Solve simultaneous equation dx/dt=3x+5y , dy/dt=-x-y , x=5 , y=-3...

$\mathrm{We} \mathrm{need} \mathrm{to} \mathrm{solve} \mathrm{the} \mathrm{given} \mathrm{simultaneous} \mathrm{equations},$

$$\begin{gathered} \frac{dx}{dt}=3x+5y \\ \frac{dy}{dt}=-x-y \end{gathered}$$

$\mathrm{at} x=5, y=-3$

$\mathrm{Let}, \frac{d}{dt}=D$

$\mathrm{Thus} \mathrm{the} \mathrm{equation} \mathrm{transforms} \mathrm{to}:$

$$\begin{gathered} \frac{dx}{dt}=3x+5y \\ \Rightarrow Dx=3x+5y \\ \Rightarrow Dx-3x-5y=0 \\ \Rightarrow \left(D-3\right)x-5y=0 ---\left(1\right) \end{gathered}$$

$$\begin{gathered} \frac{dy}{dt}=-x-y \\ \Rightarrow Dy=-x-y \\ \Rightarrow x+Dy+y=0 \\ \Rightarrow x+\left(D+1\right)y=0 ---\left(2\right) \end{gathered}$$

$\mathrm{Solving} \mathrm{equation} \left(1\right), \mathrm{we} \mathrm{get}$

$$\begin{gathered} \Rightarrow \left(D-3\right)x-5y=0 \\ \Rightarrow x=\frac{5y}{D-3} \end{gathered}$$

$\mathrm{By} \mathrm{substituting} \mathrm{the} \mathrm{value} \mathrm{of} x \mathrm{in} \mathrm{equation} \left(2\right), \mathrm{we} \mathrm{get}$

$$\begin{gathered} x+\left(D+1\right)y=0 \\ \Rightarrow \frac{5y}{\left(D-3\right)}+\left(D+1\right)y=0 \\ \Rightarrow 5y+\left(D-3\right)\left(D+1\right)y=0 \\ \Rightarrow 5y+\left(D^2-2D-3\right)y=0 \\ \Rightarrow \left(D^2-2D-3+5\right)y=0 \\ \Rightarrow \left(D^2-2D+2\right)y=0 \\ \Rightarrow y\text{'}\text{'}\left(t\right)-2y\text{'}\left(t\right)+2y\left(t\right)=0 \end{gathered}$$

$$\begin{gathered} \mathrm{Auxilliary} \mathrm{equation} \mathrm{for} \mathrm{above} \mathrm{differential} \mathrm{equation} \mathrm{is}: \\ {m}^2-2m+2=0 \\ \Rightarrow m=\frac{-\left(-2\right)\pm \sqrt{{\left(-2\right)}^2-4\left(1\right)\left(2\right)}}{2\left(1\right)}=\frac{2\pm \sqrt{-4}}{2} \\ \Rightarrow m=1\pm i \end{gathered}$$

$\Rightarrow \mathit{y}\left(t\right)\mathbf{=}\left(C_{1}sin\left(t\right)+C_{2}cos\left(t\right)\right){\mathit{e}}^{\mathbf{t}}$

$\mathrm{Using} \mathrm{equation} \left(2\right), \mathrm{we} \mathrm{can} \mathrm{write},$

$x=-\frac{dy}{dt}-y ---\left(3\right)$

$\mathrm{We} \mathrm{have},$

$\Rightarrow y\left(t\right)=\left(C_1\sin \left(t\right)+C_2\cos \left(t\right)\right)e^t$

$$\begin{gathered} \Rightarrow \frac{dy}{dt}=\frac{d}{dt}\left[\left(C_1\sin \left(t\right)+C_2\cos \left(t\right)\right)e^t\right] \\ \Rightarrow \frac{dy}{dt}=e^t\frac{d}{dt}\left[\left(C_1\sin \left(t\right)+C_2\cos \left(t\right)\right)\right]+\left(C_1\sin \left(t\right)+C_2\cos \left(t\right)\right)\frac{d}{dt}\left[e^t\right] \\ \Rightarrow \frac{dy}{dt}=\left(C_1\cos \left(t\right)-C_2\sin \left(t\right)\right)e^t+\left(C_1\sin \left(t\right)+C_2\cos \left(t\right)\right)e^t \\ \Rightarrow \frac{dy}{dt}=\left(C_1\left(\cos \left(t\right)+\sin \left(t\right)\right)+C_2\left(\cos \left(t\right)-\sin \left(t\right)\right)\right)e^t \end{gathered}$$

$\mathrm{substituting} \mathrm{the} \mathrm{value} \mathrm{of} y \mathrm{and} \frac{dy}{dt} \mathrm{in} \mathrm{equation} \left(3\right), \mathrm{we} \mathrm{get}$

$$\begin{gathered} \Rightarrow x=-\left(C_1\left(\cos \left(t\right)+\sin \left(t\right)\right)+C_2\left(\cos \left(t\right)-\sin \left(t\right)\right)\right)e^t-\left(C_1\sin \left(t\right)+C_2\cos \left(t\right)\right)e^t \\ \Rightarrow \mathit{x}\left(t\right)\mathbf{=}\mathbf{-}\left(C_1\left(cos\left(t\right)+2sin\left(t\right)\right)+C_2\left(2cos\left(t\right)-sin\left(t\right)\right)\right){\mathit{e}}^{\mathbf{t}} \end{gathered}$$

$$\begin{gathered} \mathrm{We} \mathrm{have} \mathrm{obtained} x\left(t\right) \mathrm{and} y\left(t\right). \\ \mathbf{NOTE}\mathbf{:}\mathbf{ }\mathrm{We} \mathrm{are} \mathrm{given} \mathrm{the} \mathrm{initial} \mathrm{condition} \mathrm{as} x=5,y=-3. \mathrm{But} \mathrm{we} \mathrm{would} \mathrm{need} \mathrm{these} x and y \mathrm{value} \mathrm{at} \mathrm{a} \mathrm{particular} \mathrm{value} \\ \mathrm{of} t \mathrm{in} \mathrm{order} \mathrm{to} \mathrm{find} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{constants} C_1 \mathrm{and} C_2. \end{gathered}$$