Step 1: Understanding the Question:
We are given a first-order differential equation and an initial value condition (it passes through the origin, meaning $x=0, y=0$). We need to find the specific particular solution.
Step 2: Key Formula or Approach:
The given equation can be arranged into the standard form of a Linear Differential Equation:
$\frac{dy}{dx} + P(x)y = Q(x)$.
We solve this by finding the Integrating Factor (IF), which is $e^{\int P(x) dx}$.
The general solution is then given by: $y \times \text{IF} = \int (Q(x) \times \text{IF}) dx + C$.
Step 3: Detailed Explanation:
Rewrite the given equation by dividing entirely by $(1 + x^2)$:
$$\frac{dy}{dx} + \left( \frac{2x}{1 + x^2} \right) y = \frac{4x^2}{1 + x^2}$$
Here, $P(x) = \frac{2x}{1+x^2}$ and $Q(x) = \frac{4x^2}{1+x^2}$.
Find the Integrating Factor (IF):
$$\text{IF} = e^{\int \frac{2x}{1+x^2} dx}$$
Since the numerator is the exact derivative of the denominator, the integral is $\ln|1+x^2|$.
$$\text{IF} = e^{\ln(1+x^2)} = 1 + x^2$$
Write the general solution equation:
$$y(1 + x^2) = \int \left( \frac{4x^2}{1 + x^2} \times (1 + x^2) \right) dx + C$$
The $(1+x^2)$ terms brilliantly cancel out on the right side!
$$y(1 + x^2) = \int 4x^2 \, dx + C$$
$$y(1 + x^2) = \frac{4x^3}{3} + C$$
Now, apply the initial condition: the curve passes through the origin $(0,0)$.
$$0(1 + 0^2) = \frac{4(0)^3}{3} + C \implies 0 = 0 + C \implies C = 0$$
Substitute $C=0$ back into the solution:
$$y(1 + x^2) = \frac{4x^3}{3}$$
Multiply both sides by 3 to clear the fraction:
$$3y(1 + x^2) = 4x^3$$
Step 4: Final Answer:
The equation of the curve is $3y(1 + x^2) = 4x^3$, which matches option (c).