Question

Solve the following differential equation and integrate: $ \frac{dy}{dx} + \frac{2x}{1 + x^2} \cdot y = x $

• A. \( y = \frac{x}{1 + x^2} \)
• B. \( y = \frac{x^2}{1 + x^2} \)
• C. \( y = x \)
• D. \( y = \ln(1 + x^2) \)

Hint:

For solving linear first-order differential equations, always find the integrating factor first, and remember that the general solution involves multiplying both sides of the equation by this factor.

Answer 1

The Correct Option is A

The given differential equation is: \[ \frac{dy}{dx} + \frac{2x}{1 + x^2} \cdot y = x \] This is a first-order linear differential equation of the form: \[ \frac{dy}{dx} + P(x)y = Q(x) \] where \( P(x) = \frac{2x}{1 + x^2} \) and \( Q(x) = x \). The integrating factor \( \mu(x) \) is: \[ \mu(x) = e^{\int P(x) dx} = e^{\int \frac{2x}{1 + x^2} dx} \] The integral \( \int \frac{2x}{1 + x^2} dx \) is \( \ln(1 + x^2) \). So, the integrating factor becomes: \[ \mu(x) = e^{\ln(1 + x^2)} = 1 + x^2 \] Multiplying both sides of the equation by \( \mu(x) = 1 + x^2 \), we get: \[ (1 + x^2)\frac{dy}{dx} + 2xy = x(1 + x^2) \] The left-hand side is the derivative of \( (1 + x^2)y \), so: \[ \frac{d}{dx} \left( (1 + x^2)y \right) = x(1 + x^2) \] Now integrate both sides: \[ (1 + x^2)y = \int x(1 + x^2) dx = \frac{x^2}{2} + \frac{x^4}{4} + C \]
Thus, we have: \[ y = \frac{x}{1 + x^2} \] Therefore, the correct solution is (A).