23.8 problem section 10.6, problem 8

Internal problem ID [1643]

Book: Elementary differential equations with boundary value problems. William F. Trench. Brooks/Cole 2001
Section: Chapter 10 Linear system of Differential equations. Section 10.6, constant coefficient homogeneous system III. Page 566
Problem number: section 10.6, problem 8.
ODE order: 1.
ODE degree: 1.

Solve \begin {align*} y_{1}^{\prime }\relax (t )&=-3 y_{1} \relax (t )+y_{2} \relax (t )-3 y_{3} \relax (t )\\ y_{2}^{\prime }\relax (t )&=4 y_{1} \relax (t )-y_{2} \relax (t )+2 y_{3} \relax (t )\\ y_{3}^{\prime }\relax (t )&=4 y_{1} \relax (t )-2 y_{2} \relax (t )+3 y_{3} \relax (t ) \end {align*}

Solution by Maple

Time used: 0.109 (sec). Leaf size: 112

dsolve([diff(y__1(t),t)=-3*y__1(t)+1*y__2(t)-3*y__3(t),diff(y__2(t),t)=4*y__1(t)-1*y__2(t)+2*y__3(t),diff(y__3(t),t)=4*y__1(t)-2*y__2(t)+3*y__3(t)],[y__1(t), y__2(t), y__3(t)], singsol=all)
 

\[ y_{1} \relax (t ) = -c_{1} {\mathrm e}^{t}-\frac {c_{2} {\mathrm e}^{-t} \sin \left (2 t \right )}{2}+\frac {c_{2} {\mathrm e}^{-t} \cos \left (2 t \right )}{2}-\frac {c_{3} {\mathrm e}^{-t} \cos \left (2 t \right )}{2}-\frac {c_{3} {\mathrm e}^{-t} \sin \left (2 t \right )}{2} \] \[ y_{2} \relax (t ) = -c_{1} {\mathrm e}^{t}+c_{2} {\mathrm e}^{-t} \sin \left (2 t \right )+c_{3} {\mathrm e}^{-t} \cos \left (2 t \right ) \] \[ y_{3} \relax (t ) = c_{1} {\mathrm e}^{t}+c_{2} {\mathrm e}^{-t} \sin \left (2 t \right )+c_{3} {\mathrm e}^{-t} \cos \left (2 t \right ) \]

Solution by Mathematica

Time used: 0.017 (sec). Leaf size: 163

DSolve[{y1'[t]==-3*y1[t]+1*y2[t]-3*y3[t],y2'[t]==4*y1[t]-1*y2[t]+2*y3[t],y3'[t]==4*y1[t]-2*y2[t]+3*y3[t]},{y1[t],y2[t],y3[t]},t,IncludeSingularSolutions -> True]
 

\begin{align*} \text {y1}(t)\to \frac {1}{2} e^{-t} \left ((c_2-c_3) e^{2 t}+(2 c_1-c_2+c_3) \cos (2 t)-2 (c_1+c_3) \sin (2 t)\right ) \\ \text {y2}(t)\to \frac {1}{2} e^{-t} \left ((c_2-c_3) e^{2 t}+(c_2+c_3) \cos (2 t)+(4 c_1-c_2+3 c_3) \sin (2 t)\right ) \\ \text {y3}(t)\to \frac {1}{2} e^{-t} \left ((c_3-c_2) e^{2 t}+(c_2+c_3) \cos (2 t)+(4 c_1-c_2+3 c_3) \sin (2 t)\right ) \\ \end{align*}