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The solution of $x d x+y d y=x^2 y d y-x y^2 d x$ is
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Verified Answer
The correct answer is:
$x^2-1=C\left(1+y^2\right)$
We have,
$\begin{aligned} x d x+y d y & =x^2 y d y-x y^2 d x \\ \Rightarrow \quad x d x+x y^2 d x & =x^2 y d y-y d y \\ \Rightarrow \quad\left(1+y^2\right) x d x & =-\left(1-x^2\right) y d y\end{aligned}$
On integrating both sides,
$\begin{aligned} \int \frac{2 x d x}{-1+x^2} & =\int \frac{2 y d y}{1+y^2} \\ \Rightarrow \quad \log \left(-1+x^2\right) & =\log \left(1+y^2\right)+\log c \\ \Rightarrow \quad x^2-1 & =C\left(1+y^2\right)\end{aligned}$
$\begin{aligned} x d x+y d y & =x^2 y d y-x y^2 d x \\ \Rightarrow \quad x d x+x y^2 d x & =x^2 y d y-y d y \\ \Rightarrow \quad\left(1+y^2\right) x d x & =-\left(1-x^2\right) y d y\end{aligned}$
On integrating both sides,
$\begin{aligned} \int \frac{2 x d x}{-1+x^2} & =\int \frac{2 y d y}{1+y^2} \\ \Rightarrow \quad \log \left(-1+x^2\right) & =\log \left(1+y^2\right)+\log c \\ \Rightarrow \quad x^2-1 & =C\left(1+y^2\right)\end{aligned}$
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