# private solutions

• Dec 8th 2010, 04:55 AM
transgalactic
private solutions
y''-4y+4y=xe^2x

what is the general private solution

usuallu
its
y_p=(ax+b)e^2x
but here we have
a double root of 2
• Dec 8th 2010, 05:24 AM
TheEmptySet
Quote:

Originally Posted by transgalactic
y''-4y+4y=xe^2x

what is the general private solution

usuallu
its
y_p=(ax+b)e^2x
but here we have
a double root of 2

I think you mean particular solution.

From just looking at the right hand side we would "guess" that

$y_p=(ax+b)^{2x}$

This is correct, but since the complimentary solution is $c_1e^{2x}+c_2xe^{2x}$ as you said a double root we have to multiply our guess by $x^2$ This is justified by using reduction of order. so we get

$y_p=(ax^3+bx^2)e^{-2x}$

Now just plug in to find a and b.