2.2.1 Problem 1

Internal problem ID [10329]
Book : First order enumerated odes
Section : section 2 (system of first order odes)
Problem number : 1
Date solved : Thursday, November 27, 2025 at 10:34:22 AM
CAS classification : system_of_ODEs

\begin{align*} x^{\prime }+y^{\prime }-x&=y+t \\ x^{\prime }+y^{\prime }&=2 x+3 y+{\mathrm e}^{t} \\ \end{align*}

The system is

\begin{align*} x^{\prime }+y^{\prime }&=x+y+t\tag {1}\\ x^{\prime }+y^{\prime }&=2 x+3 y+{\mathrm e}^{t}\tag {2} \end{align*}

Since the left side is the same, this implies

\begin{align*} x+y+t&=2 x+3 y+{\mathrm e}^{t}\\ y&=-\frac {x}{2}-\frac {{\mathrm e}^{t}}{2}+\frac {t}{2}\tag {3} \end{align*}

Taking derivative of the above w.r.t. \(t\) gives

\begin{align*} y^{\prime }&=-\frac {x^{\prime }}{2}-\frac {{\mathrm e}^{t}}{2}+\frac {1}{2}\tag {4} \end{align*}

Substituting (3,4) in (1) to eliminate \(y,y^{\prime }\) gives

\begin{align*} \frac {x^{\prime }}{2}-\frac {{\mathrm e}^{t}}{2}+\frac {1}{2} &= \frac {x}{2}-\frac {{\mathrm e}^{t}}{2}+\frac {3 t}{2}\\ x^{\prime } &= x+3 t -1\tag {5} \end{align*}

Which is now solved for \(x\). Solve In canonical form a linear first order is

\begin{align*} x^{\prime } + q(t)x &= p(t) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(t) &=-1\\ p(t) &=3 t -1 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dt}}\\ &= {\mathrm e}^{\int \left (-1\right )d t}\\ &= {\mathrm e}^{-t} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}}\left ( \mu x\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}}\left ( \mu x\right ) &= \left (\mu \right ) \left (3 t -1\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \left (x \,{\mathrm e}^{-t}\right ) &= \left ({\mathrm e}^{-t}\right ) \left (3 t -1\right ) \\ \mathrm {d} \left (x \,{\mathrm e}^{-t}\right ) &= \left (\left (3 t -1\right ) {\mathrm e}^{-t}\right )\, \mathrm {d} t \\ \end{align*}

Integrating gives

\begin{align*} x \,{\mathrm e}^{-t}&= \int {\left (3 t -1\right ) {\mathrm e}^{-t} \,dt} \\ &=\left (-3 t -2\right ) {\mathrm e}^{-t} + c_1 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{-t}\) gives the final solution

\[ x = c_1 \,{\mathrm e}^{t}-3 t -2 \]

Given now that we have the solution

\begin{align*} x&=c_1 \,{\mathrm e}^{t}-3 t -2 \tag {6} \end{align*}

Then substituting (6) into (3) gives

\begin{align*} y&=-\frac {c_1 \,{\mathrm e}^{t}}{2}+2 t +1-\frac {{\mathrm e}^{t}}{2} \tag {7} \end{align*}

✓ Maple. Time used: 0.082 (sec). Leaf size: 30

ode:=[diff(x(t),t)+diff(y(t),t)-x(t) = t+y(t), diff(x(t),t)+diff(y(t),t) = 2*x(t)+3*y(t)+exp(t)]; 
dsolve(ode);
 

\begin{align*} x \left (t \right ) &= -3 t -2+{\mathrm e}^{t} c_1 \\ y \left (t \right ) &= 2 t +1-\frac {{\mathrm e}^{t} c_1}{2}-\frac {{\mathrm e}^{t}}{2} \\ \end{align*}

✓ Mathematica. Time used: 0.02 (sec). Leaf size: 73

ode={D[x[t],t]+D[y[t],t]-x[t]==y[t]+t,D[x[t],t]+D[y[t],t]==2*x[t]+3*y[t]+Exp[t]}; 
ic={}; 
DSolve[{ode,ic},{x[t],y[t]},t,IncludeSingularSolutions->True]
 

\begin{align*} x(t)&\to 2 \int _1^te^{t-K[1]} K[1]dK[1]-t+(1+2 c_1) e^t\\ y(t)&\to -\int _1^te^{t-K[1]} K[1]dK[1]+t-(1+c_1) e^t \end{align*}

✗ Sympy

from sympy import * 
t = symbols("t") 
x = Function("x") 
y = Function("y") 
ode=[Eq(-t - x(t) - y(t) + Derivative(x(t), t) + Derivative(y(t), t),0),Eq(-2*x(t) - 3*y(t) - exp(t) + Derivative(x(t), t) + Derivative(y(t), t),0)] 
ics = {} 
dsolve(ode,func=[x(t),y(t)],ics=ics)