Question

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

Answer 1

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

Explanation:

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`