I have found this problem of ODE in my textbook but i can't find any way to solve it $$x^2(y+1)dx+y^2(x-1)dy = 0$$ I tried to solve it using reducible exact equation rules but i couldn't solve it. Please help me out.
Solving Ordinary Differential Equation with exact form
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0You couldn't solve it using exact equation rules because your ODE isn't exact. – 2017-01-10
1 Answers
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This is easy because it is a seperable equation: $$ x^2(y+1)dx = y^2(1-x)dy \\ \frac{x^2}{1-x}dx = \frac{y^2}{1+y}dy \\ \int \frac{x^2}{1-x}dx = \int \frac{y^2}{1+y}dy \\ \frac12(x-1)(x+3)-\log(1-x) +C = \frac12(y+1)(y-3) + \log(1+y) $$ and you can simplify that however you wish.