20.16.15 problem 15
Internal
problem
ID
[3848]
Book
:
Differential
equations
and
linear
algebra,
Stephen
W.
Goode
and
Scott
A
Annin.
Fourth
edition,
2015
Section
:
Chapter
9,
First
order
linear
systems.
Section
9.4
(Nondefective
coefficient
matrix),
page
607
Problem
number
:
15
Date
solved
:
Monday, January 27, 2025 at 08:03:22 AM
CAS
classification
:
system_of_ODEs
\begin{align*} \frac {d}{d t}x_{1} \left (t \right )&=x_{1} \left (t \right )+2 x_{2} \left (t \right )+3 x_{3} \left (t \right )+4 x_{4} \left (t \right )\\ \frac {d}{d t}x_{2} \left (t \right )&=4 x_{1} \left (t \right )+3 x_{2} \left (t \right )+2 x_{3} \left (t \right )+x_{4} \left (t \right )\\ \frac {d}{d t}x_{3} \left (t \right )&=4 x_{1} \left (t \right )+5 x_{2} \left (t \right )+6 x_{3} \left (t \right )+7 x_{4} \left (t \right )\\ \frac {d}{d t}x_{4} \left (t \right )&=7 x_{1} \left (t \right )+6 x_{2} \left (t \right )+5 x_{3} \left (t \right )+4 x_{4} \left (t \right ) \end{align*}
✓ Solution by Maple
Time used: 0.063 (sec). Leaf size: 85
dsolve([diff(x__1(t),t)=x__1(t)+2*x__2(t)+3*x__3(t)+4*x__4(t),diff(x__2(t),t)=4*x__1(t)+3*x__2(t)+2*x__3(t)+x__4(t),diff(x__3(t),t)=4*x__1(t)+5*x__2(t)+6*x__3(t)+7*x__4(t),diff(x__4(t),t)=7*x__1(t)+6*x__2(t)+5*x__3(t)+4*x__4(t)],singsol=all)
\begin{align*}
x_{1} \left (t \right ) &= c_{2} +c_3 \,{\mathrm e}^{16 t}+c_4 \,{\mathrm e}^{-2 t} \\
x_{2} \left (t \right ) &= \frac {13 c_3 \,{\mathrm e}^{16 t}}{17}-c_4 \,{\mathrm e}^{-2 t}-\frac {41 c_{2}}{51}+c_{1} \\
x_{3} \left (t \right ) &= \frac {35 c_3 \,{\mathrm e}^{16 t}}{17}+c_4 \,{\mathrm e}^{-2 t}-\frac {71 c_{2}}{51}-2 c_{1} \\
x_{4} \left (t \right ) &= \frac {31 c_3 \,{\mathrm e}^{16 t}}{17}+\frac {61 c_{2}}{51}-c_4 \,{\mathrm e}^{-2 t}+c_{1} \\
\end{align*}
✓ Solution by Mathematica
Time used: 0.015 (sec). Leaf size: 383
DSolve[{D[x1[t],t]==x1[t]+2*x2[t]+3*x3[t]+4*x4[t],D[x2[t],t]==4*x1[t]+3*x2[t]+2*x3[t]+x4[t],D[x3[t],t]==4*x1[t]+5*x2[t]+6*x3[t]+7*x4[t],D[x4[t],t]==7*x1[t]+6*x2[t]+5*x3[t]+4*x4[t]},{x1[t],x2[t],x3[t],x4[t]},t,IncludeSingularSolutions -> True]
\begin{align*}
\text {x1}(t)\to \frac {1}{96} e^{-2 t} \left (c_1 \left (-9 e^{2 t}+17 e^{18 t}+88\right )+c_2 \left (-57 e^{2 t}+17 e^{18 t}+40\right )-9 c_3 e^{2 t}+17 c_3 e^{18 t}+39 c_4 e^{2 t}+17 c_4 e^{18 t}-8 c_3-56 c_4\right ) \\
\text {x2}(t)\to \frac {1}{96} e^{-2 t} \left (c_1 \left (75 e^{2 t}+13 e^{18 t}-88\right )+c_2 \left (123 e^{2 t}+13 e^{18 t}-40\right )-21 c_3 e^{2 t}+13 c_3 e^{18 t}-69 c_4 e^{2 t}+13 c_4 e^{18 t}+8 c_3+56 c_4\right ) \\
\text {x3}(t)\to \frac {1}{96} e^{-2 t} \left (c_1 \left (-123 e^{2 t}+35 e^{18 t}+88\right )+5 c_2 \left (-15 e^{2 t}+7 e^{18 t}+8\right )+69 c_3 e^{2 t}+35 c_3 e^{18 t}+21 c_4 e^{2 t}+35 c_4 e^{18 t}-8 c_3-56 c_4\right ) \\
\text {x4}(t)\to \frac {1}{96} e^{-2 t} \left (c_1 \left (57 e^{2 t}+31 e^{18 t}-88\right )+c_2 \left (9 e^{2 t}+31 e^{18 t}-40\right )-39 c_3 e^{2 t}+31 c_3 e^{18 t}+9 c_4 e^{2 t}+31 c_4 e^{18 t}+8 c_3+56 c_4\right ) \\
\end{align*}