What is the general solution of the differential equation  2(y-4x^2)dx+xdy = 0 ?

Jul 9, 2017

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

Explanation:

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

Which we can re-arrange as follows:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 2 \left(y - 4 {x}^{2}\right)}{x}$
$\text{ } = 8 x - \frac{2 y}{x}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{2 y}{x} = 8 x$ ..... [A]

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

$\frac{\mathrm{dy}}{\mathrm{dx}} + P \left(x\right) y = Q \left(x\right)$

Then the integrating factor is given by;

$I = {e}^{\int P \left(x\right) \mathrm{dx}}$
$\setminus \setminus = \exp \left(\int \setminus \frac{2}{x} \setminus \mathrm{dx}\right)$
$\setminus \setminus = \exp \left(2 \ln x\right)$
$\setminus \setminus = \exp \left(\ln {x}^{2}\right)$
$\setminus \setminus = {x}^{2}$

And if we multiply the DE [A] by this Integrating Factor, $I$, we will have a perfect product differential;

${x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + {x}^{2} x y = 8 {x}^{3}$

$\therefore \frac{d}{\mathrm{dx}} \left({x}^{2} y\right) = 8 {x}^{3}$

Which we can now directly integrate to get:

${x}^{2} y = \int \setminus 8 {x}^{3} \setminus \mathrm{dx}$

$\therefore {x}^{2} y = 2 {x}^{4} + c$

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