Consider the system of differential equations dxdt=βˆ’4ydydt=βˆ’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?

Accepted Solution

[tex]\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}[/tex]Substituting this into the other ODE gives[tex]-\dfrac14\dfrac{\mathrm d^2y}{\mathrm dt^2}=-4y\implies y''-16y=0[/tex]Since [tex]x(t)=-\dfrac14y'(t)[/tex], it follows that [tex]x(0)=-\dfrac14y'(0)=4\implies y'(0)=-16[/tex]. The ODE in [tex]y[/tex] has characteristic equation[tex]r^2-16=0[/tex]with roots [tex]r=\pm4[/tex], admitting the characteristic solution[tex]y_c=C_1e^{4t}+C_2e^{-4t}[/tex]From the initial conditions we get[tex]y(0)=5\implies 5=C_1+C_2[/tex][tex]y'(0)=16\implies-16=4C_1-4C_2[/tex][tex]\implies C_1=\dfrac12,C_2=\dfrac92[/tex]So we have[tex]\boxed{y(t)=\dfrac12e^{4t}+\dfrac92e^{-4t}}[/tex]Take the derivative and multiply it by -1/4 to get the solution for [tex]x(t)[/tex]:[tex]-\dfrac14y'(t)=\boxed{x(t)=-\dfrac12e^{4t}+\dfrac92e^{-4t}}[/tex]