Question

\({\bf{1 - 8}}\)Solve the differential equation.

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

Short answer

The solution is\(z = - \ln \left( {{e^t} - C} \right)\)

Step-by-step solution

Definition

A differential equation is an equation with one or more derivatives of a function.

Separate variables

Consider the differential equation

\(\begin{aligned}\frac{{dz}}{{dt}} + {e^{t + z}} &= 0\\\frac{{dz}}{{dt}} &= - {e^{t + z}}\\\frac{{dz}}{{dt}} &= - \left( {{e^t} \cdot {e^z}} \right)\\{e^{ - z}}dz &= - {e^t}dt\end{aligned}\)

Integrate

Integrating we have:

\(\begin{aligned}\int {{e^{ - z}}} dz &= - \int {{e^t}} dt\\ - {e^{ - z}} &= - {e^t} + C\\{e^{ - z}} = {e^t} - C\\ - z &= \ln \left( {{e^t} - C} \right)\\z &= - \ln \left( {{e^t} - C} \right)\end{aligned}\)

Therefore the solution of differential equation\(\frac{{dz}}{{dt}} + {e^{t + z}} = 0\) is given by\(z = - \ln \left( {{e^t} - C} \right)\).