The question states $t(\dfrac{dy}{dt}) - 3y = t^4$
As a first step I am told to divide through by $t^4$ - can anyone explain the purpose of this?
Following this I get $t^{-3} (\dfrac{dy}{dt}) - 3yt^{-4} = 1$
Then it says you can write this as $\dfrac{d}{dt}(t^{-3} y)=1$
Why can you then convert it to this? Can anyone explain with more steps this happening?
And then resulting from this how is the answer $y(t)=t^3(t+c)$ achieved?
At first one we divide by $t$ and we find $$y'=\frac3t y+t^3$$ the solutions of the homogeneous equation $$y'=\frac3t y$$ are $$y_h(t)=\lambda\exp\left(\int\frac3tdt\right)=\lambda t^3$$ and by the constant variation method a particular solution has the form $$y_p(t)=\lambda(t)t^3$$ where $$\lambda'(t)t^3=t^3\Rightarrow \lambda(t)=t$$ hence the solution of the given differential equation are $$y(t)=\lambda t^3+t^4,\qquad \lambda\in\Bbb R$$
Lets take a look at this line from your post: $$t^{-3}\frac{dy}{dt}-3t^{-4}y = 1$$ If we say $u=t^{-3}$ and $v=y$ then we get, $$u\cdot v' + u'\cdot v = 1$$ Which looks like our product rule for derivatives so we can rewrite this as $\frac{d}{dt}\left(u\cdot v\right)$. Substituting back in we are left with, $$\frac{d}{dt}\left( t^{-3}y \right) = 1$$ To find the solution just integrate both sides, $$\int \frac{d}{dt}\left( t^{-3}y \right)dt = \int 1 dt$$ $$t^{-3}y = t + c$$ $$y = t^3(t+c)$$ Hope this helps.
