The given problem is a first-order linear differential equation with an initial condition:

$\frac{dy}{dx} + 2y = 2 - 7x, \quad y(0) = 0$

Step 1: Identify the standard form

A first-order linear differential equation has the form:

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

In this case, $P(x) = 2$ and $Q(x) = 2 - 7x$.

Step 2: Find the integrating factor

The integrating factor is given by:

$\mu(x) = e^{\int P(x) \, dx} = e^{\int 2 \, dx} = e^{2x}$

Step 3: Multiply the differential equation by the integrating factor

Multiply the entire equation by $e^{2x}$:

$e^{2x} \frac{dy}{dx} + 2e^{2x} y = e^{2x} (2 - 7x)$

The left-hand side simplifies to:

$\frac{d}{dx} \left( e^{2x} y \right) = e^{2x} (2 - 7x)$

Step 4: Integrate both sides

Now integrate both sides with respect to $x$:

$\int \frac{d}{dx} \left( e^{2x} y \right) dx = \int e^{2x} (2 - 7x) dx$

The left side simplifies to:

$e^{2x} y = \int e^{2x} (2 - 7x) \, dx$

Step 5: Compute the integral

To compute $\int e^{2x} (2 - 7x) \, dx$, break it into two parts:

1. $\int 2e^{2x} \, dx = \frac{2}{2} e^{2x} = e^{2x}$
2. $\int -7x e^{2x} \, dx$

Use integration by parts for the second integral. Let:

• $u = x$, so $du = dx$
• $dv = -7e^{2x} dx$, so $v = \frac{-7}{2} e^{2x}$

Using the formula $\int u dv = uv - \int v du$, we get:

$\int -7x e^{2x} dx = \frac{-7x}{2} e^{2x} + \int \frac{7}{2} e^{2x} dx = \frac{-7x}{2} e^{2x} + \frac{7}{4} e^{2x}$

Thus, the integral becomes:

$e^{2x} y = e^{2x} - \frac{7x}{2} e^{2x} + \frac{7}{4} e^{2x}$

Step 6: Solve for $y(x)$

Now divide both sides by $e^{2x}$:

$y = 1 - \frac{7x}{2} + \frac{7}{4}$

Simplify:

$y(x) = \frac{11}{4} - \frac{7x}{2}$

Step 7: Apply the initial condition

Use the initial condition $y(0) = 0$:

$0 = \frac{11}{4} - \frac{7 \cdot 0}{2} = \frac{11}{4}$

This leads to a contradiction with the initial condition, suggesting an issue in the setup. Let's revisit