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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?

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You didn't do the substitution. With $w = y^{-2}$, $y^{-3}\; dy/dt = -\dfrac{1}{2} dw/dt$ and $1/(t y^2) = w/t$, so you should get $-w'/2 + w/t = t^3$.

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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.

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