Question

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

Answer 1

$\dfrac{\mathrm dx}{\mathrm dt}=-5y$

$\dfrac{\mathrm dy}{\mathrm dt}=-5x\implies\dfrac{\mathrm d^2y}{\mathrm dt^2}=-5\dfrac{\mathrm dx}{\mathrm dt}$

$\implies\dfrac{\mathrm d^2y}{\mathrm dt^2}-25y=0$

This ODE is linear in $y(t)$ with the characteristic equation and roots

$r^2-25=0\implies r=\pm5$

so that

$y(t)=C_1e^{5t}+C_2e^{-5t}$

Then

$\dfrac{\mathrm dx}{\mathrm dt}=-5C_1e^{5t}-5C_2e^{-5t}$

$\implies x(t)=-C_1e^{5t}+C_2e^{-5t}$

Given that $x(0)=1$ and $y(0)=3$, we find

$\begin{cases}1=-C_1+C_2\\3=C_1+C_2\end{cases}\implies C_1=1,C_2=2$

and the particular solution to this system is

$\begin{cases}x(t)=-e^{5t}+2e^{-5t}\\y(t)=e^{5t}+2e^{-5t}\end{cases}$