I'm learning Calculus on my own this summer and I've encountered a simple problem which i can't seem to get the right answer to. Here is the problem and my solution:
Find y'' given x + xy + y =2
first derivative
1 + y + x(y') + y' = 0 took derivative of each term
1 + y + y'(x+1) = 0 simplified a bit
(-1-y)/(x+1) = y' simple algebra.
second derivative
[-y'(x+1) - 1(-1-y)]/[(x+1)^2] quotient rule
[-y'(x+1)]/[(x+1)^2] - (-1-y)/[(x+1)^2] = 0 restated in a more useful manner.
[-y'(x+1)]/[(x+1)^2] = (-1-y)/[(x+1)^2] simple algebra
-y'(x+1) = -1-y simple algebra
y'(x+1) = 1+y multiplied both sides by -1
y' = (1+y) / (x+1)
The answer the books gives is: [2(1+y)]/[(1=x)^2]
Thanks in advance for help.