Question

The equation of the curve passing through origin and satisfying $(1 + x^2) \frac{dy}{dx} + 2xy = 4x^2$ is ______.

• A. $y(1 + x^2) = 4x^3$
• B. $4(1 + x^2) = 4 + y^2$
• C. $3y(1 + x^2) = 4x^3$
• D. $1 + y^2 = 4x^3 + 1$

Hint:

In many LDE problems structured like $f(x) \frac{dy}{dx} + f'(x)y = Q(x)$, you can bypass the IF formula entirely! Recognize that the LHS is exactly the product rule expansion of $\frac{d}{dx}[y \cdot f(x)]$. Integrating both sides directly gives $y \cdot f(x) = \int Q(x) dx$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$. We solve it using an Integrating Factor (I.F.).

Step 2: Formula Application:
Standardize the equation: $\frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{4x^2}{1+x^2}$. $I.F. = e^{\int P(x) dx} = e^{\int \frac{2x}{1+x^2} dx} = e^{\ln(1+x^2)} = 1+x^2$.

Step 3: Explanation:
Solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) \, dx$. $y(1+x^2) = \int \frac{4x^2}{1+x^2} (1+x^2) dx = \int 4x^2 \, dx = \frac{4x^3}{3} + C$. Since it passes through origin $(0, 0)$: $0(1+0) = 0 + C \implies C = 0$. So, $y(1+x^2) = \frac{4x^3}{3} \implies 3y(1+x^2) = 4x^3$.

Step 4: Final Answer:
The equation is $3y(1 + x^2) = 4x^3$.