Question

solve this differential equation guys help me to solve it and please try to write it step wise
(spam answers will be deleted)​

Answer 1

Step-by-step explanation:

We have,

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

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

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

$\implies v + x \frac{dv}{dx} = \frac{ {x}^{2} + {x}^{2} {v}^{2} }{ {x}^{2} + {x}^{2}v }$

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

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

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

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

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

Integrating both sides,

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

$Let \: \: v = \cos(2 \alpha ) \\ dv = - 2 \sin(2 \alpha ) d \alpha$

$\implies - \int\frac{(1 + \cos(2 \alpha ) )}{(1 - \cos( 2\alpha ) )} .2 \sin( 2\alpha ) d \alpha = x + c$

$\implies - 4 \int\frac{ \cos^{2} ( \alpha ) }{ \sin ^{2} ( \alpha ) } .\sin( \alpha ) \cos( \alpha ) d \alpha = x + c$

$\implies - 4 \int\frac{ (1 - \sin^{2} ( \alpha ) )}{ \sin ( \alpha ) } . \cos( \alpha ) d \alpha = x + c$

Putting sin(α) = t => cos(α) dα= dt

$\implies - 4 \int\frac{ 1 - t^{2} }{ t} d t = x + c$

$\implies - 4 \int\frac{ 1}{ t} d t + 4 \int \: t \: dt= x + c$

$\implies - 4 ln(t) + 2 {t}^{2} = x + c$

$\implies - 4 ln( \sin( \alpha ) ) + 2 { \sin }^{2} ( \alpha ) = x + c$

$\implies - 4 ln( \sqrt{\frac{1 - v}{2} }) + (1 - v) = x + c$

$\implies - 2 ln( \frac{1 - v}{2} ) + (1 - v) = x + c$

$\implies - 2 ln( 1 - v ) + 2 ln(2) + (1 - v) = x + c$

$\implies - 2 ln( 1 - v ) + (1 - v) = x + c$

$\implies - 2 ln( 1 - \frac{y}{x} ) + (1 - \frac{y}{x} ) = x + c$

$\implies - 2 ln( x- y ) + 2 ln(x) + \frac{x - y}{x} = x + c$