- #1
eric_999
- 8
- 0
I have this equation: dy/dx = 1-y^2, so then dy/(1-y^2) = dx, so ∫dy/(1-y^2) = dx --->
∫(A/(1+y) + B/(1-y))dy = x + C. I rewrite it again: (A - Ay + B + By)/(1+y)(1-y) = 1/(1+y)(1-y) so I get A+B = 1, and B-A = 0, so B = A, and therefore 2A = 1. so A & B = 1/2.
So 1/2∫(dy/(1+y) + 1/2∫dy/(1-y) = x + C ---> 1/2(ln|1+y| -ln|1-y|) = x + C ---> ln(|1+y|/|1-y|) = 2x + 2c = 2x + C, since C is arbitrary. At last I get |1+y|/|1-y| = e^(2x + c) ?=? |(1+y)/(1-y)| = e^(2x + C). How do I solve for y? Do Ihave to split it up into two separate cases, one when (1+y)/1-y) = e^(2x +C) and one where -(1+y)/(1-y) = e^(2x + C), or is this whole method simply just wrong?
Please help me out!
∫(A/(1+y) + B/(1-y))dy = x + C. I rewrite it again: (A - Ay + B + By)/(1+y)(1-y) = 1/(1+y)(1-y) so I get A+B = 1, and B-A = 0, so B = A, and therefore 2A = 1. so A & B = 1/2.
So 1/2∫(dy/(1+y) + 1/2∫dy/(1-y) = x + C ---> 1/2(ln|1+y| -ln|1-y|) = x + C ---> ln(|1+y|/|1-y|) = 2x + 2c = 2x + C, since C is arbitrary. At last I get |1+y|/|1-y| = e^(2x + c) ?=? |(1+y)/(1-y)| = e^(2x + C). How do I solve for y? Do Ihave to split it up into two separate cases, one when (1+y)/1-y) = e^(2x +C) and one where -(1+y)/(1-y) = e^(2x + C), or is this whole method simply just wrong?
Please help me out!