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

1 Answer
Jul 6, 2018

y(x) = (A+Bx)e^-x+C+Dxy(x)=(A+Bx)ex+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)