# How to you find the general solution of sqrt(1-4x^2)y'=x?

Nov 8, 2016

$y = - \frac{1}{4} \sqrt{1 - 4 {x}^{2}} + C$

#### Explanation:

Changing the notation from Lagrange's notation to Leibniz's notation we have:

$\sqrt{1 - 4 {x}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}} = x$ which is a First Order separable DE which we can rearrange as follows:

$\mathrm{dy} = \frac{x}{\sqrt{1 - 4 {x}^{2}}} \mathrm{dx} \implies \int \mathrm{dy} = \int \frac{x}{\sqrt{1 - 4 {x}^{2}}} \mathrm{dx}$

To integrate the RHS we need to use a substitution

Let $u = 1 - 4 {x}^{2} \implies \frac{\mathrm{du}}{\mathrm{dx}} = - 8 x \implies - \frac{1}{8} \mathrm{du} = x \mathrm{dx}$

Se we can now substitute and integrate to get our DE solution:

$y = \int \frac{- \frac{1}{8}}{\sqrt{u}} \mathrm{du}$
$\therefore y = - \frac{1}{8} \int {u}^{- \frac{1}{2}} \mathrm{du}$
$\therefore y = - \frac{1}{8} {u}^{\frac{1}{2}} / \left(\frac{1}{2}\right) + C$
$\therefore y = - \frac{1}{4} {u}^{\frac{1}{2}} + C$
$\therefore y = - \frac{1}{4} \sqrt{1 - 4 {x}^{2}} + C$