2.2.3 problem 3
Internal
problem
ID
[8483]
Book
:
First
order
enumerated
odes
Section
:
section
2
(system
of
first
order
odes)
Problem
number
:
3
Date
solved
:
Sunday, November 10, 2024 at 03:54:54 AM
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 +\sin \left (t \right )+\cos \left (t \right )\\ \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 +4 \sin \left (t \right )+2 \cos \left (t \right )-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 +4 \sin \left (t \right )+2 \cos \left (t \right )-1\right ) \\
\frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \left (x \,{\mathrm e}^{-t}\right ) &= \left ({\mathrm e}^{-t}\right ) \left (3 t +4 \sin \left (t \right )+2 \cos \left (t \right )-1\right ) \\
\mathrm {d} \left (x \,{\mathrm e}^{-t}\right ) &= \left (\left (3 t +4 \sin \left (t \right )+2 \cos \left (t \right )-1\right ) {\mathrm e}^{-t}\right )\, \mathrm {d} t \\
\end{align*}
Integrating gives
\begin{align*} x \,{\mathrm e}^{-t}&= \int {\left (3 t +4 \sin \left (t \right )+2 \cos \left (t \right )-1\right ) {\mathrm e}^{-t} \,dt} \\ &=-3 \,{\mathrm e}^{-t} t -2 \,{\mathrm e}^{-t}-3 \,{\mathrm e}^{-t} \cos \left (t \right )-{\mathrm e}^{-t} \sin \left (t \right ) + \textit {\_C} \end{align*}
Dividing throughout by the integrating factor \({\mathrm e}^{-t}\) gives the final solution
\[ x \left (t \right ) = \textit {\_C} \,{\mathrm e}^{t}-\sin \left (t \right )-3 \cos \left (t \right )-3 t -2 \]
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 +\sin \left (t \right )+\cos \left (t \right )\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 +\sin \left (t \right )+\cos \left (t \right )&=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}+\frac {\sin \left (t \right )}{2}+\frac {\cos \left (t \right )}{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}+\frac {\cos \left (t \right )}{2}-\frac {\sin \left (t \right )}{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 {\cos \left (t \right )}{2}-\frac {\sin \left (t \right )}{2} &= \frac {x \left (t \right )}{2}-\frac {{\mathrm e}^{t}}{2}+\frac {3 t}{2}+\frac {3 \sin \left (t \right )}{2}+\frac {3 \cos \left (t \right )}{2}\\ \frac {d}{d t}x \left (t \right ) &= x \left (t \right )+3 t +4 \sin \left (t \right )+2 \cos \left (t \right )-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}-\sin \left (t \right )-3 \cos \left (t \right )-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}+\sin \left (t \right )+2 \cos \left (t \right )+2 t +1-\frac {{\mathrm e}^{t}}{2} \tag {7} \end{align*}
Maple step by step solution
Maple dsolve solution
Solving time : 0.198
(sec)
Leaf size : 44
dsolve([diff(x(t),t)+diff(y(t),t)-x(t) = y(t)+t+sin(t)+cos(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 ) &= -\sin \left (t \right )-3 \cos \left (t \right )+c_{1} {\mathrm e}^{t}-3 t -2 \\
y \left (t \right ) &= \sin \left (t \right )+2 \cos \left (t \right )-\frac {c_{1} {\mathrm e}^{t}}{2}+2 t +1-\frac {{\mathrm e}^{t}}{2} \\
\end{align*}
Mathematica DSolve solution
Solving time : 0.038
(sec)
Leaf size : 54
DSolve[{{D[x[t],t]+D[y[t],t]-x[t]==y[t]+t+Sin[t]+Cos[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+e^t-\sin (t)-3 \cos (t)+2 c_1 e^t-2 \\
y(t)\to 2 t-e^t+\sin (t)+2 \cos (t)-c_1 e^t+1 \\
\end{align*}