Question

(x^2+y^2+x)dx+xydy=0​

Answer 1

Step-by-step explanation:

We have,

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

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

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

$Let \: \: {y}^{2} = v \\ \implies2y \frac{dy}{dx} = \frac{dv}{dx} \\ \implies \: y \frac{dy}{dx} = \frac{1}{2} \frac{dv}{dx}$

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

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

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

Now,

$v. {x}^{2} = \int {x}^{2}( - 2x - 2)dx$

$\implies \: v. {x}^{2} = - \int {x}^{2}( 2x + 2)dx$

$\implies \: v. {x}^{2} = - 2\int( {x}^{3} + {x}^{2}) dx$

$\implies \: v. {x}^{2} = - \frac{1}{2} {x}^{4} - \frac{2}{3} {x}^{3} + C$

$\implies \: {(xy)}^{2} = - \frac{1}{2} {x}^{4} - \frac{2}{3} {x}^{3} + C$