Find the solution of the differential equation $(e^{y-x}) dy = (e^x - e^y) dx$.

Let $g$ be a differentiable function such that $\int_0^x g(t) dt = x - \int_0^x tg(t) dt$ for $x \geq 0$. Let $y = y(x)$ satisfy the differential equation $\frac{dy}{dx} - y \tan x = 2(x+1) \sec x g(x)$ for $x \in [0, \frac{\pi}{2})$. If $y(0) = 0$,then $y(\frac{\pi}{3})$ is equal to

Let $f:[1, \infty) \rightarrow[2, \infty)$ be a differentiable function. If $10 \int_1^{x} f(t) dt = 5x f(x) - x^5 - 9$ for all $x \geq 1$,then the value of $f(3)$ is:

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}=(y+1)((y+1)e^{x^{2}/2}-x)$,with $y(2)=0$. Then $y'(1)$ is equal to . . . .

The integrating factor of the differential equation $x \frac{dy}{dx} + 2y = x^2 \log x$ is

Let $\alpha$ be a non-zero real number. Suppose $f: R \rightarrow R$ is a differentiable function such that $f(0)=2$ and $\lim _{x \rightarrow-\infty} f(x)=1$. If $f^{\prime}(x)=\alpha f(x)+3$ for all $x \in R$,then $f(-\log _e 2)$ is equal to . . . . . . . . .