general solution of the equation $\frac{dy}{dx} =\exp(y/x)$ How can i get the general solution of the equation
a) $\frac{dy}{dx} = \exp(y/x)$
b) $\frac{dy}{dx} = \exp(x-y)$ and $y=2$ when $x = 0$
I tried b) first:
This is a first-order nonlinear ordinary differential equation, which is separable. 
General solution: 
$y(x) = \ln(C+e^x)$ 
Finding C , we have that:
$$2 = \ln(C + e^0)$$ 
$$ 2 = \ln(C + 1) $$ 
$$e^2 = C+1 $$ 
$$C = e^2 - 1 $$ 
Particular solution: 
$$y(x) = \ln(e^2 -1 + e^x)$$
Is that correct the solution for b)? , I stuck with a), some help please.
 A: a)
Approach $1$:
Let $u=\dfrac{y}{x}$ ,
Then $y=xu$
$\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$
$\therefore x\dfrac{du}{dx}+u=e^u$
$x\dfrac{du}{dx}=e^u-u$
$\dfrac{dx}{x}=\dfrac{du}{e^u-u}$
$\int\dfrac{dx}{x}=\int\dfrac{du}{e^u-u}$
$\ln x=\int^u\dfrac{dt}{e^t-t}+c$
$x=Ce^{\int^\frac{y}{x}\frac{dt}{e^t-t}}$
Approach $2$:
$\dfrac{dy}{dx}=e^\frac{y}{x}$
$\dfrac{dx}{dy}=e^{-\frac{y}{x}}$
Let $u=\dfrac{x}{y}$ ,
Then $x=yu$
$\dfrac{dx}{dy}=y\dfrac{du}{dy}+u$
$\therefore y\dfrac{du}{dy}+u=e^{-\frac{1}{u}}$
$y\dfrac{du}{dy}=e^{-\frac{1}{u}}-u$
$\dfrac{dy}{y}=\dfrac{du}{e^{-\frac{1}{u}}-u}$
$\int\dfrac{dy}{y}=\int\dfrac{du}{e^{-\frac{1}{u}}-u}$
$\ln y=\int^u\dfrac{dt}{e^{-\frac{1}{t}}-t}+c$
$y=Ce^{\int^\frac{x}{y}\frac{dt}{e^{-\frac{1}{t}}-t}}$
b)
$\dfrac{dy}{dx}=e^{x-y}$
$\dfrac{dy}{dx}=e^xe^{-y}$
$e^y~dy=e^x~dx$
$\int_2^ye^y~dy=\int_0^xe^x~dx$
$[e^y]_2^y=[e^x]_0^x$
$e^y-e^2=e^x-1$
$e^y=e^x+e^2-1$
$y=\ln(e^x+e^2-1)$
A: Set $y=xu$, then $y'=xu'+u$ and the equation a) transforms into
$$xu'+u=e^u \qquad \Longrightarrow \qquad  xu'=e^u-u.$$
This is obviously separable, hence the solution is implicitly given by
quadratures
$$\int^{y/x}\frac{du}{e^u-u}=\int\frac{dx}{x}=\ln x+\operatorname{const}.$$
However the integral on the left does not seem to be expressible in terms of elementary or classical special functions.
