Solve the differential equation 1/4x times dy/dx = y times √(2x^2-1).

Question

Solve the differential equation 

A.

B.

C.

D. None of these

Answer 1

dy = 4xdx * sqrt(2x²-1) * y

1/y dy = 4x sqrt2x²-1) dx

∫ 1/y dy = ∫ 4x sqrt(2x²-1) dx

On the left side:

ln |y| (There is a technically a +C here but we don't have to attach it here)

On the right side:

∫ 4x sqrt(2x²-1) dx

Using u-sub:

u = 2x²-1

du = 4x

∫ u^1/2 du = 2/3 u^3/2

2/3 u^3/2 = 2/3 (2x²-1)^3/2 + C

Equating the two:

ln |y| = 2/3 (2x²-1)^3/2 + C

e^ln|y| = e ^{2/3 (2x^2 - 1)^(3/2) + C}

y = e ^{2/3 (2x^2 - 1)^(3/2) + C}

So the answer is D) None of those.

If still confused/have questions feel free to reach out to me.