12.22.22 problem section 10.5, problem 22

Internal problem ID [2275]
Book : Elementary differential equations with boundary value problems. William F. Trench. Brooks/Cole 2001
Section : Chapter 10 Linear system of Differential equations. Section 10.5, constant coefficient homogeneous system II. Page 555
Problem number : section 10.5, problem 22
Date solved : Monday, January 27, 2025 at 05:44:27 AM
CAS classification : system_of_ODEs

\begin{align*} \frac {d}{d t}y_{1} \left (t \right )&=4 y_{1} \left (t \right )-8 y_{2} \left (t \right )-4 y_{3} \left (t \right )\\ \frac {d}{d t}y_{2} \left (t \right )&=-3 y_{1} \left (t \right )-y_{2} \left (t \right )-4 y_{3} \left (t \right )\\ \frac {d}{d t}y_{3} \left (t \right )&=y_{1} \left (t \right )-y_{2} \left (t \right )+9 y_{3} \left (t \right ) \end{align*}

With initial conditions

\begin{align*} y_{1} \left (0\right ) = -4\\ y_{2} \left (0\right ) = 1\\ y_{3} \left (0\right ) = -3 \end{align*}

✓ Solution by Maple

Time used: 0.043 (sec). Leaf size: 61

dsolve([diff(y__1(t),t) = 4*y__1(t)-8*y__2(t)-4*y__3(t), diff(y__2(t),t) = -3*y__1(t)-y__2(t)-4*y__3(t), diff(y__3(t),t) = y__1(t)-y__2(t)+9*y__3(t), y__1(0) = -4, y__2(0) = 1, y__3(0) = -3], singsol=all)
 

\begin{align*} y_{1} \left (t \right ) &= \frac {22 \,{\mathrm e}^{9 t}}{13}-\frac {50 \,{\mathrm e}^{7 t}}{11}-\frac {164 \,{\mathrm e}^{-4 t}}{143} \\ y_{2} \left (t \right ) &= \frac {22 \,{\mathrm e}^{9 t}}{13}+\frac {5 \,{\mathrm e}^{7 t}}{11}-\frac {164 \,{\mathrm e}^{-4 t}}{143} \\ y_{3} \left (t \right ) &= -\frac {11 \,{\mathrm e}^{9 t}}{2}+\frac {5 \,{\mathrm e}^{7 t}}{2} \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.008 (sec). Leaf size: 57

DSolve[{D[ y1[t],t]==4*y1[t]-8*y2[t]-4*y3[t],D[ y2[t],t]==-3*y1[t]-1*y2[t]-3*y3[t],D[ y3[t],t]==1*y1[t]-1*y2[t]+9*y3[t]},{y1[0]==-4,y2[0]==1,y3[0]==-3},{y1[t],y2[t],y3[t]},t,IncludeSingularSolutions -> True]
 

\begin{align*} \text {y1}(t)\to e^{8 t} (8 t-3)-e^{-4 t} \\ \text {y2}(t)\to 2 e^{8 t}-e^{-4 t} \\ \text {y3}(t)\to -e^{8 t} (8 t+3) \\ \end{align*}