Separable equation with $u$-substitution

Question

I'm trying to solve this separable equation. Answer: x^3-y^3=c (Apparantly my calculus book only have answer for every second assignment) $$\frac{dy}{dx}=\frac{3y-1}{x}\\ \frac{dy}{dx}=\frac{3y-1}{x} \Rightarrow \frac{1}{3y-1}dy=\frac{1}{x}dx\\ u-substition\ where\ u=3y-1\ and\ \frac{du}{3}=dy\\ \int\frac{1}{3}\frac{1}{u}=\int\frac{1}{x}dx\Rightarrow \frac{1}{3}\int\frac{1}{u}=\int\frac{1}{x}dx \\ \frac{1}{3}ln(u)+c=ln(x)+c\Rightarrow \frac{1}{3}ln(3y-1)+c=ln(x)+c\Rightarrow \sqrt[3]{3y-1}=x+c\Rightarrow x^3-3y=c$$

I know I did something wrong, but I can't see what. Can you tell me where I broke the rule of maths?

EDIT: Is this a valid answer? $$\frac{1}{3}ln(3y-1)+c=ln(x)+c\Rightarrow ln(3y-1)+c=3ln(x)+3c\Rightarrow\\ 3y-1=e^{3ln(x)}e^{3c} \Rightarrow 3y-1=x^3c\Rightarrow 3y=x^3c+1\Rightarrow y=\frac{x^3c}{3}+\frac{1}{3}$$

Answer 1

You didn't break the rule of mathematics. Your result is first integral and it is proper solution. There's no mistake in your reasoning and it's done in a proper way. To be sure that your answer is good you can check if the obtained result is a first integral.

I remind that the first integral is a function $\phi\in C^1$ which fulfills

$$ f\dfrac{\partial \phi}{\partial x} + g\dfrac{\partial \phi}{\partial y} =0. $$

Answer 2

You didn't break any rules. However, you've added the integration constant $+c$ on both sides after integration, which cancels out..

From $\ln(3y-1)^{\frac{1}{3}}=\ln(x)+c$ we obtain $(3y-1)^{\frac{1}{3}}=xe^{c}$ and thus $3y-1=x^{3}e^{3c},$ which gives the solution $y=c_{1}x^{3}+\frac{1}{3}$ where $c_{1}=e^{3c}=constant.$

Then it's always good to check that the solution satisfies the given equation. In this case we have $\frac{dy}{dx}=3c_{1}x^{2}$ and $\frac{3y-1}{x}=\frac{3(c_{1}x^{3}+\frac{1}{3})-1}{x}=\frac{3c_{1}x^{3}}{x}=3c_{1}x^{2}$ as required.