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
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
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.