Integrating factor of the differential equation $\frac{dy}{dx} + y = \frac{1+y}{x}$ is

The solution of the differential equation $\frac{dy}{dx} + \frac{y}{2} \sec x = \frac{\tan x}{2y}$,where $0 \le x < \frac{\pi}{2}$,and $y(0) = 1$,is given by

The integrating factor of the differential equation $x \cdot \frac{dy}{dx} + 2y = x^2$ is $(x \neq 0)$.

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:

Let $y=y(x)$ be the solution of the differential equation,$x y^{\prime}-y=x^{2}(x \cos x+\sin x), x>0$. If $y(\pi)=\pi$,then $y^{\prime \prime}\left(\frac{\pi}{2}\right)+y\left(\frac{\pi}{2}\right)$ is equal to

If $x y^{\prime}+y-e^x=0$ and $y(a)=b$, then $\lim _{x \rightarrow 1} y(x)$ is