76.28.6 problem 7

Internal problem ID [17873]
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 6.6 (Nonhomogeneous Linear Systems). Problems at page 436
Problem number : 7
Date solved : Tuesday, January 28, 2025 at 11:04:21 AM
CAS classification : system_of_ODEs

\begin{align*} \frac {d}{d t}x_{1} \left (t \right )&=-\frac {x_{1} \left (t \right )}{2}+\frac {x_{2} \left (t \right )}{2}-\frac {x_{3} \left (t \right )}{2}+1\\ \frac {d}{d t}x_{2} \left (t \right )&=-x_{1} \left (t \right )-2 x_{2} \left (t \right )+x_{3} \left (t \right )+t\\ \frac {d}{d t}x_{3} \left (t \right )&=\frac {x_{1} \left (t \right )}{2}+\frac {x_{2} \left (t \right )}{2}-\frac {3 x_{3} \left (t \right )}{2}+11 \,{\mathrm e}^{-3 t} \end{align*}

✓ Solution by Maple

Time used: 0.198 (sec). Leaf size: 96

dsolve([diff(x__1(t),t)=-1/2*x__1(t)+1/2*x__2(t)-1/2*x__3(t)+1,diff(x__2(t),t)=-1*x__1(t)-2*x__2(t)+1*x__3(t)+t,diff(x__3(t),t)=1/2*x__1(t)+1/2*x__2(t)-3/2*x__3(t)+11*exp(-3*t)],singsol=all)
 

\begin{align*} x_{1} \left (t \right ) &= -{\mathrm e}^{-2 t} c_{2} -\frac {11 \,{\mathrm e}^{-3 t}}{4}+\frac {t}{4}+\frac {7}{8}+c_{3} {\mathrm e}^{-t} \\ x_{2} \left (t \right ) &= 2 \,{\mathrm e}^{-2 t} c_{2} +\frac {11 \,{\mathrm e}^{-3 t}}{2}+\frac {t}{2}-\frac {3}{4}-2 c_{3} {\mathrm e}^{-t}+{\mathrm e}^{-t} c_{1} \\ x_{3} \left (t \right ) &= -{\mathrm e}^{-2 t} c_{2} -\frac {33 \,{\mathrm e}^{-3 t}}{4}-\frac {1}{8}-c_{3} {\mathrm e}^{-t}+\frac {t}{4}+{\mathrm e}^{-t} c_{1} \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.989 (sec). Leaf size: 165

DSolve[{D[x1[t],t]==-1/2*x1[t]+1/2*x2[t]-1/2*x3[t]+1,D[x2[t],t]==-1*x1[t]-2*x2[t]+1*x3[t]+t,D[x3[t],t]==1/2*x1[t]+1/2*x2[t]-3/2*x3[t]+11*Exp[-3*t]},{x1[t],x2[t],x3[t]},t,IncludeSingularSolutions -> True]
 

\begin{align*} \text {x1}(t)\to \frac {1}{8} e^{-3 t} \left (e^{3 t} (2 t+7)-4 (c_1+c_2-c_3) e^t+4 (3 c_1+c_2-c_3) e^{2 t}-22\right ) \\ \text {x2}(t)\to \frac {1}{4} e^{-3 t} \left (e^{3 t} (2 t-3)-4 (c_1-c_3) e^{2 t}+4 (c_1+c_2-c_3) e^t+22\right ) \\ \text {x3}(t)\to \frac {1}{8} e^{-3 t} \left (e^{3 t} (2 t-1)-4 (c_1+c_2-c_3) e^t+4 (c_1+c_2+c_3) e^{2 t}-66\right ) \\ \end{align*}