Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 505: 17

$$ y =\ln(4t^5+c)$$

We have $$\frac{dy}{dt}-20t^4e^{-y}=0\Longrightarrow e^{y}dy=20t^4\\ \Longrightarrow\int e^{y}dy=\int 20t^4dt+c\\ \Longrightarrow e^{y} =4t^5+c \Longrightarrow y =\ln(4t^5+c).$$ Hence, the general solution is $$ y =\ln(4t^5+c)$$