# Calculus - Second Order Differential Equations

Solve the initial-value problem.
y'' - 2y' + y = 0 , y(2) = 0 , y'(2) = 1
r^2-2r+1=0, r1=r2=1
y(x)=c1*e^x+c2*x*e^x
y(2)=c1*e^2+c2*2*e^2=0
c1=-(2*c2*exp(2))/exp(2)
c1=-2*c2
y'(x)=-2*c2*e^x+c2*e^x*(x-1)
y'(2)=-2*c2*e^2+c2*e^2*(2-1)=1
c2(-2e^2+e^2)=1
c2=1/(-2e^2+e^2)=1/(-e^2)
c1=-2/(-e^2)=2/(e^2)
y(x)=(2/e^2)(e^x)-(1/e^2)xe^x
y(x)=(2e^x-xe^x)/e^2

What is wrong with this answer? Did I miss or mess up a step? Thanks.

Where did you get the (x-1) in y'?

Well I don't know how I got the (x-1) but I corrected it and I get the following answer,

y(x) = (2/e^2)*e^x - (1/e^2)*x*e^x

which is still incorrect. Any thoughts???

Nevermind I got it...thanks.

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asked by COFFEE

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