Using substitution to make an equation into a separable differentiable equation

Question

I have the question:

By making the substitution $y = t^nz$ and making a cunning choice of n, show that the following equations can be reduced to separable equations and solve them. $$\dfrac{dy}{dt} = \dfrac{1-ty^2}{2t^2y}$$

In my answer book, the first step it changes the equation to the following:

$$t^n\dfrac{dz}{dt}+nt^{n-1}z = \dfrac{1-t^{2n+1}z^2}{2t^{n+2}z}$$

I understand how the rhs was derived with the obvious substitution but can someone please explain where the lhs came from? would appreciate it done in small steps if needed! Thank you

Answer 1

Apply the product rule to $\frac{dy}{dt}$ with $y=t^nz$: $$\dfrac{dy}{dt}=\dfrac{d}{dt}\left[t^nz\right]=t^n\dfrac{dz}{dt}+z\dfrac{d}{dt}\left[t^n\right]=t^n\dfrac{dz}{dt}+nt^{n-1}z$$

Answer 2

Hint:

$$ y=t^nz \Rightarrow \dfrac{dy}{dt}=nt^{n-1}z+\dfrac{dz}{dt}t^n $$ and: $$ 1-ty^2=1-t(t^nz)^2=1-t^{2n+1}z^2 $$ $$ 2t^2y=2t^2(t^nz)=2t^{n+2}z $$