| You're right about the C = 1/2, but you didn't take it all the way.
dy/dx = (y-1)/x^2
dy/(y-1) = 1/x^2 dx
ln |y-1| = -1/x + C
C = 1/2 (as indicated above)
ln |y-1| = -1/x + 1/2
|y-1| = e^(-1/x + 1/2)
y - 1 = e^(-1/x + 1/2) (because of initial condition (0,2))
y = e^(-1/x + 1/2) + 1
y = e^(-1/x)*e^(1/2) + 1
And technically, you can simplify and put in the initial conditions, but you have to be a lot more careful with the algebra.
Last edited by TheMathProf : 05-09-2008 at 01:39 PM.
Reason: Dropped a minus sign in the typing of the solution.
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