How to solve for $y$ here?

Question

I’ve been working on an initial value problem that I want to solve for $y$ in. I’ve reached: $$\frac{y-4}{y+1}=(e^{-t}+C)^5$$ But I’m not sure where to go from here. How can I solve for $y$? Thanks.

Note: if you all need the steps I took to get to this point, I can provide that, but either way I don’t know how to solve for $y$ from here.

Answer 1

$$\frac{y-4}{y+1}=(e^{-t}+C)^5$$ Multiply with $y+1$: $$y-4=(y+1)(e^{-t}+C)^5\\ y-4=y(e^{-t}+C)^5+(e^{-t}+C)^5\\ y-y(e^{-t}+C)^5=(e^{-t}+C)^5+4\\ y=\frac{(e^{-t}+C)^5+4}{1-(e^{-t}+C)^5}$$

Answer 2

$$\frac{y-4}{y+1}=(e^{-t}+C)^5$$ $$\frac{y+1-5}{y+1}=(e^{-t}+C)^5$$ $$1-\frac{5}{y+1}=(e^{-t}+C)^5$$ $$\frac{5}{y+1}=1-(e^ {-t}+C)^5$$ $$\frac{y+1}5=\dfrac 1{1-(e^{-t}+C)^5}$$ $${y+1}=\dfrac 5{1-(e^{-t}+C)^5}$$ $$\implies y(t)=\dfrac 5{1-(e^{-t}+C)^5}-1$$