8.26.3 problem 3

Internal problem ID [2776]
Book : Differential equations and their applications, 4th ed., M. Braun
Section : Chapter 3. Systems of differential equations. Section 3.13 (Solving systems by Laplace transform). Page 370
Problem number : 3
Date solved : Tuesday, September 30, 2025 at 05:51:59 AM
CAS classification : system_of_ODEs

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

With initial conditions

\begin{align*} x_{1} \left (0\right )&=2 \\ x_{2} \left (0\right )&=1 \\ \end{align*}
Maple. Time used: 0.202 (sec). Leaf size: 48
ode:=[diff(x__1(t),t) = 3*x__1(t)-2*x__2(t)+t, diff(x__2(t),t) = 2*x__1(t)-2*x__2(t)+3*exp(t)]; 
ic:=[x__1(0) = 2, x__2(0) = 1]; 
dsolve([ode,op(ic)]);
 
\begin{align*} x_{1} \left (t \right ) &= -\frac {4 \,{\mathrm e}^{-t}}{3}+\frac {{\mathrm e}^{2 t}}{3}+3 \,{\mathrm e}^{t}-t \\ x_{2} \left (t \right ) &= -\frac {8 \,{\mathrm e}^{-t}}{3}+\frac {{\mathrm e}^{2 t}}{6}+3 \,{\mathrm e}^{t}+\frac {1}{2}-t \\ \end{align*}
Mathematica. Time used: 0.223 (sec). Leaf size: 59
ode={D[x1[t],t]==3*x1[t]-2*x2[t]+t,D[ x2[t],t]==2*x1[t]-2*x2[t]+3*Exp[t]}; 
ic={x1[0]==2,x2[0]==1}; 
DSolve[{ode,ic},{x1[t],x2[t]},t,IncludeSingularSolutions->True]
 
\begin{align*} \text {x1}(t)&\to \frac {1}{3} \left (-3 t-4 e^{-t}+9 e^t+e^{2 t}\right )\\ \text {x2}(t)&\to \frac {1}{6} \left (-6 t-16 e^{-t}+18 e^t+e^{2 t}+3\right ) \end{align*}
Sympy. Time used: 0.124 (sec). Leaf size: 48
from sympy import * 
t = symbols("t") 
x__1 = Function("x__1") 
x__2 = Function("x__2") 
ode=[Eq(-t - 3*x__1(t) + 2*x__2(t) + Derivative(x__1(t), t),0),Eq(-2*x__1(t) + 2*x__2(t) - 3*exp(t) + Derivative(x__2(t), t),0)] 
ics = {} 
dsolve(ode,func=[x__1(t),x__2(t)],ics=ics)
 
\[ \left [ x^{1}{\left (t \right )} = \frac {C_{1} e^{- t}}{2} + 2 C_{2} e^{2 t} - t + 3 e^{t}, \ x^{2}{\left (t \right )} = C_{1} e^{- t} + C_{2} e^{2 t} - t + 3 e^{t} + \frac {1}{2}\right ] \]