ODE for y" + 3y' = 28 cosh 4x

Question

I have this Equation: $y" + 3y' = 28 cosh(4x)$

So far this is what I have:

$ m^2+3m = 0 $

$ m(m + 3) = 0 $

$ m_1 = 0; $

$ m_2 = -3 $

$ Y_h = C_1 e^0 + C_2 e^{-3x} $

$ Y_h = C_1 + C_2 e^{-3x} $

$ r(x) = 28 cosh(4x) $

$ Y_p = K cosh(4x) + M sinh(4x) $

$ Y_p' = -4K sinh(4x) + 4M cosh(4x) $

$ Y_p'' = -16K cosh(4x) - 16M sinh(4x) $

Im stuck here:

$ -16K cosh(4x) - 16M sinh(4x) - 12K sinh(4x) + 12M cosh(4x) = 28 cosh(4x) $

I can seperate $cos$ and $sin$ to get $M$ and $K$

Do I need to use identities here to eliminate either $sinh$ or $cosh$?

Answer 1

I believe your next step is just to equate coefficients.

$$12M-16K=28$$ $$-16M-12K=0$$

Answer 2

Hint: express $\cosh(4x)$ in terms of $\exp$.

EDIT: $\cosh(4x) = \dfrac{1}{2} e^{4x} + \dfrac{1}{2} e^{-4x}$.
When $r$ is not a root of the characteristic polynomial, $a y'' + b y' + c y = e^{r x}$ has a solution $\dfrac{e^{rx}}{a r^2 + b r + c}$