Internal problem ID [11928]
Internal file name [OUTPUT/11938_Saturday_April_13_2024_10_26_37_PM_14908028/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: 2.
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 )-2 t\\ x^{\prime }\left (t \right )+y^{\prime }\left (t \right )&=t^{2}+3 x \left (t \right )+y \left (t \right ) \end {align*}
The system is \begin {align*} x^{\prime }\left (t \right )+y^{\prime }\left (t \right )&=x \left (t \right )-2 t\tag {1}\\ x^{\prime }\left (t \right )+y^{\prime }\left (t \right )&=t^{2}+3 x \left (t \right )+y \left (t \right )\tag {2} \end {align*}
Since the left side is the same, this implies \begin {align*} x \left (t \right )-2 t&=t^{2}+3 x \left (t \right )+y \left (t \right )\\ y \left (t \right )&=-t^{2}-2 x \left (t \right )-2 t\tag {3} \end {align*}
Taking derivative of the above w.r.t. \(t\) gives \begin {align*} y^{\prime }\left (t \right )&=-2 t -2 x^{\prime }\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*} -x^{\prime }\left (t \right )-2 t -2 &= x \left (t \right )-2 t\\ x^{\prime }\left (t \right ) &= -x \left (t \right )-2\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}\\ -\ln \left (x +2\right )&=t +c_{1} \end {align*}
Solving for \(x\) gives these solutions \begin {align*} x_1&={\mathrm e}^{-c_{1} -t}-2\\ &=\frac {{\mathrm e}^{-t}}{c_{1}}-2 \end {align*}
Given now that we have the solution \begin {align*} x \left (t \right )&=\frac {{\mathrm e}^{-t}}{c_{1}}-2 \tag {6} \end {align*}
Then substituting (6) into (3) gives \begin {align*} y \left (t \right )&=-\frac {\left (t^{2} c_{1} {\mathrm e}^{t}+2 t c_{1} {\mathrm e}^{t}-4 \,{\mathrm e}^{t} c_{1} +2\right ) {\mathrm e}^{-t}}{c_{1}} \tag {7} \end {align*}
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 33
dsolve([diff(x(t),t)+diff(y(t),t)-x(t)=-2*t,diff(x(t),t)+diff(y(t),t)-3*x(t)-y(t)=t^2],singsol=all)
\begin{align*} x \left (t \right ) &= -2+{\mathrm e}^{-t} c_{1} \\ y \left (t \right ) &= -t^{2}+4-2 \,{\mathrm e}^{-t} c_{1} -2 t \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.056 (sec). Leaf size: 42
DSolve[{x'[t]+y'[t]-x[t]==-2*t,x'[t]+y'[t]-3*x[t]-y[t]==t^2},{x[t],y[t]},t,IncludeSingularSolutions -> True]
\begin{align*} x(t)\to -2-\frac {1}{4} c_1 e^{-t} \\ y(t)\to -t^2-2 t+\frac {c_1 e^{-t}}{2}+4 \\ \end{align*}