To solve the differential equation:

\[
\frac{dy}{dt} = \frac{1}{e^{-y}} + e^y
\]
with the initial condition \( y(0) = 0 \), let's follow these steps:

### Step 1: Simplify the equation
First, notice that:

\[
\frac{1}{e^{-y}} = e^y
\]

Thus, the equation becomes:

\[
\frac{dy}{dt} = e^y + e^y = 2e^y
\]

### Step 2: Separate variables
We can now separate variables by dividing both sides by \( 2e^y \):

\[
\frac{1}{2e^y} dy = dt
\]

### Step 3: Integrate both sides
Now, integrate both sides:

\[
\int \frac{1}{2e^y} dy = \int dt
\]

The left side is a simple exponential integral, and the right side is just \( t \). So we get:

\[
\frac{1}{2} \int e^{-y} dy = t
\]

The integral of \( e^{-y} \) is \( -e^{-y} \), so:

\[
-\frac{1}{2} e^{-y} = t + C
\]

### Step 4: Solve for \( y \)
Multiply both sides by -2:

\[
e^{-y} = 2(-t - C)
\]

Taking the natural logarithm of both sides:

\[
-y = \ln(2(-t - C))
\]

Thus:

\[
y = -\ln(2(-t - C))
\]

### Step 5: Apply the initial condition \( y(0) = 0 \)
Substitute \( t = 0 \) and \( y(0) = 0 \) into the equation:

\[
0 = -\ln(2(0 - C))
\]

So, \( \ln(-2C) = 0 \), which means \( -2C = 1 \) or \( C = -\frac{1}{2} \).

### Final solution:
Substitute \( C = -\frac{1}{2} \) into the equation:

\[
y = -\ln(2(-t + \frac{1}{2}))
\]

This is the solution to the differential equation.

Would you like further details or clarifications?

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Here are 5 related questions to expand on this topic:
1. How would the solution change if the initial condition were different?
2. What is the behavior of the solution as \( t \to \infty \)?
3. Can this method be applied to non-exponential differential equations?
4. How do you check the validity of the solution in this context?
5. What happens to the solution if the equation includes an additional term?

**Tip:** Always check your solution by substituting it back into the original differential equation to verify its accuracy.

Explore the step-by-step solution to the differential equation dy/dt = 1/e^-y + e^y with the initial condition y(0)=0. This guide covers concepts like separation of variables and exponential function integration, suitable for undergraduate calculus students.

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