particular solution to $y^4 -2y'' +y = xe^x $

Question

I am trying to find the particular solution to the $y^4 -2y'' +y = xe^x $ and currently am misunderstanding what to do.

My steps:

the polynomial operator concerned is $p(s)= s^4 -2s^2 + 1 $ which 0 at 1: $p(1) = 0 $

so now i know that the solution will be something like:

$y_p = x^2(Ax + B)e^x$ where the parentheses show that it is a linear operator on the last coefficient.

I am told to use the exponential shift rule on $e^x$ so i believe that it is this:

$y_p = x^2e^x(Ax + B + 1)$

but i'm not sure if this is correct or where to go from here. because now the operator is operating on a coefficient of 1?

thanks for your help

Answer 1

$$y^4 -2y'' +y = xe^x$$ The caracteristic polynomial is $$\implies r^4-2r^2+1=0$$ $$(r^2-1)^2=0 $$ $$(r-1)^2(r+1)^2=0 \implies r=1,-1$$ The solution to the homogeneous equation is $$y_h=c_1e^x+c_2xe^x+c_3e^{-x}+c_4xe^{-x}$$ for the particular solution try $$y_p=(Ax^3+Bx^2)e^x$$

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Another method

$$y^4 -2y'' +y = xe^x$$ $$y^4 -y''-y'' +y = xe^x$$ $$(y'' -y)''-(y'' -y) = xe^x$$ Substitute $z=y''-y$ $$z''-z=xe^x$$ it's linear of second order