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

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Answer 1 · 2016-11-08T11:21:21

$y = -1/4 sqrt(1-4x^2) + C$

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

$sqrt(1-4x^2)dy/dx = x$ which is a First Order separable DE which we can rearrange as follows:

$dy = x/sqrt(1-4x^2) dx => int dy = int x/sqrt(1-4x^2) dx$

To integrate the RHS we need to use a substitution

Let $u=1-4x^2 => (du)/dx = -8x => -1/8du = xdx$

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

$y = int (-1/8)/sqrt(u) du$
$:. y = -1/8 int u^(-1/2) du$
$:. y = -1/8 u^(1/2)/(1/2) + C$
$:. y = -1/4 u^(1/2) + C$
$:. y = -1/4 sqrt(1-4x^2) + C$

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