The general integral of the partial differential equation $y^{2} p - x y q = x (z - 2 y)$ is

ϕ (x^{2} + y^{2}, y^{2} - y z) = 0

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ϕ (x^{2} - y^{2}, y^{2} + y z) = 0

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ϕ (x y, y z) = 0

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ϕ (x + y, I n x - z) = 0

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Solution

The correct option is A $ϕ (x^{2} + y^{2}, y^{2} - y z) = 0$
$y^{2} p - x y q = x (z - 2 y)$
On comparision with $P . p + Q q = R$
$P = y^{2}, Q = - x y, R = x (z - 2 y)$
By Lagrange's Auxiliary equation
$\frac{d x}{P} = \frac{d y}{Q} = \frac{d z}{R}$
i.e., $\frac{d x}{y^{2}} (I) = \frac{d y}{- x y} (I I) = \frac{d z}{x (z - 2 y)} (I I I) = \frac{x d x + y d y}{0} (I V)$
Taking $I$ and $I V$ , $\frac{d x}{y^{2}} = \frac{x d x + y d y}{0}$
$\Rightarrow x d x + y d y = 0$
$i . e ., x^{2} + y^{2} + C_{1}$
Again By $I I$ and $I I I$ , $\frac{d y}{- x y} = \frac{d z}{x (z - 2 y)}$
$z d y - 2 y d y = - y d z$
or $y d z + z d y = 2 y d y o r d (y z) = d (y^{2})$
i.e., $y z - y^{2} = C$ or $y^{2} - y z = C_{2}$
So General solution is $ϕ (C_{1}, C_{2}) = 0$
$\Rightarrow ϕ (x^{2} + y^{2}, y^{2} - y z) = 0$

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