Abstract
I was given this question in class, and I am quite confused whether the proper solution for it even exists. The content of the task is as follows:
Given differential equation $y''+2ay'+b^{2}y = \cos(x)$ $[*]$ with $0 \leqslant a < b < 1$. Find the unique periodic solution of this equation, whose period is $2\pi$. For what value of $a$ the amplitude of this solution takes the greatest value?
I have found a similar question on MathExchange here, but it didn't seem to be helpful, as I have already found the general solution for given $a$ and $b$.
General solution for $a$ and $b$
By given $0 \leqslant a < b < 1$, thus, for homogeneous case of $[*]$ characteristic polynomial, $\lambda^{2}+2a\lambda+b^{2}=0, D_{\lambda} = 4a^{2} – 4b^{2}<0$, hence $$y_{h} = C_{1}e^{x(-a+i\sqrt{b^{2}-a^{2}})} + C_{2}e^{x(-a-i\sqrt{b^{2}-a^{2}})}.$$
Thus: (partial solution derivation is omitted)$$y(x) = C_{1}e^{x(-a+i\sqrt{b^{2}-a^{2}})} + C_{2}e^{x(-a-i\sqrt{b^{2}-a^{2}})} +\small\frac{(b^{2}-1)\cos(x)+2a\sin(x)}{4a^{2}+b^{4}-2b^{2}+1}.$$
But then, $y(x)$ is $2\pi$-periodic only if either $(a,b) = (0,0)$ or $(0,1)$, as otherwise $e^{-ax}$ will undermine periodicity, however such cases are restricted by the given conditions on $a$ and $b$.
At the moment I have no clue, whether the task has incorrect conditions on $a$ and $b$,or I am just blind…
Thank you in advance for the help.
Best Answer
To bring this problem to a conclusion, note that the general solution $$y=C_1 y_1+C_2 y_2+y_p$$ represents the full two-dimensional space of solutions of the inhomogeneous second-order linear ODE $$y''+2ay'+b^2 y=\cos x.$$ Every choice of $C_1,C_2$ corresponds to a valid solution of the ODE for some initial condition. In particular, $y_p$ is itself a solution. So if the particular solution happens to be periodic, then this is the desired example.
A side question is how one finds such a periodic particular solution. In the present case, the method of undetermined coefficients suffices: we guess a solution of the form $y_p = A \cos t+B\sin t$ and then determine what $A,B$ will work. (If the forcing function were more complicated, then we could instead use variation of parameters or the Fourier transform method.)
Lastly, this solution is unique: If two periodic solutions $y_{p1},y_{p2}$ existed, then $y_{p1}-y_{p2}$ would be a $2\pi$-periodic solution to the homogeneous ODE. But the only such solution is $y=0$. So once a periodic particular solution is found, then this must be the unique periodic solution.