Find the general solution of the differential equation: $(x^2 + 1) \frac{dy}{dx} + 2xy = \sqrt{x^2 + 4}$

$y(x^2 + 1) = \frac{x}{2}\sqrt{x^2 + 4} + 2 \ln|x + \sqrt{x^2 + 4}| + C$

The correct answer is Option (1) → $y(x^2 + 1) = \frac{x}{2}\sqrt{x^2 + 4} + 2 \ln|x + \sqrt{x^2 + 4}| + C$ ##

The given differential equation is $(x^2 + 1) \frac{dy}{dx} + 2xy = \sqrt{x^2 + 4}$

Rewrite the differential equation:

$\frac{dy}{dx} + \frac{2xy}{(x^2 + 1)} = \frac{\sqrt{x^2 + 4}}{(x^2 + 1)}$

The integrating factor of the differential equation is

$e^{\int \frac{2x}{x^2 + 1} dx} = e^{\log(x^2 + 1)} = (x^2 + 1)$

Multiply the integrating factor with both sides:

$(x^2 + 1)dy + 2xy dx = \sqrt{x^2 + 4} dx$

Integrating both sides:

$\int (x^2 + 1)dy + \int 2xy dx = \int \sqrt{x^2 + 4} dx$

$y(x^2 + 1) = \frac{x}{2} \sqrt{x^2 + 4} + \frac{4}{2} \log |x + \sqrt{x^2 + 4}| + c$

$y(x^2 + 1) = \frac{x}{2} \sqrt{x^2 + 4} + 2 \log |x + \sqrt{x^2 + 4}| + c$

The general solution of the differential equation is $y(x^2 + 1) = \frac{x}{2} \sqrt{x^2 + 4} + 2 \log |x + \sqrt{x^2 + 4}| + c$.