Consider the system of differential equations dxdt=−5ydydt=−5x. Convert this system to a second order differential equation in y by differentiating the second equation with respect to t and substituting for x from the first equation. Solve the equation you obtained for y as a function of t; hence find x as a function of t. If we also require x(0)=1 and y(0)=3, what are x and y? x(t)= equation editorEquation Editor y(t)= equation editorEquation Editor

[tex]\dfrac{\mathrm dx}{\mathrm dt}=-5y[/tex][tex]\dfrac{\mathrm dy}{\mathrm dt}=-5x\implies\dfrac{\mathrm d^2y}{\mathrm dt^2}=-5\dfrac{\mathrm dx}{\mathrm dt}[/tex][tex]\implies\dfrac{\mathrm d^2y}{\mathrm dt^2}-25y=0[/tex]This ODE is linear in [tex]y(t)[/tex] with the characteristic equation and roots[tex]r^2-25=0\implies r=\pm5[/tex]so that[tex]y(t)=C_1e^{5t}+C_2e^{-5t}[/tex]Then[tex]\dfrac{\mathrm dx}{\mathrm dt}=-5C_1e^{5t}-5C_2e^{-5t}[/tex][tex]\implies x(t)=-C_1e^{5t}+C_2e^{-5t}[/tex]Given that [tex]x(0)=1[/tex] and [tex]y(0)=3[/tex], we find[tex]\begin{cases}1=-C_1+C_2\\3=C_1+C_2\end{cases}\implies C_1=1,C_2=2[/tex]and the particular solution to this system is[tex]\begin{cases}x(t)=-e^{5t}+2e^{-5t}\\y(t)=e^{5t}+2e^{-5t}\end{cases}[/tex]