How do you solve ydx=(ye^(y^3) - 2x)dy?

Question

ODE solution

Answer 1

This one is a little tricky. This one is easy to solve as y being the dependent variable.
ydx = (ye^(y^3) - 2x)dy
divide both sides by dy
y(dx/dy) = ye^(y^3) - 2x
Add 2x to both sides
yx' + 2x = ye^(y^3)
divide both sides by y
x' + (2/y)x = e^(y^3)

Now to find the integrating factor
μ = e^∫(2/y) = e^(2lny) = e^(ln y^2) = y²

Now multiply both sides by μ
y²x' + (2y)x = y²e^(y^3)

The left hand side can be combined using the product rule:
(y²x)' = y² e^(y^3)dy
Integrate both sides:
y²x = ∫y² e^(y^3)dy
Multiply the right-hand side by 3 and (1/3)
y²x = (1/3)∫3y² e^(y^3)dy

To integrate right-hand side, make a u-substitution:
u = y^3
du = 3y^2 dy
Now substitute into the integral
y²x = (1/3)∫e^(u)du
y²x = (1/3)e^(u) + C
but u = y^3
y²x = (1/3)e^(y^3) + C
Divide both sides by y²:
x = (1/(3y²))e^(y^3) + (C/y²), where C is a constant

Answer 2

The trick is to solve for x, not y. So letting x' be dx/dy, we have:

yx' = ye^(y³) - 2x
yx' + 2x = ye^(y³)
x' + 2/y x = e^(y³)

The integrating factor will be e^(∫2/y dy) = e^(2 ln y) = y², so we have:

y²x' + 2yx = y²e^(y³)

Now just integrate both sides with respect to y:

y²x = ∫y²e^(y³) dy = e^(y³)/3 + C
x = e^(y³)/(3y²) + C/y²

And we are done. Now if you absolutely have to have this in terms of y, I don't think it's possible to get anything more than an implicit solution in this case.