Help with solving the differential equation $y'' - 12y' = \cos(2x)$ where $y(0) = y'(0) = 0$

Question

I'm currently struggling with solving the following differential equation:

$$y'' - 12y' = \cos(2x)$$

with initial conditions $y(0) = 0$ and $y'(0) = 0$.

I have already found the homogeneous solution, but I'm having trouble guessing the particular solution. I know that if the equation also contained y, the particular solution would be $y_p = A \cos x + B \sin x$. However, since $y$ is not included in the equation, I'm not sure how to proceed.

I've tried guessing a solution using $e^{2ix}$, but then $d$ still needs to be $y$, and I'm not sure what to do next.

Can anyone provide any insight or guidance on how to find the particular solution? Any help would be greatly appreciated.

Thank you!

Answer 1

HINT

What about integrating both sides?

\begin{align*} y'' - 12y' = \cos(2x) & \Longleftrightarrow y' - 12y = \frac{\sin(2x)}{2} + C\\\\ & \Longleftrightarrow (e^{-12x}y)' = \frac{e^{-12x}\sin(2x)}{2} + Ce^{-12x} \end{align*}

Can you take it from here?

Answer 2

Let $y_p=A\cos(2x)+B\sin(2x)$. Work out $y_p^{\prime},y_p^{\prime \prime}$ and express $y_p^{\prime \prime}-12y_p^{\prime }$ in the form $C\cos(2x)+D\sin(2x)$ where $C\text{ and }D$ are functions of $A\text{ and }B$. Solve the euations $C=1,D=0$ for $A\text{ and }B$.

Answer 3

$$y'' - 12y' = \cos(2x)=\Re \{ e^{2ix}\}$$ $$y'' - 12y' = e^{2ix}$$ Try $y_p=Ae^ {2ix}$ $$Ae^{2ix}(-4-24i)=e^{2ix}$$ $$ A =-\dfrac 1 4\dfrac 1 {1+6i}$$ $$ A =-\dfrac 1 4\dfrac {1-6i} {37}$$ $$ \implies y_p =-\dfrac {1-6i}{148} e^{2ix}$$ We need the real part of the particular solution $u_p$ so we use $e^{2ix}=\cos(2x)+i\sin(2x)$: $$\Re \{ y_p\}=-\dfrac 1 {148}\cos (2x)-\dfrac 6 {148} \sin (2x)$$ $$\Re \{y_p\}=-\dfrac 1 {148}\cos (2x)-\dfrac 3 {74} \sin (2x)$$