Find general solution to the equation $xy^{'}=2x^2\sqrt{y}+4y$ This is Bernoulli's differential equation:
$$xy^{'}-4y=2x^2y^{\frac{1}{2}}$$
Substitution $y=z^{\frac{1}{1-\alpha}},\alpha=\frac{1}{2},y^{'}=z^{'2}$ gives $$xz^{'2}-4z^2=2x^2z$$
Is this correct? What is the method for solving this equation?

We are given
$$xy^{'}-4y=2x^2y^{\frac{1}{2}}$$
Let $z = y^{1-\frac{1}{2}} = y^{\frac 12}.$ Then
$$z' = \frac12 y^{-\frac12}y' \implies y' = 2\frac{z'}{y^{-\frac12}}.$$With this, we get
$$2x\frac{z'}{y^{-\frac12}}-4y=2x^2y^{\frac{1}{2}}$$
Now multiply by $y^{-\frac12}$. We get
$$2x{z'}-4y^{\frac12}=2x^2$$ which is the same as
$$2x{z'}-4z=2x^2$$
Now solve this last equation (linear first order).
