Can we use the Lambert W solution $y=-3W(K_2x^{-4/3})$ instead of $y =-3W(\frac 1 3\sqrt[3]{-\frac{K_1}{x^4}})$ if we choose an appropriate constant?

Question

During the process of solving the separable differential equation $4y - x(y-3)y' = 0$, our solution acquires a constant when we go from $\frac 4 x = \frac{y-3}{y}y'$ to $\ln x + C_1 = y - 3 \ln y$. We then do: $$ e^{-\frac{4}{3}(\ln x) -\frac{C_1}{3}} = e^{\ln y - \frac y 3}$$ $$ x^{-\frac 4 3}e^{-\frac{C_1}3}= ye^{-\frac y 3}$$ $$-\frac 1 3 x^{-\frac 4 3}e^{-\frac{C_1}3}= -\frac y 3 e^{-\frac y 3}$$ I substitute $- \frac 1 3 e^{\frac {C_1}3}$ by the constant $K_1$ to get $$ K_1x^{-\frac 4 3}=-\frac y 3 e^{-\frac y 3}$$ applying the Lambert W function I get: $$-3W(K_1x^{-\frac 4 3}) = y$$

Instead, wolframalpha suggests me the solution $y = -3W\left( \frac 1 3\sqrt[3]{-\frac{K_2}{x^4}}\right)$.

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My question is: is the process through which I substitute in the constant $K_1$ legitimate? Since otherwise, I'm unable to determine why my answer doesn't correspond to the wolframalpha one. The reason why I thought I was allowed to substitute $K_1$ was because, in the step where we go from $\frac 4 x = \frac{y-3}{y}y'$ to $\ln x + C_1 = y - 3 \ln y$, we obviously see that taking the derivative of any constant makes our equation true, and therefore we would be free to choose whichever $K_1$ makes our equation look nice. But is this reasoning correct, given that wolframalpha provides a more complicated solution?

Answer 1

Your answer is correct, and comparing the two constants we have $K_{1}=-\frac{1}{3}K_{2}^\frac{1}{3}$.

After integration you can also define the constant of integration as follows

$$\frac{4}{x}=\frac{y-3}{y}y'$$ $$4\ln(x)-\ln(K_2)=y-3\ln(y)$$ $$\dots$$ $$-\frac{y}{3}e^{-\frac{y}{3}}=\frac{1}{3}\sqrt[3]{-\frac{K_2}{x^4}}$$

Or instead of setting $K_{1}=-\frac{1}{3}e^{-\frac{C_1}{3}}$, set $K_{2}=e^{-C_{1}}$, to obtain the solution from WA.