The given differential equation is:

$y' + y = e^x$

This is a first-order linear differential equation of the form:

$y' + P(x)y = Q(x)$

Here, $P(x) = 1$ and $Q(x) = e^x$.

To solve this, we use the method of integrating factors. The integrating factor $\mu(x)$ is given by:

$\mu(x) = e^{\int P(x) \, dx} = e^{\int 1 \, dx} = e^x$

Now, multiply both sides of the original equation by $e^x$:

$e^x y' + e^x y = e^x e^x$

This simplifies to:

$\frac{d}{dx} \left( e^x y \right) = e^{2x}$

Next, integrate both sides with respect to $x$:

$e^x y = \int e^{2x} \, dx$

The right-hand side is:

$\int e^{2x} \, dx = \frac{1}{2} e^{2x} + C$

Thus, we have:

$e^x y = \frac{1}{2} e^{2x} + C$

Finally, solve for $y$ by dividing both sides by $e^x$:

$y = \frac{1}{2} e^x + Ce^{-x}$

This is the general solution to the differential equation.

Would you like a more detailed explanation of any steps?

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Here are some related questions:

1. How do you apply the integrating factor method to non-homogeneous equations?
2. How does the solution change if $Q(x)$ is a different function?
3. What is the significance of the constant $C$ in the solution?
4. How can we verify the solution by substituting it back into the original equation?
5. How can the method of integrating factors be extended to higher-order differential equations?

Tip: Always check your solution by substituting it back into the original differential equation to ensure it satisfies the equation.