The solution of the differential equation $(3 x y + y^{2}) d x + (x^{2} + x y) d y = 0$ is

x^{2} (2 x y + y^{2}) = c^{2}

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x^{2} (2 x y - y^{2}) = c^{2}

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x^{2} (y^{2} - 2 x y) = c^{2}

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None of these

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Solution

The correct option is A $x^{2} (2 x y + y^{2}) = c^{2}$
Homogeneous equation can be written in the form of
$\frac{d y}{d x} = - \frac{3 x y + y^{2}}{x^{2} + x y}$
Put $y = v x$ and $\frac{d y}{d x} = v + x \frac{d v}{d x}$ , we get
$v + x \frac{d v}{d x} = - \frac{3 x^{2} v + x^{2} v^{2}}{x^{2} + x^{2} v}$
$\Rightarrow x \frac{d v}{d x} = \frac{- 2 v (v + 2)}{v + 1}$
$\Rightarrow \frac{1}{x} d x = - \frac{(v + 1)}{2 v (v + 2)} d v$
$\Rightarrow - \frac{2}{x} = - [\frac{1}{2 (v + 2)} + \frac{1}{2 v}] d v$
On integrating, we get
$- 2 {log}_{e} x = \frac{1}{2} log (v + 2) + \frac{1}{2} log v - log c$
$\Rightarrow v (v + 2) x^{4} = c^{2}$
$\Rightarrow \frac{y}{x} (\frac{y}{x} + 2) x^{4} = c^{2} (∵ v = \frac{y}{x})$
Hence, required solution is $(y^{2} + 2 x y) x^{2} = c^{2}$ .

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Similar questions

(1 + y^{2}) d x - x y d y = 0

\sqrt{1 + x^{2} + y^{2} + x^{2} y^{2}} + x y \frac{d y}{d x} = 0

(3 x y + y^{2}) d x + (x^{2} + x y) d y = 0

\sqrt{1 + x^{2} + y^{2} + x^{2} y^{2}} + x y \frac{d y}{d x} = 0

C

C_{1}

2 x y \frac{d y}{d x} = y^{2} - x^{2},

x > 0.

C_{2}

\frac{2 x y}{x^{2} - y^{2}} = \frac{d y}{d x} .

(1, 1),

C_{1}

C_{2}

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