what is the general solution of the differential equation? sqrt(1-4x^2)y'=x?

Question

I need help on trying to at least start this problem.

Answer 1

y' = x / sqrt( 1-4x^2) = (1/4) (1/2)(8x) / sqrt(1-4x^2)
so y = constant - (1/4) sqrt( 1-4x^2)
I dont think there any general method to bring out here ... you should just spot that y = sqrt(1-4x^2) would differentiate to -8x/sqrt(1-4x^2)

Answer 2

You need to integrate xdx/sqrt(1-4x^2)
Let u = 4x^2
Then du=8xdx --> xdx = du/8
So integral becomes 1/8integral [du/sqrt(1-u)]
= (1/8)[{2(-u-2)/3}{sqrt(1-u)}] +C
= [-(u+2)/12][sqrt(1-u)] +C
= [-(4x^2+2)][sqrt(1-4x^2] + C

Answer 3

Divide both sides by the square root:
dy/dx = x/sqr[1-4x^2]
dy = {x/sqr[1-4x^2]}dx
Integrate both sides

Answer 4

rearange to get:

dy/dx = x/(1-2x) = x-0.5

therefore:

y = 1*x^0 -0

therefore:

y = 1