Internal problem ID [7389]
Internal file name [OUTPUT/6635_Monday_November_27_2023_11_02_14_PM_64086885/index.tex]
Book: First order enumerated odes
Section: section 2 (system of first order ode’s)
Problem number: 3.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "system of linear ODEs"
Solve \begin {align*} x^{\prime }\left (t \right )+y^{\prime }\left (t \right )&=x \left (t \right )+y \left (t \right )+t +\sin \left (t \right )+\cos \left (t \right )\\ x^{\prime }\left (t \right )+y^{\prime }\left (t \right )&=2 x \left (t \right )+3 y \left (t \right )+{\mathrm e}^{t} \end {align*}
The system is \begin {align*} x^{\prime }\left (t \right )+y^{\prime }\left (t \right )&=x \left (t \right )+y \left (t \right )+t +\sin \left (t \right )+\cos \left (t \right )\tag {1}\\ x^{\prime }\left (t \right )+y^{\prime }\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*} y^{\prime }\left (t \right )&=-\frac {x^{\prime }\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 \(y \left (t \right ),y^{\prime }\left (t \right )\) gives \begin {align*} \frac {x^{\prime }\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}\\ x^{\prime }\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 \(x \left (t \right )\).
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} x^{\prime }\left (t \right ) + p(t)x \left (t \right ) &= q(t) \end {align*}
Where here \begin {align*} p(t) &=-1\\ q(t) &=3 t +4 \sin \left (t \right )+2 \cos \left (t \right )-1 \end {align*}
Hence the ode is \begin {align*} x^{\prime }\left (t \right )-x \left (t \right ) = 3 t +4 \sin \left (t \right )+2 \cos \left (t \right )-1 \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\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 ) &= \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 ({\mathrm e}^{-t} x\right ) &= \left ({\mathrm e}^{-t}\right ) \left (3 t +4 \sin \left (t \right )+2 \cos \left (t \right )-1\right )\\ \mathrm {d} \left ({\mathrm e}^{-t} x\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*} {\mathrm e}^{-t} x &= \int {\left (3 t +4 \sin \left (t \right )+2 \cos \left (t \right )-1\right ) {\mathrm e}^{-t}\,\mathrm {d} t}\\ {\mathrm e}^{-t} x &= -3 \,{\mathrm e}^{-t} t -2 \,{\mathrm e}^{-t}-3 \,{\mathrm e}^{-t} \cos \left (t \right )-{\mathrm e}^{-t} \sin \left (t \right ) + c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{-t}\) results in \begin {align*} x \left (t \right ) &= {\mathrm e}^{t} \left (-3 \,{\mathrm e}^{-t} t -2 \,{\mathrm e}^{-t}-3 \,{\mathrm e}^{-t} \cos \left (t \right )-{\mathrm e}^{-t} \sin \left (t \right )\right )+c_{1} {\mathrm e}^{t} \end {align*}
which simplifies to \begin {align*} x \left (t \right ) &= -2+c_{1} {\mathrm e}^{t}-3 t -\sin \left (t \right )-3 \cos \left (t \right ) \end {align*}
Given now that we have the solution \begin {align*} x \left (t \right )&=-2+c_{1} {\mathrm e}^{t}-3 t -\sin \left (t \right )-3 \cos \left (t \right ) \tag {6} \end {align*}
Then substituting (6) into (3) gives \begin {align*} y \left (t \right )&=1-\frac {c_{1} {\mathrm e}^{t}}{2}+2 t +\sin \left (t \right )+2 \cos \left (t \right )-\frac {{\mathrm e}^{t}}{2} \tag {7} \end {align*}
✓ Solution by Maple
Time used: 0.078 (sec). Leaf size: 45
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)],singsol=all)
\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*}
✓ Solution by Mathematica
Time used: 0.055 (sec). Leaf size: 54
DSolve[{x'[t]+y'[t]-x[t]==y[t]+t+Sin[t]+Cos[t],x'[t]+y'[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*}