The general solution of the differential equation (x - y ^ 2 )dx +...

Given, $\left(x - {y}^{2}\right)dx + y\left(5x + {y}^{2}\right)dy = 0$

After rearranging we get,

$\dfrac{dy}{dx} = \dfrac{{y}^{2} - x}{y\left(5x + {y}^{2}\right)}$

Now, let ${y}^{2} = v ⇒ 2y\dfrac{dy}{dx} = \dfrac{dv}{dx}$

$\dfrac{1}{2y}\dfrac{dv}{dx} = \dfrac{v - x}{y\left(5x + v\right)}$

$\dfrac{dv}{dx} = 2\left(\dfrac{v - x}{5x + v}\right)$

Now let $v = kx$

$⇒ k + x\dfrac{dk}{dx} = 2\left(\dfrac{kx - x}{5x + kx}\right)$

$⇒ x\dfrac{dk}{dx} = - \dfrac{\left({k}^{2} + 3k + 2\right)}{k + 5}$

$⇒ ∫ \dfrac{\left(5 + k\right)}{\left(k + 1\right)\left(k + 2\right)}dk = ∫ - \dfrac{dx}{x}$

$⇒ ∫ \left(\dfrac{4}{k + 1} - \dfrac{3}{k + 2}\right)dk = - ∫ \dfrac{dx}{x}$

$⇒ 4\ln \left(k + 1\right) - 3\ln \left(k + 2\right) = - \ln x + lnc$

$⇒ \dfrac{{\left(k + 1\right)}^{4}}{{\left(k + 2\right)}^{3}} = \dfrac{c}{x}$

$⇒ \dfrac{{\left(\dfrac{v}{x} + 1\right)}^{4}}{{\left(\dfrac{v}{x} + 2\right)}^{3}} = \dfrac{c}{x}$

$⇒ c{\left({y}^{2} + 2x\right)}^{3} = {\left({y}^{2} + x\right)}^{4}$