**Problem 01**

$y(2xy + 1) \, dx - x \, dy = 0$

**Solution 01**

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$y(2xy + 1) \, dx - x \, dy = 0$

$2xy^2 \, dx + y \, dx - x \, dy = 0$

$2xy^2 \, dx + (y \, dx - x \, dy) = 0$

Divide by y^{2}

$2x \, dx + \dfrac{y \, dx - x \, dy}{y^2} = 0$

$2x \, dx + d\left( \dfrac{x}{y} \right) = 0$

$\displaystyle 2\int x \, dx + \int d\left( \dfrac{x}{y} \right) = 0$

$x^2 + \dfrac{x}{y} = c$

Multiply by y

$x^2y + x = cy$

$x(xy + 1) = cy$ *answer*