Question

Consider the system of differential equations dxdt=−4ydydt=−4x. 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)=4 and y(0)=5, what are x and y?

Answer 1

$\dfrac{\mathrm dy}{\mathrm dt}=-4x\implies x=-\dfrac14\dfrac{\mathrm dy}{\mathrm dt}\implies\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac14\dfrac{\mathrm d^2y}{\mathrm dt^2}$

Substituting this into the other ODE gives

$-\dfrac14\dfrac{\mathrm d^2y}{\mathrm dt^2}=-4y\implies y''-16y=0$

Since $x(t)=-\dfrac14y'(t)$, it follows that $x(0)=-\dfrac14y'(0)=4\implies y'(0)=-16$. The ODE in $y$ has characteristic equation

$r^2-16=0$

with roots $r=\pm4$, admitting the characteristic solution

$y_c=C_1e^{4t}+C_2e^{-4t}$

From the initial conditions we get

$y(0)=5\implies 5=C_1+C_2$

$y'(0)=16\implies-16=4C_1-4C_2$

$\implies C_1=\dfrac12,C_2=\dfrac92$

So we have

$\boxed{y(t)=\dfrac12e^{4t}+\dfrac92e^{-4t}}$

Take the derivative and multiply it by -1/4 to get the solution for $x(t)$:

$-\dfrac14y'(t)=\boxed{x(t)=-\dfrac12e^{4t}+\dfrac92e^{-4t}}$