Thread: 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

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



This is correct, but since the complimentary solution is as you said a double root we have to multiply our guess by This is justified by using reduction of order. so we get



Now just plug in to find a and b.