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 : Monday, January 27, 2025 at 03:22:53 AM
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*}

✓ Solution by Maple

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

dsolve([diff(x__1(t),t)=0*x__1(t)+0*x__2(t)+1*x__3(t)+0*x__4(t),diff(x__2(t),t)=0*x__1(t)+0*x__2(t)+0*x__3(t)+1*x__4(t),diff(x__3(t),t)=-2*x__1(t)+2*x__2(t)-3*x__3(t)+1*x__4(t),diff(x__4(t),t)=2*x__1(t)-2*x__2(t)+1*x__3(t)-3*x__4(t)],singsol=all)
 

\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*}

✓ Solution by Mathematica

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

DSolve[{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]},{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*}