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.