Karl wrote:Substitute u^3 = y. Hence 3u^2 (du/dx) = dy/dx. Equation becomes

3xu^2 (du/dx) = 6u^3 + 12x^4u^2

Divide out 3u^2 and the equation is linear and can be solved for u in the usual way for linear first order equations. Then backsubstitute u to get y.

Aren't you the same Karl that runs Karl's Calculus Tutor?

Anyway, I'm just making sure that I've done everything correctly as you said:

u^3 = y

3u^2 (du/dx) = dy/dx

3xu^2 (du/dx) = 6u^3 + 12x^4u^2

Now, the next part, dividing out 3u^2 as you said I should leaves me with:

x(du/dx) = 2u + 4x^2

Dividing both sides by x gives:

du/dx = (2/x)u + 4x

(du/dx) - (2/x)u = 4x

I suppose the next step is to solve for the integrating factor (which I'll let be p(x)):

p(x) = e^integral[(-2/x)] dx

p(x) = e^(-2ln|x| + C)

p(x) = Ce^ln(1/x^2)

p(x) = C/x^2

Multiplying the ODE by this integrating factor (omitting the C since it's sort of extraneous anyway, so now it's just x^-2) gives me:

(x^-2)(du/dx) - (2x^-3)u = 4x^-1

The left side can be converted to:

(d/dx)[ux^-2] = 4x^-1

Multiplying both sides by dx:

d(ux^-2) = 4x^-1 dx

Integrating both sides:

ux^-2 = 4ln|x| + C

Dividing both sides by x^-2 leaves:

u = 4x^2ln|x| + Cx^2

u^3 = (4x^2ln|x| + Cx^2)^3

y = (4x^2ln|x| + Cx^2)^3

That should be the solution, from what I calculated.

Right?