<s>[INST] Calculate the general solution and the specific solution of the following differential equation: $(3y - 1)^2) (y'^2) = 4y$ Calculate the general solution and the specific solution of the following differential equation: $(3y - 1)^2  (y')^2 = 4y.$
The general solution is $(x + C)^2 = y(y - 1)^2$ and the specific is $y = 0$.
I have tried rewriting it like $y' = \dfrac{2y}{3y + 1}$ but I don't know what to do next.
 [/INST] You have got $\displaystyle y'=\frac{2\sqrt y}{3y-1}$
Therefore $\displaystyle \frac{dy}{dx}=\frac{2\sqrt y}{3y-1}$
$\displaystyle\implies\int \frac{3y-1}{2\sqrt y}dy=\int dx$
$\implies\displaystyle \frac{3}{2}\int\sqrt y-\frac{1}{2}\int\frac{dy}{\sqrt y}=x+C$
$\implies \displaystyle y^{\frac{3}{2}}-y^{\frac{1}{2}}=x+C$
Squaring both sides we get
$y^3+y-2y^2=(x+C)^2\implies y(y-1)^2=(x+C)^2$
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