# Calculus - Second Order Differential Equations

Solve the boundary-value problem.
y''+5y'-6y=0, y(0)=0, y(2)=1
r^2+5r-6=0, r1=1, r2=-6
y=c1*e^x + c2*e^-6x
y(x)=c1*e^x+c2*e^-6x
y'(x)=c1*e^x-6*c2*e^-6x
y(0)=c1+c2=0, c1=-c2
y(2)=c1*e^2+c2*e^(-12)=1
-c2*e^2-6c2*e^(-12)=1
-c2(e^2-6*e^-12)=1
c2=-1/(e^2-6e^-12)
c1=1/(e^2-6e^-12)
y(x)=(e^x)/(e^2-6*e^-12)+(6*e^-6x)/(e^2-6*e^-12)

Is this correct? Is there a way to further simplify this for y(x)? Thanks.

No. The second term has 6e^-6x in it, I don't see where the coefficent six came from. Testing it to the initial conditions, at x=0, y does not equal zero with that six in.

Oh, thank you for catching my mistake. I got the correct answer now. Thanks!

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