10.18.14 problem 18

Internal problem ID [1441]
Book : Elementary differential equations and boundary value problems, 10th ed., Boyce and DiPrima
Section : Chapter 7.9, Nonhomogeneous Linear Systems. page 447
Problem number : 18
Date solved : Monday, January 27, 2025 at 04:57:23 AM
CAS classification : system_of_ODEs

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

With initial conditions

\begin{align*} x_{1} \left (0\right ) = \alpha _{1}\\ x_{2} \left (0\right ) = \alpha _{2} \end{align*}

✓ Solution by Maple

Time used: 0.037 (sec). Leaf size: 92

dsolve([diff(x__1(t),t) = -2*x__1(t)+x__2(t)+2*exp(-t), diff(x__2(t),t) = x__1(t)-2*x__2(t)+3*t, x__1(0) = alpha__1, x__2(0) = alpha__2], singsol=all)
 

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

✓ Solution by Mathematica

Time used: 0.062 (sec). Leaf size: 122

DSolve[{D[ x1[t],t]==-2*x1[t]+1*x2[t]+2*Exp[-t],D[ x2[t],t]==1*x1[t]-2*x2[t]+3*t},{x1[0]==a1,x2[0]==a2},{x1[t],x2[t]},t,IncludeSingularSolutions -> True]
 

\begin{align*} \text {x1}(t)\to \frac {1}{6} e^{-3 t} \left (3 \text {a1} \left (e^{2 t}+1\right )+3 \text {a2} \left (e^{2 t}-1\right )+12 e^{2 t}-8 e^{3 t}+6 e^{2 t} t+6 e^{3 t} t-4\right ) \\ \text {x2}(t)\to \frac {1}{6} e^{-3 t} \left (3 \text {a1} \left (e^{2 t}-1\right )+3 \text {a2} \left (e^{2 t}+1\right )+6 e^{2 t} (t+1)+2 e^{3 t} (6 t-5)+4\right ) \\ \end{align*}