The general solution of y' = 6 x^4 y^1/3 − 3y/x is:

Rearrange to get:

y' + 3y/x = 6(x^4)*[y^(1/3)]

This is the form of a Bernoulli differential equation

so make u = y^(1-[1/3]) = y^(2/3), y = u^(3/2) = u^1.5

u'= (2/3)*[y^(-1/3)]*(y') (differentiate with respect to x)

y' = (3/2)(y^[1/3]) * (u')

Now substitute everything in.

[ (3/2)([u^1.5]^[1/3]) * (u') ] + 3*[u^1.5]/x = 6(x^4)*([u^1.5]^[1/3])

Looks messy, but [u^1.5]^[1/3] = u^0.5, and divide everything by u^0.5 to get:

(3/2)(u') + 3*u/x = 6(x^4)

u' + 2u/x = 4(x^4)

Integrating factor is x^2

(x^2)u' + 2xu = 4(x^6)

([x^2]*u)' = 4(x^6)

Integrate both sides with respect to x

(x^2)u = (4/7)(x^7) + C

u = (4/7)(x^5) + C/(x^2)

y^(2/3) = (4/7)(x^5) + C/(x^2)


y = [ (4/7)(x^5) + C/(x^2) ]^(3/2) <-------Final Answer

All you have to do is reduce the expression 6 x^4 y^1/3 by removing a factor of 3 from the numerator and the denominator leaving you with: 2x^4y - 3y/x