The equation in the image is:

$\frac{dy}{dx} + 2xy = \sin x + 3x^2$

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

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

• $P(x) = 2x$,
• $Q(x) = \sin x + 3x^2$.

We solve this using the method of integrating factors.

Step 1: Find the integrating factor $\mu(x)$

The integrating factor is given by:

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

1. Compute $\int 2x \, dx$: $\int 2x \, dx = x^2.$

2. Therefore: $\mu(x) = e^{x^2}.$

Step 2: Multiply through by the integrating factor

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

$e^{x^2} \frac{dy}{dx} + 2x e^{x^2} y = e^{x^2} (\sin x + 3x^2).$

Notice the left-hand side is the derivative of $e^{x^2} y$:

$\frac{d}{dx} \left( e^{x^2} y \right) = e^{x^2} \sin x + 3x^2 e^{x^2}.$

Step 3: Integrate both sides

Integrate both sides with respect to $x$:

$\int \frac{d}{dx} \left( e^{x^2} y \right) \, dx = \int e^{x^2} \sin x \, dx + \int 3x^2 e^{x^2} \, dx.$

The left-hand side simplifies to:

$e^{x^2} y.$

Thus:

$e^{x^2} y = \int e^{x^2} \sin x \, dx + \int 3x^2 e^{x^2} \, dx + C,$ where $C$ is the constant of integration.

Step 4: Solve for $y$

Divide through by $e^{x^2}$:

$y = \frac{1}{e^{x^2}} \left[ \int e^{x^2} \sin x \, dx + \int 3x^2 e^{x^2} \, dx + C \right].$