2.1 problem 1

Internal problem ID [8035]
Book : First order enumerated odes
Section : section 2 (system of first order odes)
Problem number : 1
Date solved : Monday, October 21, 2024 at 04:44:33 PM
CAS classification : system_of_ODEs

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

In canonical form a linear first order is

\begin{align*} \frac {d}{d t}x \left (t \right ) + q(t)x \left (t \right ) &= 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} + \textit {\_C} \end{align*}

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

\[ x \left (t \right ) = \textit {\_C} \,{\mathrm e}^{t}-3 t -2 \]

The system is

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

Since the left side is the same, this implies

\begin{align*} x \left (t \right )+y \left (t \right )+t&=2 x \left (t \right )+3 y \left (t \right )+{\mathrm e}^{t}\\ y \left (t \right )&=-\frac {x \left (t \right )}{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*} \frac {d}{d t}y \left (t \right )&=-\frac {\frac {d}{d t}x \left (t \right )}{2}-\frac {{\mathrm e}^{t}}{2}+\frac {1}{2}\tag {4} \end{align*}

Substituting (3,4) in (1) to eliminate \(y \left (t \right ),\frac {d}{d t}y \left (t \right )\) gives

\begin{align*} \frac {\frac {d}{d t}x \left (t \right )}{2}-\frac {{\mathrm e}^{t}}{2}+\frac {1}{2} &= \frac {x \left (t \right )}{2}-\frac {{\mathrm e}^{t}}{2}+\frac {3 t}{2}\\ \frac {d}{d t}x \left (t \right ) &= x \left (t \right )+3 t -1\tag {5} \end{align*}

Which is now solved for \(x \left (t \right )\). Given now that we have the solution

\begin{align*} x \left (t \right )&=\textit {\_C} \,{\mathrm e}^{t}-3 t -2 \tag {6} \end{align*}

Then substituting (6) into (3) gives

\begin{align*} y \left (t \right )&=-\frac {\textit {\_C} \,{\mathrm e}^{t}}{2}+2 t +1-\frac {{\mathrm e}^{t}}{2} \tag {7} \end{align*}

2.1.1 Maple step by step solution

2.1.2 Maple dsolve solution

Solving time : 0.059 (sec)
Leaf size : 30

dsolve([diff(x(t),t)+diff(y(t),t)-x(t) = y(t)+t, diff(x(t),t)+diff(y(t),t) = 2*x(t)+3*y(t)+exp(t)] 
       ,{op([x(t), y(t)])})
 

\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*}

2.1.3 Mathematica DSolve solution

Solving time : 0.026 (sec)
Leaf size : 37

DSolve[{{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]},{}}, 
       {x[t],y[t]},t,IncludeSingularSolutions->True]
 

\begin{align*} x(t)\to -3 t+(1+2 c_1) e^t-2 \\ y(t)\to 2 t-(1+c_1) e^t+1 \\ \end{align*}