The integrating factor of the differential equation $(1+x^{2}) \frac{dy}{dx} + y = e^{\tan^{-1} x}$ is

Let $y=y(x)$ be the solution of the differential equation $x \frac{dy}{dx}+y=x \log x, (x > 1)$. If $2(y(2))=\log 4-1$,then the value of $y(e)$ is:

The general solution of the differential equation $(\sec x + \tan x) \frac{dy}{dx} + (\sec^2 x + \sec x \tan x) y = 1$ is

The integrating factor of the differential equation $\frac{dy}{dx}(1+x) - xy = 1-x$ is . . . . . . .

If $\cos x \frac{dy}{dx} = y \sin x - 1$, where $x \neq (2n+1) \frac{\pi}{2}, n \in Z$, is the differential equation corresponding to the curve $y = f(x)$ and $f(0) = 1$, then $f(x) =$

The solution of the differential equation $2 \frac{dy}{dx} - \frac{y}{x} = \frac{y^2}{x^2}$,given that $y = 2$ when $x = 1$,is