Internal
problem
ID
[1003]
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
50
Date
solved
:
Tuesday, March 04, 2025 at 12:07:14 PM
CAS
classification
:
system_of_ODEs
ode:=[diff(x__1(t),t) = 9*x__1(t)+13*x__2(t)-13*x__6(t), diff(x__2(t),t) = -14*x__1(t)+19*x__2(t)-10*x__3(t)-20*x__4(t)+10*x__5(t)+4*x__6(t), diff(x__3(t),t) = -30*x__1(t)+12*x__2(t)-7*x__3(t)-30*x__4(t)+12*x__5(t)+18*x__6(t), diff(x__4(t),t) = -12*x__1(t)+10*x__2(t)-10*x__3(t)-9*x__4(t)+10*x__5(t)+2*x__6(t), diff(x__5(t),t) = 6*x__1(t)+9*x__2(t)+6*x__4(t)+5*x__5(t)-15*x__6(t), diff(x__6(t),t) = -14*x__1(t)+23*x__2(t)-10*x__3(t)-20*x__4(t)+10*x__5(t)]; dsolve(ode);
ode={D[ x1[t],t]==9*x1[t]+13*x2[t]-13*x6[t],D[ x2[t],t]==-14*x1[t]+19*x2[t]-10*x3[t]-20*x4[t]+10*x5[t]+4*x6[t],D[ x3[t],t]==-30*x1[t]+12*x2[t]-7*x3[t]-30*x4[t]+12*x5[t]+18*x6[t],D[ x4[t],t]==-12*x1[t]+10*x2[t]-10*x3[t]-9*x4[t]+10*x5[t]+2*x6[t],D[ x5[t],t]==6*x1[t]+9*x2[t]+6*x4[t]+5*x5[t]-15*x6[t],D[ x6[t],t]==-14*x1[t]+23*x2[t]-10*x3[t]-20*x4[t]-10*x5[t]}; ic={}; DSolve[{ode,ic},{x1[t],x2[t],x3[t],x4[t],x5[t],x6[t]},t,IncludeSingularSolutions->True]
Too large to display
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") x__5 = Function("x__5") x__6 = Function("x__6") ode=[Eq(-9*x__1(t) - 13*x__2(t) + 13*x__6(t) + Derivative(x__1(t), t),0),Eq(14*x__1(t) - 19*x__2(t) + 10*x__3(t) + 20*x__4(t) - 10*x__5(t) - 4*x__6(t) + Derivative(x__2(t), t),0),Eq(30*x__1(t) - 12*x__2(t) + 7*x__3(t) + 30*x__4(t) - 12*x__5(t) - 18*x__6(t) + Derivative(x__3(t), t),0),Eq(12*x__1(t) - 10*x__2(t) + 10*x__3(t) + 9*x__4(t) - 10*x__5(t) - 2*x__6(t) + Derivative(x__4(t), t),0),Eq(-6*x__1(t) - 9*x__2(t) - 6*x__4(t) - 5*x__5(t) + 15*x__6(t) + Derivative(x__5(t), t),0),Eq(14*x__1(t) - 23*x__2(t) + 10*x__3(t) + 20*x__4(t) - 10*x__5(t) + Derivative(x__6(t), t),0)] ics = {} dsolve(ode,func=[x__1(t),x__2(t),x__3(t),x__4(t),x__5(t),x__6(t)],ics=ics)