The given equation is:

$\frac{dy}{dx} = e^x - 3y$

This is a first-order linear differential equation. To solve it, we can use the method of integrating factors. Here's the step-by-step approach:

Step 1: Rewrite the equation in standard form

The standard form of a first-order linear differential equation is:

$\frac{dy}{dx} + P(x)y = Q(x)$

Rewriting the given equation:

$\frac{dy}{dx} + 3y = e^x$

Here, $P(x) = 3$ and $Q(x) = e^x$.

Step 2: Find the integrating factor

The integrating factor $\mu(x)$ is given by:

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

Since $P(x) = 3$:

$\mu(x) = e^{\int 3 \, dx} = e^{3x}$

Step 3: Multiply both sides by the integrating factor

Multiplying the entire equation by $e^{3x}$:

$e^{3x} \frac{dy}{dx} + 3e^{3x}y = e^{3x} e^x$

Simplifying the right-hand side:

$e^{3x} \frac{dy}{dx} + 3e^{3x}y = e^{4x}$

The left-hand side is the derivative of $y \cdot e^{3x}$:

$\frac{d}{dx}(y \cdot e^{3x}) = e^{4x}$

Step 4: Integrate both sides

Integrating both sides with respect to $x$:

$y \cdot e^{3x} = \int e^{4x} \, dx$

The integral of $e^{4x}$ is:

$y \cdot e^{3x} = \frac{e^{4x}}{4} + C$

Step 5: Solve for $y$

Finally, solve for $y$ by dividing both sides by $e^{3x}$:

$y = \frac{e^{4x}}{4e^{3x}} + \frac{C}{e^{3x}} = \frac{e^x}{4} + Ce^{-3x}$

Final Solution:

The general solution to the differential equation is:

$y = \frac{e^x}{4} + Ce^{-3x}$

Would you like more details or have any further questions?

Related questions:

1. How do you solve non-linear first-order differential equations?
2. What is the method of separation of variables in solving differential equations?
3. How do you determine a particular solution given initial conditions?
4. What are homogeneous and non-homogeneous differential equations?
5. Can you explain the method of undetermined coefficients for solving non-homogeneous equations?

Tip:

When solving first-order linear differential equations, always check if the equation is already in standard form before applying the integrating factor method.