The differential equation $x^2(y+1) d x+y^2(x-1) d y=0$ has the general solution given by (where $C$ is a constant of integration.)

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Aicharya · Accepted Answer

$\begin{aligned} & x^2(y+1) d x+y^2(x-1) d y=0 \ & \Rightarrow \int \frac{x^2}{1-x} d x=\int \frac{y^2}{y-1} d y \ & \Rightarrow \int\left(-x-1+\frac{1}{1+x}ight) d x=\int\left(y-1+\frac{1}{y+1}ight) d y \ & \Rightarrow-\frac{x^2}{2}-x-\log \left|1-x+C^1=\frac{y^2}{2}-y+\log ight| y+1 \mid \ & \Rightarrow C^1=+x+-y+\log |1-x|+\log |y+1| \ & \Rightarrow 2 C^1=x^2+2 x+y^2-2 y+2 \log |(1-x)(y+1)| \ & \Rightarrow 2 C^1+2=(x+1)^2+(y-1)^2+2 \log |(1-x)(y+1)|\end{aligned}$ $\Rightarrow(x+1)^2+(y-1)^2+2 \log |(x-1)(y+1)|=C$

The differential equation $x^2(y+1) d x+y^2(x-1) d y=0$ has the general solution given by (where $C$ is a…

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