CUET Preparation Today
The integrating factor of the differential equation $x\frac{dy}{dx}-y = 2x^2 $ is : |
$e^{-y}$ $e^{-x}$ x $\frac{1}{x}$ |
$\frac{1}{x}$ |
The correct answer is Option (4) → $\frac{1}{x}$ Dividing given equation by "x" $\frac{dy}{dx}-\frac{y}{x}=2x$ So $I.F.=e^{\int -\frac{1}{x}dx}=e^{-\log x}=\frac{1}{x}$ |
