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I have a brief question. I am trying to solve the following simple differential equation:

[itex] x\dfrac{dy}{dx}= y^2-1[/itex]

I manage to solve it one way, but when I try to solve it in the following way, I cannot get the correct answer. I would really appreciate if someone could point out my mistake :)

[itex] \dfrac{dy}{y^2-1}= \dfrac{dx}{x}[/itex]

Rewriting the fraction on the left as the sum of two fractions

[itex] \dfrac{1}{2}(\dfrac{-1}{y+1}+\dfrac{1}{y-1})dy= \dfrac{dx}{x}[/itex]

Integrating both sides:

[itex] \dfrac{1}{2}∫(\dfrac{-1}{y+1}+\dfrac{1}{y-1})dy= ∫\dfrac{dx}{x}[/itex]

[itex] \dfrac{1}{2}∫(-1ln(y+1)+ln (y-1))dy= ln(x)+c[/itex]

[itex] \dfrac{1}{2} ln(\dfrac{y-1}{y+1})= ln(x)+c[/itex]

It is given that when x=1, y=0, thus c=0

[itex] \dfrac{1}{2} ln(\dfrac{y-1}{y+1})= ln(x)[/itex]

[itex] \dfrac{y-1}{y+1}= x^2[/itex]

Which gives:

[itex] y= \dfrac{1+x^2}{1-x^2}[/itex]

But, he correct result is:

[itex] y= \dfrac{1-x^2}{1+x^2}[/itex]

I would really appreciate if anyone could point out the mistake. Thank you. :)

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# Please tell me what I did wrong when solving this DE

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