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### Answers (1)

Answer : $(a)\left(x^{2}-1\right)=C\left(y^{2}+1\right)$

Hint: Use variable separable method i.e. separate x and y terms by taking x terms in one side and y terms to the other.

Given : $x d x+y d x=x^{2} y d y-y^{2} x d x$

Explanation  : $x d x+y^{2} x d x=x^{2} y d y-y d y$

\begin{aligned} &\Rightarrow x\left(1+y^{2}\right) d x=y\left(x^{2}-1\right) d y \\ &\Rightarrow \frac{x}{x^{2}-1} d x=\frac{y}{y^{2}+1} d y \end{aligned}

Integrate both sides

\begin{aligned} &\Rightarrow \int \frac{x}{x^{2}-1} d x=\int \frac{y}{y^{2}+1} d y\\ &\text { Let } x^{2}-1=t \text { and } y^{2}+1=v \end{aligned}

\begin{aligned} &\Rightarrow 2 x d x=d t \text { and } 2 y d y=d v \\ &\Rightarrow x d x=\frac{d t}{2} \text { and } y d y=\frac{d v}{2} \end{aligned}

Put in (i)

\begin{aligned} &\Rightarrow \frac{1}{2} \int \frac{d t}{t}=\frac{1}{2} \int \frac{d v}{v} \\ &\Rightarrow \frac{1}{2} \log t=\frac{1}{2} \log v+\frac{1}{2} \log C \end{aligned}

\begin{aligned} &\Rightarrow \log t=\log v+\log C \\ &\Rightarrow \log t=\log C v \end{aligned}

\begin{aligned} &\Rightarrow t=C v \\ &\Rightarrow\left(x^{2}-1\right)=C\left(y^{2}+1\right) \end{aligned}

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