(1-4x^2)^1/2 y'=x?

Answer 1

(1 - 4x^2)^1/2 (dy/dx)=x

Divide both sides to isolate y'.

(dy/dx) = x/(1 - 4x^2)^(1/2)

Since this is in the topic of differential equations, let's multiply both sides by dx.

dy = x/(1 - 4x^2)^(1/2) dx

Integrate both sides;

y = ∫ (x/(1 - 4x^2)^(1/2) dx)

And this has just become a Calculus II question.
Use substitution. I'm going to rearrange it to make the substitution obvious.

y = ∫ (1/(1 - 4x^2)^(1/2) x dx)

Let u = 1 - 4x^2. Then
du = -8x dx, so
(-1/8) du = x dx.
{Note: x dx is the tail end of our integral, so (-1/8) du will be the tail end after the substitution.}

y = ∫ ( 1/u^(1/2) (-1/8) du )

Pull out the factor (-1/8),

y = (-1/8) ∫ ( 1/u^(1/2) du )

Change that to u^(-1/2),

y = (-1/8) ∫ (u^(-1/2) du )

And now we can just use the reverse power rule.

y = (-1/8) ( 2u^(1/2) ) + C
y = (-1/4) u^(1/2) + C

But u = 1 - 4x^2, so back-substituting, we get

y = (-1/4) [1 - 4x^2]^(1/2) + C

Answer 2

of course y and z can not the two be even via fact this makes LHS even and RHS strange. Can y, z be equivalent? if so then x*y^2 = 2y^2 + a million ----> (x - 2)*(y^2) = a million and clearly x = 3, y = a million is the only answer with integers. from now on i will assume that y is the extra desirable of y, z as reversed recommendations are fairly the comparable. Can z = a million be a answer devoid of y = a million? xy = y^2 + 2 ----> y^2 - xy + 2 = 0 ----> y = [x + sqrt(x^2 - 8)]/2 x = 3 ends up in y = 2 so x = 3, y = 2, z = a million is a answer with all 3 different. Can z = 2? 2xy = y^2 + 5 ----> y^2 - 2xy + 5 = 0 ----> y = [2x + sqrt(4x^2 - 20)]/2 = x + sqrt(x^2 - 5) This has answer x = 3, y = 5, z = 2. For sqrt(x^2 - 5) to be an integer we want x^2 - 5 = ok^2 yet 2^2 and 3^2 are the only squares differing with the help of 5 so this ends up in no different recommendations. I even have discovered, yet won't placed the evidence here, that z (the smaller of y, z submit to in innovations) will not be able to be 3 or 4. This makes me suspect that there are not the different recommendations. EDIT. That final line is incorrect! i've got basically discovered that x = 3, y = 13, z = 5 is a answer. whether, that's finding as though x would desire to be 3.

Answer 3

dy/dx= x/(1-4x^2)^1/2
so y+C = Int xdx/(1-4x^2)^1/2
If bwe call 1-4x^2 = z -8xdx = dz
so we get the integral of
-1/8 *1/z^1/2 dz = -1/4 z^1/2 = -1/4 (1-4x^2)1/2
so
y=-1/4(1-4x^2)1/2 +C