Question

solve (3xy-2aysquare) dx+(xsquare -2axy) dy=0​

Answer 1

Step-by-step explanation:

We have,

$(3xy - 2ay ^{2} )dx + ( {x}^{2} - 2axy)dy = 0$

$\implies \frac{dy}{dx} = - \frac{(3xy - 2a {y}^{2} )}{( {x}^{2} - 2axy) }$

$Let \: y = vx$

$\implies\frac{dy}{dx}=v+x\frac{dv}{dx}$

$\implies \: v + x \frac{dv}{dx} = \frac{2a {v}^{2} {x}^{2} - 3v {x}^{2} }{ {x}^{2} - 2av {x}^{2} }$

$\implies \: x \frac{dv}{dx} = \frac{(2a {v}^{2} - 3v) }{(1 - 2av )} - v$

$\implies \: x \frac{dv}{dx} = \frac{2a {v}^{2} - 3v - v + 2a {v}^{2} }{(1 - 2av )}$

$\implies \: x \frac{dv}{dx} = \frac{4a {v}^{2} - 4v }{(1 - 2av )}$

$\implies \: \frac{(1 - 2av)}{(4a {v}^{2} - 4v)} dv = \frac{dx}{x}$

$\implies \: \frac{(1 - av) - av}{4v(av -1)} dv = \frac{dx}{x}$

Integrating both sides,

$\implies \: \int\frac{(1 - av) - av}{4v(av -1)} dv = \int \frac{dx}{x}$

$\implies \: - \int \frac{dv}{4v} - \frac{a}{4} \int \frac{dv}{(av - 1)} = \int \frac{dx}{x}$

$\implies - \frac{1}{4} ln(v) - \frac{a}{4} ln(av - 1) = ln(x) + c$

$\implies - \frac{1}{4} ln( \frac{y}{x} ) - \frac{a}{4} ln(a \frac{y}{x} - 1) = ln(x) + c$