Question
Solve the following differential equations:

$(\text{y + xy})\text{dx}+(\text{x}-\text{xy}^2)\text{dy}=0$

Answer

We have,
$(\text{y + xy})\text{dx}+(\text{x}-\text{xy}^2)\text{dy}=0$
$\Rightarrow\text{y}(1+\text{x})\text{dx = x}(\text{y}^2-1)\text{dy}$
$\Rightarrow\frac{1+\text{x}}{\text{x}}\text{dx}=\frac{\text{y}^2-1}{\text{y}}\text{dy}$
Integrating both sides, we get
$\int\frac{1+\text{x}}{\text{x}}\text{dx}=\int\frac{\text{y}^2-1}{\text{y}}\text{dy}$
$\Rightarrow\int\frac{1}{\text{x}}\text{dx}+\int\text{dx}=\int\text{y dy}-\int\frac{1}{\text{y}}\text{dy}$
$\Rightarrow\log|\text{x}|+\text{x}=\frac{\text{y}^2}{2}-\log|\text{y}|+\text{C}$
$\Rightarrow\log|\text{x}|+\text{x}-\frac{\text{y}^2}{2}+\log|\text{y}|=\text{C}$
Hence, $\log|\text{x}|+\text{x}-\frac{\text{y}^2}{2}+\log|\text{y}|=\text{C}$ is the required solution.

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