Internal problem ID [5447]
Internal file name [OUTPUT/4938_Tuesday_February_06_2024_10_14_29_PM_76667389/index.tex]
Book: Schaums Outline. Theory and problems of Differential Equations, 1st edition. Frank Ayres.
McGraw Hill 1952
Section: Chapter 21. System of simultaneous linear equations. Supplemetary problems. Page
163
Problem number: 13.
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 )-3 y \left (t \right )-1+{\mathrm e}^{-t}\\ x^{\prime }\left (t \right )+y^{\prime }\left (t \right )&=-2 x \left (t \right )-3 y \left (t \right )+{\mathrm e}^{2 t}+1 \end {align*}
The system is \begin {align*} x^{\prime }\left (t \right )+y^{\prime }\left (t \right )&=x \left (t \right )-3 y \left (t \right )-1+{\mathrm e}^{-t}\tag {1}\\ x^{\prime }\left (t \right )+y^{\prime }\left (t \right )&=-2 x \left (t \right )-3 y \left (t \right )+{\mathrm e}^{2 t}+1\tag {2} \end {align*}
Since the left side is the same, this implies \begin {align*} x \left (t \right )-3 y \left (t \right )-1+{\mathrm e}^{-t}&=-2 x \left (t \right )-3 y \left (t \right )+{\mathrm e}^{2 t}+1\\ x \left (t \right )&=\frac {\left ({\mathrm e}^{3 t}+2 \,{\mathrm e}^{t}-1\right ) {\mathrm e}^{-t}}{3}\tag {3} \end {align*}
Taking derivative of the above w.r.t. \(t\) gives \begin {align*} x^{\prime }\left (t \right )&=\frac {\left (3 \,{\mathrm e}^{3 t}+2 \,{\mathrm e}^{t}\right ) {\mathrm e}^{-t}}{3}-\frac {\left ({\mathrm e}^{3 t}+2 \,{\mathrm e}^{t}-1\right ) {\mathrm e}^{-t}}{3}\tag {4} \end {align*}
Substituting (3,4) in (1) to eliminate \(x \left (t \right ),x^{\prime }\left (t \right )\) gives \begin {align*} \frac {\left (3 \,{\mathrm e}^{3 t}+2 \,{\mathrm e}^{t}\right ) {\mathrm e}^{-t}}{3}-\frac {\left ({\mathrm e}^{3 t}+2 \,{\mathrm e}^{t}-1\right ) {\mathrm e}^{-t}}{3}+y^{\prime }\left (t \right ) &= \frac {\left ({\mathrm e}^{3 t}+2 \,{\mathrm e}^{t}-1\right ) {\mathrm e}^{-t}}{3}-3 y \left (t \right )-1+{\mathrm e}^{-t}\\ x^{\prime }\left (t \right ) &= -\frac {\left ({\mathrm e}^{3 t}+9 \,{\mathrm e}^{t} y \left (t \right )+{\mathrm e}^{t}-1\right ) {\mathrm e}^{-t}}{3}\tag {5} \end {align*}
Which is now solved for \(y \left (t \right )\).
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime }\left (t \right ) + p(t)y \left (t \right ) &= q(t) \end {align*}
Where here \begin {align*} p(t) &=3\\ q(t) &=-\frac {\left ({\mathrm e}^{3 t}+{\mathrm e}^{t}-1\right ) {\mathrm e}^{-t}}{3} \end {align*}
Hence the ode is \begin {align*} y^{\prime }\left (t \right )+3 y \left (t \right ) = -\frac {\left ({\mathrm e}^{3 t}+{\mathrm e}^{t}-1\right ) {\mathrm e}^{-t}}{3} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int 3d t} \\ &= {\mathrm e}^{3 t} \\ \end{align*} The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}}\left ( \mu y\right ) &= \left (\mu \right ) \left (-\frac {\left ({\mathrm e}^{3 t}+{\mathrm e}^{t}-1\right ) {\mathrm e}^{-t}}{3}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \left ({\mathrm e}^{3 t} y\right ) &= \left ({\mathrm e}^{3 t}\right ) \left (-\frac {\left ({\mathrm e}^{3 t}+{\mathrm e}^{t}-1\right ) {\mathrm e}^{-t}}{3}\right )\\ \mathrm {d} \left ({\mathrm e}^{3 t} y\right ) &= \left (-\frac {{\mathrm e}^{2 t} \left ({\mathrm e}^{3 t}+{\mathrm e}^{t}-1\right )}{3}\right )\, \mathrm {d} t \end {align*}
Integrating gives \begin {align*} {\mathrm e}^{3 t} y &= \int {-\frac {{\mathrm e}^{2 t} \left ({\mathrm e}^{3 t}+{\mathrm e}^{t}-1\right )}{3}\,\mathrm {d} t}\\ {\mathrm e}^{3 t} y &= -\frac {{\mathrm e}^{3 t}}{9}-\frac {{\mathrm e}^{5 t}}{15}+\frac {{\mathrm e}^{2 t}}{6} + c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{3 t}\) results in \begin {align*} y \left (t \right ) &= {\mathrm e}^{-3 t} \left (-\frac {{\mathrm e}^{3 t}}{9}-\frac {{\mathrm e}^{5 t}}{15}+\frac {{\mathrm e}^{2 t}}{6}\right )+c_{1} {\mathrm e}^{-3 t} \end {align*}
which simplifies to \begin {align*} y \left (t \right ) &= \frac {\left (-6 \,{\mathrm e}^{5 t}-10 \,{\mathrm e}^{3 t}+15 \,{\mathrm e}^{2 t}+90 c_{1} \right ) {\mathrm e}^{-3 t}}{90} \end {align*}
Given now that we have the solution \begin {align*} y \left (t \right )&=\frac {\left (-6 \,{\mathrm e}^{5 t}-10 \,{\mathrm e}^{3 t}+15 \,{\mathrm e}^{2 t}+90 c_{1} \right ) {\mathrm e}^{-3 t}}{90} \tag {6} \end {align*}
Then substituting (6) into (3) gives \begin {align*} x \left (t \right )&=\frac {\left ({\mathrm e}^{3 t}+2 \,{\mathrm e}^{t}-1\right ) {\mathrm e}^{-t}}{3} \tag {7} \end {align*}
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 42
dsolve([diff(x(t),t)-x(t)+diff(y(t),t)+3*y(t)=exp(-t)-1,diff(x(t),t)+2*x(t)+diff(y(t),t)+3*y(t)=exp(2*t)+1],singsol=all)
\begin{align*} x \left (t \right ) &= \frac {{\mathrm e}^{2 t}}{3}+\frac {2}{3}-\frac {{\mathrm e}^{-t}}{3} \\ y \left (t \right ) &= -\frac {1}{9}-\frac {{\mathrm e}^{2 t}}{15}+\frac {{\mathrm e}^{-t}}{6}+c_{1} {\mathrm e}^{-3 t} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.059 (sec). Leaf size: 62
DSolve[{x'[t]-x[t]+y'[t]+3*y[t]==Exp[-t]-1,x'[t]+2*x[t]+y'[t]+3*y[t]==Exp[2*t]+1},{x[t],y[t]},t,IncludeSingularSolutions -> True]
\begin{align*} x(t)\to \frac {1}{3} e^{-t} \left (2 e^t+e^{3 t}-1\right ) \\ y(t)\to \frac {e^{-t}}{6}-\frac {e^{2 t}}{15}+\frac {1}{16} c_1 e^{-3 t}-\frac {1}{9} \\ \end{align*}