Solve the simultaneous equations

\[ \begin{align} \frac{dx}{dt} & = 3x+5y \quad \quad \enclose{circle}{\color{black}{1}}\\ \frac{dy}{dt} & = x-y \quad \quad \enclose{circle}{\color{black}{2}} \end{align} \]

given that the initial conditions are \( x=7 \) and \( y= -1\) , where \(t=0\)

\[ \begin{align} & Rearranging \quad \enclose{circle}{\color{black}{2}} \quad gives \\ x & = \frac{dy}{dt}+y \quad \quad \quad \quad \enclose{circle}{\color{black}{3}}\\ & Then\ differentiating\ \quad \enclose{circle}{\color{black}{3}} \quad with\, respect\, to \quad t \quad gives\\ \frac{dx}{dt} & = \frac{d^2y}{dt^2}+ \frac{dy}{dt} \quad \quad \quad \quad \enclose{circle}{\color{black}{4}} \\ & Substituting \quad \enclose{circle}{\color{black}{3}} \quad and \quad \enclose{circle}{\color{black}{4}} \quad into \quad \enclose{circle}{\color{black}{1}} \quad gives \\ \frac{d^2y}{dt^2}+\frac{dy}{dt} & = 3( \frac{dy}{dt}+y) + 5y \\ 0 & =\frac{d^2y}{dt}-2\frac{dy}{dt}-8y \\ \\ & Which\ yields\ the\ auxiliary\ equation \\ 0 & = \lambda^2 - 2 \lambda - 8 \\ 0 & = ( \lambda-4)( \lambda +2) \\ \lambda & = 4 \quad or \quad \lambda= -2 \\ \\ & So\ the\ general\ solution\ for\ y\ is \\ y & = Ae^{4t} + Be^{-2t} \\ & and\ subsituting\ into \quad \enclose{circle}{\color{black}{3}} \quad gives\ the\ general\ solution\ for\ x \\ x & = \frac{dy}{dt}+y \\ & = (4Ae^{4t}-2Be^{-2t})+(Ae^{4t}+Be^{-2t}) \\ & = 5Ae^{4t} - Be^{-2t} \\ \\ & When\ t=0 \,, x=7 \implies 7= 5A - B \\ & When\ t=0 \,, y=-1 \implies -1= A + B \\ \\ & \therefore \quad A = 1 \quad and \quad B = -2 \\ \\ & So\ the\ particular\ solution\ is \\ x & = 5e^{4t}+2e^{-2t} \\ y & = e^{4t}-2e^{-2t} \\ \end{align} \]