The differential equation y^2 dx + (2xy - 1) | vedclass

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(D) Given differential equation is $y^2 dx + (2xy - 1) dy = 0$. Dividing by $y^2 dy$,we get: $\frac{dx}{dy} + \frac{2xy - 1}{y^2} = 0$ $\frac{dx}{dy}

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linear in $x$

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(D) Given differential equation is $y^2 dx + (2xy - 1) dy = 0$. Dividing by $y^2 dy$,we get: $\frac{dx}{dy} + \frac{2xy - 1}{y^2} = 0$ $\frac{dx}{dy} + \frac{2x}{y} - \frac{1}{y^2} = 0$ $\frac{dx}{dy} + \left(\frac{2}{y}\right)x = \frac{1}{y^2}$ This is of the form $\frac{dx}{dy} + Px = Q$,where $P = \frac{2}{y}$ and $Q = \frac{1}{y^2}$ are functions of $y$ only. Therefore,the given differential equation is linear in $x$.

The differential equation $y^2 dx + (2xy - 1) dy = 0$ is

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