Question

ydx+2(y^4-x)dy=0

Answer 1

The given equation can be written as $y~dx = -2(y^{4} - x)dy$. It would be easier to solve this equation in terms of $\frac{dx}{dy}.$

$\dfrac{dx}{dy} = -\dfrac{(2y^{4} - x)}{y} = -2y^3 + \dfrac{2x}{y}\\ \dfrac{dx}{dy} - \dfrac{2x}{y} = -2y^3 \\$

which is of the form

$\dfrac{dx}{dy} + P(y)x = Q(y).$

whose solution is given by $xe^{\int Pdy} = \int Q e^{\int Pdy} dy + c$, where $e^{\int Pdy}$  is called the integrating factor.

Here $P(y) = -\frac{2}{y}$ and $e^{\int Pdy} = e^{\int \frac{-2}{y}} = \frac{1}{y^{2}}$. Hence the solution of the given equation is

$\dfrac{x}{y^{2}} = \int -2y^{3} \cdot \frac{1}{y^{2}}dy + c\\ \dfrac{x}{y^{2}} = \int -2y dy + c\\ \dfrac{x}{y^{2}} = -y^{2} + c\\ \therefore x = cy^{2} - y^{4}$