# What is the general solution of the differential equation? :  (x - 4) y^4 dx - x^3 (y^2 - 3) dy =0

Jul 9, 2017

$\frac{1}{y} ^ 3 - \frac{1}{y} = \frac{2}{x} ^ 2 - \frac{1}{x} + c$

#### Explanation:

We have the following Differential Equation in differential form

$\left(x - 4\right) {y}^{4} \mathrm{dx} - {x}^{3} \left({y}^{2} - 3\right) \mathrm{dy} = 0$

Which we can re-arrange as follows:

$\frac{{y}^{2} - 3}{y} ^ 4 \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x - 4}{x} ^ 3$

This is now a separable DIfferential Equation, and so "separating the variables" gives us:

$\int \setminus \frac{{y}^{2} - 3}{y} ^ 4 \setminus \mathrm{dy} = \int \setminus \frac{x - 4}{x} ^ 3 \setminus \mathrm{dx}$

This is now a trivial integration problem, thus:

$\int \setminus \frac{1}{y} ^ 2 - \frac{3}{y} ^ 4 \setminus \mathrm{dy} = \int \setminus \frac{1}{x} ^ 2 - \frac{4}{x} ^ 3 \setminus \mathrm{dx}$

$\therefore - \frac{1}{y} + \frac{1}{y} ^ 3 = - \frac{1}{x} + \frac{2}{x} ^ 2 + c$

Hence the solution is:

$\frac{1}{y} ^ 3 - \frac{1}{y} = \frac{2}{x} ^ 2 - \frac{1}{x} + c$