Find the general solution of the following differential equation: $x \, dy - (y + 2x^2) dx = 0$

$y = 2x^2 + Cx$

The correct answer is Option (3) → $y = 2x^2 + Cx$ ##

The given differential equation can be written as:

$\frac{dy}{dx} = \frac{y + 2x^2}{x}$

$\Rightarrow \frac{dy}{dx} - \frac{1}{x}y = 2x$

Here, $P = -\frac{1}{x}, Q = 2x$

$I.F. = e^{\int P dx} = e^{\int -\frac{1}{x} dx} = e^{-\log x} = \frac{1}{x}$

The solution is:

$yx^{-1} = \int \left( 2x \times \frac{1}{x} \right) dx$

$\Rightarrow \frac{y}{x} = 2x + c$

$\Rightarrow y = 2x^2 + cx$