Question

(iii) (x2 - yx2) dy + (y2 + xy2) dx = 0.​

Answer 1

Step-by-step explanation:

We have,

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

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

$\implies \frac{dy}{dx} = - \frac{ {y}^{2} (1 + x) }{ {x}^{2}(1 - y ) )}$

$\implies \frac{(1 - y)dy}{{y}^{2} } = - \frac{ (1 + x) }{ {x}^{2} } dx$

$\implies \int\frac{(1 - y)dy}{{y}^{2} } = - \int\frac{ (1 + x) }{ {x}^{2} } dx$

$\implies \int\frac{y - 1}{ {y}^{2} } dy = \int \frac{1 + x}{ {x}^{2} } dx$

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

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