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

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Answer 1 · 2018-07-06T00:57:27

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

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

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