9.5.4 problem Example 6

Internal problem ID [1007]
Book : Differential equations and linear algebra, 4th ed., Edwards and Penney
Section : Section 7.6, Multiple Eigenvalue Solutions. Examples. Page 437
Problem number : Example 6
Date solved : Tuesday, March 04, 2025 at 12:07:21 PM
CAS classification : system_of_ODEs

\begin{align*} \frac {d}{d t}x_{1} \left (t \right )&=x_{3} \left (t \right )\\ \frac {d}{d t}x_{2} \left (t \right )&=x_{4} \left (t \right )\\ \frac {d}{d t}x_{3} \left (t \right )&=-2 x_{1} \left (t \right )+2 x_{2} \left (t \right )-3 x_{3} \left (t \right )+x_{4} \left (t \right )\\ \frac {d}{d t}x_{4} \left (t \right )&=2 x_{1} \left (t \right )-2 x_{2} \left (t \right )+x_{3} \left (t \right )-3 x_{4} \left (t \right ) \end{align*}

✓ Maple. Time used: 0.053 (sec). Leaf size: 94

ode:=[diff(x__1(t),t) = x__3(t), diff(x__2(t),t) = x__4(t), diff(x__3(t),t) = -2*x__1(t)+2*x__2(t)-3*x__3(t)+x__4(t), diff(x__4(t),t) = 2*x__1(t)-2*x__2(t)+x__3(t)-3*x__4(t)]; 
dsolve(ode);
 

\begin{align*} x_{1} \left (t \right ) &= c_2 +c_3 \,{\mathrm e}^{-2 t}+c_4 \,{\mathrm e}^{-2 t} t \\ x_{2} \left (t \right ) &= -c_3 \,{\mathrm e}^{-2 t}-c_4 \,{\mathrm e}^{-2 t} t +c_4 \,{\mathrm e}^{-2 t}+c_2 +c_1 \,{\mathrm e}^{-2 t} \\ x_{3} \left (t \right ) &= -{\mathrm e}^{-2 t} \left (2 c_4 t +2 c_3 -c_4 \right ) \\ x_{4} \left (t \right ) &= -{\mathrm e}^{-2 t} \left (-2 c_4 t +2 c_1 -2 c_3 +3 c_4 \right ) \\ \end{align*}

✓ Mathematica. Time used: 0.051 (sec). Leaf size: 210

ode={D[ x1[t],t]==0*x1[t]+0*x2[t]+1*x3[t]+0*x4[t],D[ x2[t],t]==0*x1[t]+0*x2[t]+0*x3[t]+1*x4[t],D[ x3[t],t]==-2*x1[t]+2*x2[t]-3*x3[t]+1*x4[t],D[ x4[t],t]==2*x1[t]-2*x2[t]+1*x3[t]-3*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}{4} e^{-2 t} \left (2 c_1 \left (2 t+e^{2 t}+1\right )+2 c_2 \left (-2 t+e^{2 t}-1\right )+c_3 e^{2 t}+2 c_3 t+c_4 e^{2 t}-2 c_4 t-c_3-c_4\right ) \\ \text {x2}(t)\to \frac {1}{4} e^{-2 t} \left (2 c_1 \left (-2 t+e^{2 t}-1\right )+2 c_2 \left (2 t+e^{2 t}+1\right )+c_3 e^{2 t}-2 c_3 t+c_4 e^{2 t}+2 c_4 t-c_3-c_4\right ) \\ \text {x3}(t)\to e^{-2 t} ((-2 c_1+2 c_2-c_3+c_4) t+c_3) \\ \text {x4}(t)\to e^{-2 t} ((2 c_1-2 c_2+c_3-c_4) t+c_4) \\ \end{align*}

✓ Sympy. Time used: 0.184 (sec). Leaf size: 95

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(-x__3(t) + Derivative(x__1(t), t),0),Eq(-x__4(t) + Derivative(x__2(t), t),0),Eq(2*x__1(t) - 2*x__2(t) + 3*x__3(t) - x__4(t) + Derivative(x__3(t), t),0),Eq(-2*x__1(t) + 2*x__2(t) - x__3(t) + 3*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)
 

\[ \left [ x^{1}{\left (t \right )} = C_{1} - 2 C_{3} t e^{- 2 t} - \left (2 C_{2} + C_{3} + \frac {C_{4}}{2}\right ) e^{- 2 t}, \ x^{2}{\left (t \right )} = C_{1} + 2 C_{3} t e^{- 2 t} + \left (2 C_{2} + C_{3}\right ) e^{- 2 t}, \ x^{3}{\left (t \right )} = 4 C_{3} t e^{- 2 t} + \left (4 C_{2} + C_{4}\right ) e^{- 2 t}, \ x^{4}{\left (t \right )} = - 4 C_{2} e^{- 2 t} - 4 C_{3} t e^{- 2 t}\right ] \]