Internal problem ID [11929]
Internal file name [OUTPUT/11939_Saturday_April_13_2024_10_26_38_PM_48022610/index.tex]
Book: Differential Equations by Shepley L. Ross. Third edition. John Willey. New Delhi.
2004.
Section: Chapter 7, Systems of linear differential equations. Section 7.1. Exercises page
277
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 )+3 y \left (t \right )+{\mathrm e}^{t}\\ x^{\prime }\left (t \right )+y^{\prime }\left (t \right )&=-x \left (t \right )+{\mathrm e}^{3 t} \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 )+{\mathrm e}^{t}\tag {1}\\ x^{\prime }\left (t \right )+y^{\prime }\left (t \right )&=-x \left (t \right )+{\mathrm e}^{3 t}\tag {2} \end {align*}
Since the left side is the same, this implies \begin {align*} x \left (t \right )+3 y \left (t \right )+{\mathrm e}^{t}&=-x \left (t \right )+{\mathrm e}^{3 t}\\ y \left (t \right )&=\frac {{\mathrm e}^{3 t}}{3}-\frac {2 x \left (t \right )}{3}-\frac {{\mathrm e}^{t}}{3}\tag {3} \end {align*}
Taking derivative of the above w.r.t. \(t\) gives \begin {align*} y^{\prime }\left (t \right )&={\mathrm e}^{3 t}-\frac {2 x^{\prime }\left (t \right )}{3}-\frac {{\mathrm e}^{t}}{3}\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 )}{3}+{\mathrm e}^{3 t}-\frac {{\mathrm e}^{t}}{3} &= -x \left (t \right )+{\mathrm e}^{3 t}\\ x^{\prime }\left (t \right ) &= -3 x \left (t \right )+{\mathrm e}^{t}\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) &=3\\ q(t) &={\mathrm e}^{t} \end {align*}
Hence the ode is \begin {align*} x^{\prime }\left (t \right )+3 x \left (t \right ) = {\mathrm e}^{t} \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 x\right ) &= \left (\mu \right ) \left ({\mathrm e}^{t}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}t}} \left ({\mathrm e}^{3 t} x\right ) &= \left ({\mathrm e}^{3 t}\right ) \left ({\mathrm e}^{t}\right )\\ \mathrm {d} \left ({\mathrm e}^{3 t} x\right ) &= {\mathrm e}^{4 t}\, \mathrm {d} t \end {align*}
Integrating gives \begin {align*} {\mathrm e}^{3 t} x &= \int {{\mathrm e}^{4 t}\,\mathrm {d} t}\\ {\mathrm e}^{3 t} x &= \frac {{\mathrm e}^{4 t}}{4} + c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{3 t}\) results in \begin {align*} x \left (t \right ) &= \frac {{\mathrm e}^{-3 t} {\mathrm e}^{4 t}}{4}+{\mathrm e}^{-3 t} c_{1} \end {align*}
which simplifies to \begin {align*} x \left (t \right ) &= \frac {\left ({\mathrm e}^{4 t}+4 c_{1} \right ) {\mathrm e}^{-3 t}}{4} \end {align*}
Given now that we have the solution \begin {align*} x \left (t \right )&=\frac {\left ({\mathrm e}^{4 t}+4 c_{1} \right ) {\mathrm e}^{-3 t}}{4} \tag {6} \end {align*}
Then substituting (6) into (3) gives \begin {align*} y \left (t \right )&=\frac {{\mathrm e}^{3 t}}{3}-\frac {{\mathrm e}^{-3 t} {\mathrm e}^{4 t}}{6}-\frac {2 \,{\mathrm e}^{-3 t} c_{1}}{3}-\frac {{\mathrm e}^{t}}{3} \tag {7} \end {align*}
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 37
dsolve([diff(x(t),t)+diff(y(t),t)-x(t)-3*y(t)=exp(t),diff(x(t),t)+diff(y(t),t)+x(t)=exp(3*t)],singsol=all)
\begin{align*} x \left (t \right ) &= \frac {{\mathrm e}^{t}}{4}+c_{1} {\mathrm e}^{-3 t} \\ y \left (t \right ) &= \frac {{\mathrm e}^{3 t}}{3}-\frac {{\mathrm e}^{t}}{2}-\frac {2 c_{1} {\mathrm e}^{-3 t}}{3} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.043 (sec). Leaf size: 55
DSolve[{x'[t]+y'[t]-x[t]-3*y[t]==Exp[t],x'[t]+y'[t]+x[t]==Exp[3*t]},{x[t],y[t]},t,IncludeSingularSolutions -> True]
\begin{align*} x(t)\to \frac {e^t}{4}+\frac {3}{16} c_1 e^{-3 t} \\ y(t)\to -\frac {e^t}{2}+\frac {e^{3 t}}{3}-\frac {1}{8} c_1 e^{-3 t} \\ \end{align*}