Internal problem ID [11927]
Internal file name [OUTPUT/11937_Saturday_April_13_2024_10_26_37_PM_64304397/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: 1.
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 )&=2 x \left (t \right )+4 y \left (t \right )+{\mathrm e}^{t}\\ x^{\prime }\left (t \right )+y^{\prime }\left (t \right )&=y \left (t \right )+{\mathrm e}^{4 t} \end {align*}
The system is \begin {align*} x^{\prime }\left (t \right )+y^{\prime }\left (t \right )&=2 x \left (t \right )+4 y \left (t \right )+{\mathrm e}^{t}\tag {1}\\ x^{\prime }\left (t \right )+y^{\prime }\left (t \right )&=y \left (t \right )+{\mathrm e}^{4 t}\tag {2} \end {align*}
Since the left side is the same, this implies \begin {align*} 2 x \left (t \right )+4 y \left (t \right )+{\mathrm e}^{t}&=y \left (t \right )+{\mathrm e}^{4 t}\\ y \left (t \right )&=\frac {{\mathrm e}^{4 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 )&=\frac {4 \,{\mathrm e}^{4 t}}{3}-\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}+\frac {4 \,{\mathrm e}^{4 t}}{3}-\frac {{\mathrm e}^{t}}{3} &= -\frac {2 x \left (t \right )}{3}+\frac {4 \,{\mathrm e}^{4 t}}{3}-\frac {{\mathrm e}^{t}}{3}\\ x^{\prime }\left (t \right ) &= -2 x \left (t \right )\tag {5} \end {align*}
Which is now solved for \(x \left (t \right )\). Integrating both sides gives \begin {align*} \int -\frac {1}{2 x}d x &= t +c_{1}\\ -\frac {\ln \left (x \right )}{2}&=t +c_{1} \end {align*}
Solving for \(x\) gives these solutions \begin {align*} x_1&={\mathrm e}^{-2 t -2 c_{1}}\\ &=\frac {{\mathrm e}^{-2 t}}{c_{1}^{2}} \end {align*}
Given now that we have the solution \begin {align*} x \left (t \right )&=\frac {{\mathrm e}^{-2 t}}{c_{1}^{2}} \tag {6} \end {align*}
Then substituting (6) into (3) gives \begin {align*} y \left (t \right )&=\frac {\left ({\mathrm e}^{6 t} c_{1}^{2}-{\mathrm e}^{3 t} c_{1}^{2}-2\right ) {\mathrm e}^{-2 t}}{3 c_{1}^{2}} \tag {7} \end {align*}
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 32
dsolve([diff(x(t),t)+diff(y(t),t)-2*x(t)-4*y(t)=exp(t),diff(x(t),t)+diff(y(t),t)-y(t)=exp(4*t)],singsol=all)
\begin{align*} x \left (t \right ) &= c_{1} {\mathrm e}^{-2 t} \\ y \left (t \right ) &= \frac {{\mathrm e}^{4 t}}{3}-\frac {{\mathrm e}^{t}}{3}-\frac {2 c_{1} {\mathrm e}^{-2 t}}{3} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.055 (sec). Leaf size: 52
DSolve[{x'[t]+y'[t]-2*x[t]-4*y[t]==Exp[t],x'[t]+y'[t]-y[t]==Exp[4*t]},{x[t],y[t]},t,IncludeSingularSolutions -> True]
\begin{align*} x(t)\to \frac {1}{12} (3+4 c_1) e^{-2 t} \\ y(t)\to \frac {1}{18} e^{-2 t} \left (-6 e^{3 t}+6 e^{6 t}-3-4 c_1\right ) \\ \end{align*}