# (d^4y)/(dx^4)+2d^3y/dx^3+(d^2y)/(dx^2)=cos6x??

Jul 6, 2018

$y \left(x\right) = \left(A + B x\right) {e}^{-} x + C + D x$
$q \quad q \quad q \quad + \frac{35}{49284} \cos \left(6 x\right) - \frac{1}{4107} \sin \left(6 x\right)$

#### Explanation:

Let $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = u$.

Then the equation becomes

$\frac{{d}^{2} u}{\mathrm{dx}} ^ 2 + 2 \frac{\mathrm{du}}{\mathrm{dx}} + u = \cos \left(6 x\right)$

This is a standard second order differential equation that can be solved by the usual methods to give

$u \left(x\right) = A {e}^{-} x + B x {e}^{-} x + \frac{1}{1369} \left(- 35 \cos \left(6 x\right) + 12 \sin \left(6 x\right)\right)$

Thus

$\frac{{d}^{y}}{\mathrm{dx}} ^ 2 = A {e}^{-} x + B x {e}^{-} x + \frac{1}{1369} \left(- 35 \cos \left(6 x\right) + 12 \sin \left(6 x\right)\right)$

and integration twice yields the solution

$y \left(x\right) = {e}^{-} x \left(A + B \left(2 + x\right)\right) + C + D x$
$q \quad q \quad q \quad + \frac{35}{49284} \cos \left(6 x\right) - \frac{1}{4107} \sin \left(6 x\right)$

Renaming the constant $A + 2 B$ as $A$, this can be written as

$y \left(x\right) = \left(A + B x\right) {e}^{-} x + C + D x$
$q \quad q \quad q \quad + \frac{35}{49284} \cos \left(6 x\right) - \frac{1}{4107} \sin \left(6 x\right)$