Solve Differential equation $2t\ y(t)\ y'(t)=t^2+3y(t)^2$ with substitution

Question

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.

Answer 1

Or start with setting $v=y^2$, $$ tv'=t^2+3v $$ which is linear with integrating factor $1/t^3$ for the normalized equation, $$ \left(\frac{v}{t^3}\right)'=\frac{tv'-3v}{t^4}=\frac1{t^2}\implies \frac{v}{t^3}=-\frac1t+C,~~v=Ct^3-t^2,\\~\\y=\pm t\sqrt{Ct-1}. $$

Answer 2

You make it more complicated than it is.

Once you substitute $y=tu$ the differential equation becomes $$2ttu(tu'+u)=t^2+3t^2u^2$$

Which after simplification and reordering gives $$\dfrac{2uu'}{1+u^2}=\dfrac 1t$$

And this integrates to $1+u^2=c|t|$ with $c\ge 0\qquad$ (i.e. $\ln|1+u^2|=\ln|t|+cst$).

Finally $$y=\pm t\sqrt{c|t|-1}$$

Answer 3

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