# How to you find the general solution of √1−4x^2y'=x?

How to you find the general solution of $\sqrt{1-4{x}^{2}}y\prime =x$?
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Kimberly Evans
Changing the notation from Lagrange's notation to Leibniz's notation we have:

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

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

To integrate the RHS we need to use a substitution

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

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

$y=\int \frac{-\frac{1}{8}}{\sqrt{u}}du$
$\therefore y=-\frac{1}{8}\int {u}^{-\frac{1}{2}}du$
$\therefore y=-\frac{1}{8}\frac{{u}^{\frac{1}{2}}}{\frac{1}{2}}+C$
$\therefore y=-\frac{1}{4}{u}^{\frac{1}{2}}+C$
$\therefore y=-\frac{1}{4}\sqrt{1-4{x}^{2}}+C$