Question

solve (xy-2y^(2))dx-(x^(2)-3xy)dy=0

Answer 1

Step-by-step explanation:

We have,

$(x - 2 {y}^{2}) dx - ( {x}^{2} - 3 xy)dy = 0$

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

$Let \: y = vx \\ \implies \: \frac{dy}{dx} = v + x \frac{dv}{dx}$

So,

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

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

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

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

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

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

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

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

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

$\implies \: - \frac{1 }{ v } - 3 \ln(v) = \ln x + C$

$\implies \: - \frac{x }{ y } - 3 \ln \bigg( \frac{y}{x} \bigg) = \ln x + C$

$\implies \: - \frac{x }{ y } - 3 \ln(y) + 3 \ln(x) = \ln x + C$

$\implies \: - \frac{x }{ y } - 3 \ln(y) + 2 \ln(x) = C$