Solve Differential equation $2t\ y(t)\ y'(t)=t^2+3y(t)^2$ with substitution I am trying to solve this differential equation $2t\ y(t)\ y'(t)=t^2+3y(t)^2$ using substitution, with the initial value $y(1)=0$ This is what I've done so far:
$$
\begin{align*}
y'(t) &= \frac{t^2+3y(t)^2}{2ty(t)} = \frac{t^2}{2ty(t)} + \frac{3y(t)^2}{2ty(t)} \\
&= \frac{t}{2y(t)} + \frac{3y(t)}{2t} \\
&= \frac{1}{\frac{2y(t)}{t}} + \frac{3y(t)}{2t} \\
\\
\text{Substitution:}\\
f\Big(\frac{y(t)}{t}\Big) &= y'(t) \ \ \text{with} \ \ f(z(t)) = \frac{1}{2z(t)} + \frac{3}{2}z(t) \\
z(t)&= y(t) \ \ \text{and} \ \ z_0 = z(t_0) = z(1) = 0 \\
\\
z'(t) &= \frac{1}{t} \Big( f(z(t) - z(t)\Big) = \frac{1}{t} \Big( \frac{1}{2z(t)} + \frac{3z(t)}{2} - \frac{2z(t)}{2} \Big) \\
&= \frac{1}{t} \Big( \frac{1}{2z(t)} + \frac{z(t)}{2} \Big) = \frac{1}{t} \Big( \frac{1}{2z(t)} + \frac{z(t)^2}{2z(t)} \Big) \\
&= \frac{1}{t} \frac{1+z(t)^2}{2z(t)} \\
\\
\text{Seperation of variables:}\\
g(t) &= \frac{1}{t} \ ,\ \ h(z(t)) = \frac{1+z(t)^2}{2z(t)} \\
\\
\int_{t_0=1}^t \frac{1}{t} d\tau &= \Big[\ln(\tau)\Big]_1^t = ln(t) - 0 = ln(t) \\
\\
\int_{z_0=0}^{z(t)} \frac{1+x}{2x} dx &= \int_{0}^{z(t)} \frac{1}{2x} dx + \frac{x}{2x} dx = \int_{0}^{z(t)} \frac{1}{2x} dx + \int_{0}^{z(t)} \frac{1}{2} dx \\
&= \frac{1}{2} \int_{0}^{z(t)} \frac{1}{x} dx + \frac{1}{2} \int_{0}^{z(t)} 1 \  dx = \frac{1}{2} \int_{0}^{z(t)} \frac{1}{x} dx + \frac{z(t)}{2} \\
&= \frac{1}{2} \int_{\alpha}^{z(t)} \frac{1}{x} dx + \frac{z(t)}{2} = \frac{1}{2} \Big[ \ln(x) \Big]_\alpha^{z(t)} + \frac{z(t)}{2} \ \ \ \Big( \alpha \in \big(0,z(t)\big)\Big) \\
&= \frac{1}{2} \ln(z(t)) - \frac{1}{2} \ln(\alpha) + \frac{z(t)}{2} \\
\\
\lim_{\alpha \rightarrow 0} \ &\frac{1}{2} \ln(z(t)) - \frac{1}{2} \ln(\alpha) + \frac{z(t)}{2} = \frac{1}{2} \ln(z(t)) + \frac{z(t)}{2} \\
\\
\text{Now solve for z(t):} \\
ln(t) &= \frac{1}{2} \ln(z(t)) + \frac{z(t)}{2}
\end{align*}
$$
This is the point where I don't know how to continue.

Hint
$$2t\ y(t)\ y'(t)=t^2+3y(t)^2$$
It's a Bernouilli's equation.
You can solve it this way
$$t(y^2)'=t^2+3y^2$$
It becomes then a first order linear equation ($v=y^2$)
$$tv'-3v=t^2$$
That's easy to solve with integrating factor
