Question

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

Answer 1

Step-by-step explanation:

We have,

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

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

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

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

$Let \: y^{2} = t \\ \implies2y \frac{dy}{dx} = \frac{dt}{dx}$

$\implies \: \frac{dt}{dx} - \frac{2 t }{x} = - {x}^{2}$

$I.F. = {e}^{ - \int \frac{2}{x}dx} = {e}^{ - 2 ln(x) } = \frac{1}{ {x}^{2} }$

$\implies \frac{t}{ {x}^{2} } = \int \frac{1}{ {x}^{2} } .( - {x}^{2} )dx$

$\implies \frac{t}{ {x}^{2} } = - \int \: dx$

$\implies \frac{t}{ {x}^{2} } = - x + c$

$\implies \frac{ {y}^{2} }{ {x}^{2} } = - x + c$

$\implies {y}^{2} = - {x}^{3} + c {x}^{2}$