Question

Solve the differential equation.

dz/dt + e^(t + z) = 0

Answer 1

$\frac{dz}{dt} + e^{t + z} = 0 \ \Rightarrow\ \frac{dz}{dt} = - e^{t + z} \ \Rightarrow\ \frac{dz}{dt} = - e^{t}e^z \ \Rightarrow\ \\ \\ \frac{dz}{e^z} = -e^{t} dt \ \Rightarrow\ \int e^{-z} dz=\int -e^{t} dt \ \Rightarrow\ \\ \\ -e^{-z} = -e^{t} + C \ \Rightarrow\ e^{-z} = e^{t} - C \ \Rightarrow\ -z = \ln(e^{t} - C)\ \Rightarrow\ \\ \\ z = -\ln(e^{t} - C)$

Answer 2

By solving given differential equation, Solution is $z=-ln(e^{t}-C)$

What is a differential equation?

A differential equation is an equation that contains at least one derivative of an unknown function, either an ordinary derivative or a partial derivative.

Given differential equation

$\frac{dz}{dt}+e^{t+z} = 0$

⇒ $\frac{dz}{dt} = -e^{t+z}$

⇒ $\frac{dz}{dt}=-e^{t}e^{z}$

⇒ $\frac{dz}{e^{z} }= -e^{t} dt$

Apply integration on both sides

⇒ ∫$e^{-z} dz=$ ∫ $-e^{-t} dt$

⇒ $-e^{-z}=- e^{t} +C$

⇒ $e^{-z}=e^{t} -C$

⇒ $-z = ln(e^{t} -C)$

⇒ $z=-ln(e^{t}-C)$

By solving given differential equation, Solution is $z=-ln(e^{t}-C)$

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