This ODE is separable as
1/(6x) dy/dx = y √(3x ² - 1)
→ dy / y = 6x √(3x ² - 1) dx
Integrate both sides:
∫ dy / y = ∫ 6x √(3x ² - 1) dx
The left side is trivial. For the right side, substitute u = 3x ² - 1 and du = 6x dx :
∫ dy / y = ∫ √u du
ln|y | = u ³′² + C
(that is, u is raised to the 3/2 power)
ln|y | = (3x ² - 1)³′² + C
Solve for y by taking the exponential of both sides:
exp(ln|y |) = exp((3x ² - 1)³′² + C )
y = exp((3x ² - 1)³′²) × exp(C )
y = C exp((3x ² - 1)³′²)
which can be written as
y = C exp(√((3x ² - 1)³))
which makes the answer none of these; this solution can't be expressed as either option given in A or B.
