Question

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

Answer 1

`y(x) = (A+Bx)e^-x+C+Dx`
`qquadqquadqquad+35/49284 cos(6x)-1/4107 sin(6x)`

Explanation:

Let `(d^2y)/dx^2 = u`.

Then the equation becomes

`(d^2u)/dx^2+2(du)/dx+u=cos(6x)`

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

`u(x) = Ae^-x+Bxe^-x+1/1369(-35cos(6x)+12sin(6x))`

Thus

`(d^y)/dx^2 = Ae^-x+Bxe^-x+1/1369(-35cos(6x)+12sin(6x))`

and integration twice yields the solution

`y(x) = e^-x(A+B(2+x))+C+Dx`
`qquadqquadqquad+35/49284 cos(6x)-1/4107 sin(6x)`

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

`y(x) = (A+Bx)e^-x+C+Dx`
`qquadqquadqquad+35/49284 cos(6x)-1/4107 sin(6x)`