Internal problem ID [348]
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 45.
ODE order: 1.
ODE degree: 1.
Solve \begin {align*} x_{1}^{\prime }\relax (t )&=9 x_{1} \relax (t )-7 x_{2} \relax (t )-5 x_{3} \relax (t )\\ x_{2}^{\prime }\relax (t )&=-12 x_{1} \relax (t )+7 x_{2} \relax (t )+11 x_{3} \relax (t )+9 x_{4} \relax (t )\\ x_{3}^{\prime }\relax (t )&=24 x_{1} \relax (t )-17 x_{2} \relax (t )-19 x_{3} \relax (t )-9 x_{4} \relax (t )\\ x_{4}^{\prime }\relax (t )&=-18 x_{1} \relax (t )+13 x_{2} \relax (t )+17 x_{3} \relax (t )+9 x_{4} \relax (t ) \end {align*}
✓ Solution by Maple
Time used: 0.094 (sec). Leaf size: 104
dsolve([diff(x__1(t),t)=9*x__1(t)-7*x__2(t)-5*x__3(t)+0*x__4(t),diff(x__2(t),t)=-12*x__1(t)+7*x__2(t)+11*x__3(t)+9*x__4(t),diff(x__3(t),t)=24*x__1(t)-17*x__2(t)-19*x__3(t)-9*x__4(t),diff(x__4(t),t)=-18*x__1(t)+13*x__2(t)+17*x__3(t)+9*x__4(t)],[x__1(t), x__2(t), x__3(t), x__4(t)], singsol=all)
\[ x_{1} \relax (t ) = -c_{2} {\mathrm e}^{6 t}-c_{3} {\mathrm e}^{-3 t}+2 \,{\mathrm e}^{3 t} c_{4}+c_{1} \] \[ x_{2} \relax (t ) = c_{2} {\mathrm e}^{6 t}-c_{3} {\mathrm e}^{-3 t}+{\mathrm e}^{3 t} c_{4}+2 c_{1} \] \[ x_{3} \relax (t ) = -2 c_{2} {\mathrm e}^{6 t}-c_{3} {\mathrm e}^{-3 t}+{\mathrm e}^{3 t} c_{4}-c_{1} \] \[ x_{4} \relax (t ) = c_{1}+c_{2} {\mathrm e}^{6 t}+c_{3} {\mathrm e}^{-3 t}+{\mathrm e}^{3 t} c_{4} \]
✓ Solution by Mathematica
Time used: 0.01 (sec). Leaf size: 346
DSolve[{x1'[t]==9*x1[t]-7*x2[t]-5*x3[t]+0*x4[t],x2'[t]==-12*x1[t]+7*x2[t]+11*x3[t]+9*x4[t],x3'[t]==24*x1[t]-17*x2[t]-19*x3[t]-9*x4[t],x4'[t]==-18*x1[t]+13*x2[t]+17*x3[t]+9*x4[t]},{x1[t],x2[t],x3[t],x4[t]},t,IncludeSingularSolutions -> True]
\begin{align*} \text {x1}(t)\to (-c_1+c_2+c_3) e^{-3 t}-\frac {1}{3} (4 c_2+5 c_3+3 c_4) e^{6 t}+\frac {2}{3} e^{3 t} (6 c_1 \cosh (3 t)-3 c_1+2 c_2+4 c_3+3 c_4)-c_2-2 c_3-c_4 \\ \text {x2}(t)\to \frac {1}{3} \left (3 (c_2+c_3) e^{-3 t}+(2 c_2+4 c_3+3 c_4) e^{3 t}+(4 c_2+5 c_3+3 c_4) e^{6 t}-6 c_1 \left (e^{6 t}+\cosh (3 t)-2\right )-6 (c_2+2 c_3+c_4)\right ) \\ \text {x3}(t)\to (c_2+c_3) e^{-3 t}+\frac {1}{3} (2 c_2+4 c_3+3 c_4) e^{3 t}-\frac {2}{3} (-6 c_1+4 c_2+5 c_3+3 c_4) e^{6 t}-2 c_1 \cosh (3 t)-2 c_1+c_2+2 c_3+c_4 \\ \text {x4}(t)\to \frac {1}{3} e^{-3 t} \left (-3 (c_2+2 c_3+c_4) e^{3 t}+(4 c_2+5 c_3+3 c_4) e^{9 t}+e^{6 t} (-12 c_1 \sinh (3 t)-3 c_1+2 c_2+4 c_3+3 c_4)-3 (-c_1+c_2+c_3)\right ) \\ \end{align*}