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

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- Dec 8th 2010, 04:55 AMtransgalacticprivate 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 AMTheEmptySet
I think you mean particular solution.

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

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

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

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

Now just plug in to find a and b.