14.25.13 problem 14

Internal problem ID [2770]
Book : Differential equations and their applications, 4th ed., M. Braun
Section : Section 3.12, Systems of differential equations. The nonhomogeneous equation. variation of parameters. Page 366
Problem number : 14
Date solved : Monday, January 27, 2025 at 06:13:05 AM
CAS classification : system_of_ODEs

\begin{align*} x_{1}^{\prime }\left (t \right )&=-x_{1} \left (t \right )-x_{2} \left (t \right )+1\\ x_{2}^{\prime }\left (t \right )&=-4 x_{2} \left (t \right )-x_{3} \left (t \right )+t\\ x_{3}^{\prime }\left (t \right )&=5 x_{2} \left (t \right )+{\mathrm e}^{t} \end{align*}

✓ Solution by Maple

Time used: 0.211 (sec). Leaf size: 122

dsolve([diff(x__1(t),t)=-1*x__1(t)-1*x__2(t)+0*x__3(t)+1,diff(x__2(t),t)=0*x__1(t)-4*x__2(t)-1*x__3(t)+t,diff(x__3(t),t)=0*x__1(t)+5*x__2(t)-0*x__3(t)+exp(t)],singsol=all)
 

\begin{align*} x_{1} \left (t \right ) &= -\frac {{\mathrm e}^{-2 t} \sin \left (t \right ) c_2}{2}+\frac {{\mathrm e}^{-2 t} \sin \left (t \right ) c_3}{2}+\frac {c_2 \,{\mathrm e}^{-2 t} \cos \left (t \right )}{2}+\frac {{\mathrm e}^{-2 t} \cos \left (t \right ) c_3}{2}+\frac {{\mathrm e}^{t}}{20}+\frac {4}{5}+{\mathrm e}^{-t} c_1 \\ x_{2} \left (t \right ) &= {\mathrm e}^{-2 t} \sin \left (t \right ) c_3 +c_2 \,{\mathrm e}^{-2 t} \cos \left (t \right )+\frac {1}{5}-\frac {{\mathrm e}^{t}}{10} \\ x_{3} \left (t \right ) &= -2 \,{\mathrm e}^{-2 t} \sin \left (t \right ) c_3 -{\mathrm e}^{-2 t} \cos \left (t \right ) c_3 -2 c_2 \,{\mathrm e}^{-2 t} \cos \left (t \right )+{\mathrm e}^{-2 t} \sin \left (t \right ) c_2 +\frac {{\mathrm e}^{t}}{2}-\frac {4}{5}+t \\ \end{align*}

✓ Solution by Mathematica

Time used: 0.884 (sec). Leaf size: 144

DSolve[{D[ x1[t],t]==-1*x1[t]-1*x2[t]+0*x3[t]+1,D[ x2[t],t]==0*x1[t]-4*x2[t]-1*x3[t]+t,D[ x3[t],t]==0*x1[t]+5*x2[t]-0*x3[t]+Exp[t]},{x1[t],x2[t],x3[t]},t,IncludeSingularSolutions -> True]
 

\begin{align*} \text {x1}(t)\to \frac {1}{20} e^{-2 t} \left (e^t \left (16 e^t+e^{2 t}+10 (2 c_1+c_2+c_3)\right )-10 (c_2+c_3) \cos (t)-10 (3 c_2+c_3) \sin (t)\right ) \\ \text {x2}(t)\to \frac {1}{10} \left (2-e^t\right )+c_2 e^{-2 t} \cos (t)-(2 c_2+c_3) e^{-2 t} \sin (t) \\ \text {x3}(t)\to t+\frac {e^t}{2}+c_3 e^{-2 t} \cos (t)+(5 c_2+2 c_3) e^{-2 t} \sin (t)-\frac {4}{5} \\ \end{align*}