Find the unique periodic solution of this equation whose period is $2\pi$.

214 Views Asked by At

How to find the unique periodic solution of the equation that period is $2\pi$? The equation: $$y''+2ay'+b^2y = \cos (x), \quad 0 \le a < b < 1$$

My attempt:

I've tried to solve it as non-homogeneous equation by the undetermined coefficients method but it didn't work out due to the undefined coefficients.

1

There are 1 best solutions below

0
On BEST ANSWER

We are looking for a solution of the form $$ y=c_1\cos x+c_2\sin x. $$ Then $$ y''+2ay'+b^2y=\big((b^2-1)c_1+2ac_2\big)\cos x+\big((b^2-1)c_2-2ac_1\big)\sin x=\cos x $$ Hence $c_1,c_2$ should satisfy the system $$ (b^2-1)c_1+2ac_2=1 \\ (b^2-1)c_2-2ac_1=0 $$