SOLUTION: Retransmitted Question: Find the definite solution of the following differential equation xy'= y^2, if y(1) = -1. Thank-you in advance.

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Question 398081: Retransmitted Question:
Find the definite solution of the following differential equation xy'= y^2, if y(1) = -1.
Thank-you in advance.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
x%28dy%2Fdx%29+=+y%5E2 <==> dy%2Fy%5E2+=+dx%2Fx <==> y%5E%28-2%29dy+=+dx%2Fx
==> -1%2Fy+=+ln%28abs%28x%29%29+%2B+k, after integration, for an arbitrary constant k.
If x = 1, the y = -1: ==> -1%2F%28-1%29+=+ln+%281%29+%2B+k ==> k = 1.
==> the equation is -1%2Fy+=+ln%28abs%28x%29%29+%2B+1, or
==> -%28%28y%2B1%29%2Fy%29+=+ln%28abs%28x%29%29 ==> e%5E%28-%28%28y%2B1%29%2Fy%29%29+=+abs%28x%29.