Solving bernoulli differential equation How to solve $$t \frac{dy}{dt} + y = t^4 y^3$$
First I divided by $t$ to get $$\frac{dy}{dt} + \frac{y}{t} = t^3 y^3$$
Then I multiplied through by $y^{-3}$ to get $$y^{-3} \frac{dy}{dt} + \frac1{ty^2} = t^3$$
Then I used the subsitution $w = y^{-2}$ and $w'=-2y^{-3}\frac{dy}{dt}$
Subbing that in the equation becomes $$\frac1{y^3} \frac{dy}{dt} + \frac1{ty^2} = t^3$$
How do I solve it from here as I still have a $\frac1{ty^2}$ instead of just $y$ making it linear?
Also are all the rest of my workings correct?

You almost find the solution. You've made a good substitution by letting $w=\frac{1}{y^2}$ and $w'=-\frac{2y'}{y^3}$, then
\begin{align}
\frac{1}{y^3}\frac{dy}{dt}+\frac{1}{ty^2}&=t^3\qquad\rightarrow\qquad-\frac{1}{2}w'+\frac{w}{t}=t^3\qquad\rightarrow\qquad w'-\frac{2}{t}w=-2t^3
\end{align}
The first-order nonlinear ODE becomes a first-order linear ODE. Take the integrating factor
\begin{align}
e^{-\Large\int\;\frac{2}{t}\,dt}=e^{-2\ln t+C}=e^{\ln t^{-2}}e^C=At^{-2}
\end{align}
then multiply both sides of the first-order linear ODE by $At^{-2}$.
\begin{align}
At^{-2}w'-At^{-2}\frac{2}{t}w&=-2t^3At^{-2}\\
At^{-2}w'-2At^{-3}w&=-2At\\
\frac{d}{dt}\left(At^{-2}w\right)&=-2At\\
d\left(At^{-2}w\right)&=-2At\;dt
\end{align}
The last step is integrating both sides.
