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

1 Answer
Steve M
Nov 8, 2016

# 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 #

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