How to evaluate the following ODE?

Question

Suppose that a hazard function is described by: $$ \phi(t)=\frac{-\mathcal{R}'(t)}{\mathcal{R}(t)} $$ if $\phi(t)=16t^3$, how to evaluate $\mathcal{R}(t)$?

Answer 1

Note that

$$(\ln(R))'=\frac {R'}R$$ Therefore $$\phi(t)=\frac{-\mathcal{R}'(t)}{\mathcal{R}(t)}$$ $$16t^3=\frac{-\mathcal{R}'(t)}{\mathcal{R}(t)}$$ $$\ln(R)=-16\int t^3dt$$ $$\ln(R)=-4{t^4}+K$$ $$ \implies R(t)=Ke^{- 4{t^4}}$$