14.25.12 problem 13

Internal problem ID [2769]
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 : 13
Date solved : Monday, January 27, 2025 at 06:13:04 AM
CAS classification : system_of_ODEs

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

✓ Solution by Maple

Time used: 0.059 (sec). Leaf size: 122

dsolve([diff(x__1(t),t)=1*x__1(t)+2*x__2(t)-3*x__3(t)+exp(t),diff(x__2(t),t)=1*x__1(t)+1*x__2(t)+2*x__3(t),diff(x__3(t),t)=1*x__1(t)-1*x__2(t)+4*x__3(t)-exp(t)],singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.153 (sec). Leaf size: 133

DSolve[{D[ x1[t],t]==1*x1[t]+2*x2[t]-3*x3[t]+Exp[t],D[ x2[t],t]==1*x1[t]+1*x2[t]+2*x3[t],D[ x3[t],t]==1*x1[t]-1*x2[t]+4*x3[t]-Exp[t]},{x1[t],x2[t],x3[t]},t,IncludeSingularSolutions -> True]
 

\begin{align*} \text {x1}(t)\to \frac {1}{2} e^t \left (4+e^t (-2 c_1 (t-1)-c_2 (t-4) t+c_3 (t-6) t)\right ) \\ \text {x2}(t)\to \frac {1}{2} e^t \left (-4+e^t \left ((c_2-c_3) t^2+2 (c_1-c_2+2 c_3) t+2 c_2\right )\right ) \\ \text {x3}(t)\to \frac {1}{2} e^t \left (-2+e^t \left ((c_2-c_3) t^2+2 (c_1-c_2+2 c_3) t+2 c_3\right )\right ) \\ \end{align*}