Solving the ODE $\frac{d^2 \bar{\phi}}{dx^2} - s \bar{\phi} = \frac{\cos(x)}{s}$ Using Method of Undetermined Coefficients

Question

I am trying to solve the ODE $\dfrac{d^2 \bar{\phi}}{dx^2} - s \bar{\phi} = \dfrac{\cos(x)}{s}$.

If I am not mistaken, I can solve this using the method of undetermined coefficients.

Solving the homogeneous equation

$$\dfrac{d^2 \bar{\phi}}{dx^2} - s \bar{\phi} = 0,$$

we get the characteristic equation

$$m^2 - s = 0,$$

which gives us

$$m = \pm \sqrt{s}.$$

Therefore, the complementary equation is

$$y_c(x) = Ae^{x\sqrt{s}} + Be^{-x\sqrt{s}}$$

What I'm now confused about is the form of the particular solution. Given that we have $\dfrac{\cos(x)}{s}$, how do I go about solving this?

I would appreciate it if people could please take the time to clarify this.

Answer 1

The equation is in $x$, so you can treat $s$ as a constant. Let the particular solution be

$$ \phi_p(x) = A\cos x + B\sin x $$

Then ${\phi_p}'' = -A\cos x - B\sin x = -(A\cos x + B\sin x)$, therefore

$$ {\phi_p}'' - s\phi_p = -(A\cos x + B\sin x) - s(A\cos x + B\sin x) = -(1+s)(A\cos x + B\sin x) $$

This gives

$$ A = - \frac{1}{s(1+s)}, B = 0 $$

Answer 2

For the particular solution, consider $$ A \cos x + B \sin x $$ and find your $A$ and $B$.