48 t} + C_{4} e^{64 t}\right ]
\]