76.29.5 problem 5

Internal problem ID [17881]
Book : Differential equations. An introduction to modern methods and applications. James Brannan, William E. Boyce. Third edition. Wiley 2015
Section : Chapter 6. Systems of First Order Linear Equations. Section 6.7 (Defective Matrices). Problems at page 444
Problem number : 5
Date solved : Tuesday, January 28, 2025 at 11:09:39 AM
CAS classification : system_of_ODEs

\begin{align*} \frac {d}{d t}x_{1} \left (t \right )&=-7 x_{1} \left (t \right )+9 x_{2} \left (t \right )-6 x_{3} \left (t \right )\\ \frac {d}{d t}x_{2} \left (t \right )&=-8 x_{1} \left (t \right )+11 x_{2} \left (t \right )-7 x_{3} \left (t \right )\\ \frac {d}{d t}x_{3} \left (t \right )&=-2 x_{1} \left (t \right )+3 x_{2} \left (t \right )-x_{3} \left (t \right ) \end{align*}

✓ Solution by Maple

Time used: 0.087 (sec). Leaf size: 77

dsolve([diff(x__1(t),t)=-7*x__1(t)+9*x__2(t)-6*x__3(t),diff(x__2(t),t)=-8*x__1(t)+11*x__2(t)-7*x__3(t),diff(x__3(t),t)=-2*x__1(t)+3*x__2(t)-1*x__3(t)],singsol=all)
 

\begin{align*} x_{1} \left (t \right ) &= {\mathrm e}^{-t} c_{1} +c_{2} {\mathrm e}^{2 t}+c_{3} {\mathrm e}^{2 t} t \\ x_{2} \left (t \right ) &= \frac {2 \,{\mathrm e}^{-t} c_{1}}{3}+\frac {5 c_{2} {\mathrm e}^{2 t}}{3}+\frac {5 c_{3} {\mathrm e}^{2 t} t}{3}-\frac {c_{3} {\mathrm e}^{2 t}}{3} \\ x_{3} \left (t \right ) &= \frac {{\mathrm e}^{2 t} \left (3 c_{3} t +3 c_{2} -2 c_{3} \right )}{3} \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.007 (sec). Leaf size: 158

DSolve[{D[x1[t],t]==-7*x1[t]+9*x2[t]-6*x3[t],D[x2[t],t]==-8*x1[t]+11*x2[t]-7*x3[t],D[x3[t],t]==-2*x1[t]+3*x2[t]-1*x3[t]},{x1[t],x2[t],x3[t]},t,IncludeSingularSolutions -> True]
 

\begin{align*} \text {x1}(t)\to \frac {1}{3} e^{-t} \left (c_1 \left (7-2 e^{3 t} (3 t+2)\right )+3 \left (c_2 \left (e^{3 t} (3 t+2)-2\right )-c_3 e^{3 t} (3 t+1)+c_3\right )\right ) \\ \text {x2}(t)\to \frac {1}{9} e^{-t} \left (-2 c_1 \left (e^{3 t} (15 t+7)-7\right )+3 c_2 \left (e^{3 t} (15 t+7)-4\right )-3 c_3 \left (e^{3 t} (15 t+2)-2\right )\right ) \\ \text {x3}(t)\to e^{2 t} ((-2 c_1+3 c_2-3 c_3) t+c_3) \\ \end{align*}