For the linear DE y'-(1/2)y=2cost the solution y(t)=(-4/5)[cost-2sint]+ce^(t/2). for y(0)=a this is the equilibrium solution which I'm guessing separates solutions of different behavior? y(0)=a=4/5. level curves are m=(1/2)y+2cost, y=2m-4cost

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- September 11th 2013, 04:48 AMmathlover10equilibrium equation?
For the linear DE y'-(1/2)y=2cost the solution y(t)=(-4/5)[cost-2sint]+ce^(t/2). for y(0)=a this is the equilibrium solution which I'm guessing separates solutions of different behavior? y(0)=a=4/5. level curves are m=(1/2)y+2cost, y=2m-4cost

- September 12th 2013, 02:56 AMchiroRe: equilibrium equation?
Hey mathlover10.

Equilibrium solutions occur when the first derivative is 0. is then The idea is when this is the case, then the behavior doesn't change and remains constant. If your solution is as it is then I don't think you will ever have a true equilibrium solution since y' does not only depend on y but it also depends on t.