Approximating solutions for the ODE $y'=\exp(y/x)$ I am currently trying to solve excercise 1-38 from Mathews and Walker. In this excercise I am asked to consider the differential equation:
$$\frac{\mathrm{d}y}{\mathrm{d}x}=\exp(y/x)$$
for two different cases.
(a) Suppose y(1)=0. Give a series expansion which is valid for x near 1. Neglect terms of order $(x-1)^4$.
(b) Suppose $y(x_0)=+\infty$ ($x_0>0$). Give an approximate expression for $y(x)$ which is useful for x slightly less than $x_0$.
Now due to this highly nonlinear term I'm not sure how to proceed. For the (a) question I know I have to substitute:
$$y(x)=\sum\limits_{n=1}^\infty a_n(x-1)^2,$$
where I have already taken the initial condition into account. But when I fill this in, I get into trouble with the exponential. Even if I expand the exponential as
$$\exp(x)=\sum\limits_{n=0}^\infty\frac{x^n}{n!},$$
because this leads to a double sum and gives me a troublesome time to determine $a_1$ trough $a_3$. I don't know if there are any tricks to overcome this ?
For question (b) I tried making the substitution $u(x)=1/y(x)$ (so that u=0 in $x_0$), leading to the differential equation
$$\frac{\mathrm{d}u}{\mathrm{d}x}=u^2\exp(\frac{1}{ux}).$$
Again this differential equation seems not so trivial to solve.
 A: Calculate the first few derivatives analytically
$$y' = e^{\frac{y}{x}}$$
$$y'' = e^{\frac{y}{x}}\left(\frac{y'}{x} - \frac{y}{x^2}\right)$$
$$y''' = e^{\frac{y}{x}}\left(\frac{y'}{x} - \frac{y}{x^2}\right)^2 + e^{\frac{y}{x}}\left(\frac{y''}{x} - \frac{2y'}{x^2} + \frac{2y}{x^3}\right)$$
which gives us $y'(1) = 1$, $y''(1) = 1$ and $y'''(1) = 0$. Now Taylors theorem gives us
$$y(x) = \sum \frac{y^{(n)}(1)}{n!}(x-1)^n = (x-1) + \frac{(x-1)^2}{2} + \mathcal{O}[(x-1)^4]$$
For the second part note that
$$y' = x\left(\frac{y}{x}\right)' + \frac{y}{x}$$
so if the second term could be neglected close to the singularity at $x=x_0$ then the ODE would read
$$\frac{d}{dx}e^{-\frac{y}{x}} = -\left(\frac{y}{x}\right)'e^{-\frac{y}{x}} \approx -\frac{1}{x}$$
which has the solution
$$y \approx -x\log\left[\log\left(\frac{x_0}{x}\right)\right]$$
It remains to check that this is a good solution. We do this by calulcating
$$\frac{y'}{e^{y/x}} = 1 - z \log z,~~~~\text{where}~~~~z =\log\left(\frac{x_0}{x}\right)$$
and as $x\to x_0$ we have $z\to 0$ and since $\lim_{z\to 0} z\log z = 0$ we have that our solution is a good approximation (in the sense that $\frac{y'}{e^{y/x}} \approx 1$) close to $x=x_0$ .
A: The analytical solution of the ODE can be expressed on a parametric form :
$$\frac{\mathrm{d}y}{\mathrm{d}x}=\exp(y/x)$$
Let $t=\exp(y/x)$ or $y=x\ln(t)$
$$\ln(t)+\frac{x}{t}\frac{\mathrm{d}t}{\mathrm{d}x}=t$$
$$\frac{1}{x}\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{1}{t\left(t-ln(t)\right)}$$
$$\ln(x)=\int{\frac{dt}{t\left(t-ln(t)\right)}}$$
There is no closed form for this integral. 
The parametric form of analytic solution is :
$$\begin{cases} x=e^{\int{\frac{dt}{t\left(t-ln(t)\right)}}} \\ y=\ln(t)e^{\int{\frac{dt}{t\left(t-ln(t)\right)}}}  \end{cases}$$
The arbitrary constant which is included in the undefined integral must be the same for both (this is equivalent to an arbitrary coefficient mutiplying the exponential of a definded integral, instead of the exponential of the undefined integral).
The figure below shows two approximate formulas compared to the analytical result (red curve) 
The green curve is dranw with a very simple formula derived from my parametric solution. This is a first approximate of the analytic solution.
The blue curve is drawn with the formula similar to the Winther's formula (but with different parameter $x_0$, i.e.: $1+\ln(\frac{x_0}{x}) = \ln (\frac{e x_0}{x})$ instead of $\ln (\frac{x_0}{x})$ , so it is not a valuable comparison with the Winther's results).

A: This is the answer to the question raised by Nick. 
My first answer was not about the approximates as asked for, but was the analytical solving of the ODE.
Now, three approximates formulas for different ranges of $x$ are derivated from the parametric solution : 
$$\begin{cases} x=x_0 e^{\int_1^t {\frac{d\theta}{\theta\left(\theta-ln(\theta)\right)}}} \\ y=x_0 \ln(t)e^{\int_1^t{\frac{d\theta}{\theta\left(\theta-ln(\theta)\right)}}}  \end{cases}$$
$x_0$ is defined as the value of $x$ so that $y(x_0)=0$
the ODE is of the homogeneous kind. So all the curves corresponding to the solutions of the ODE are of identical shape, but homothetic, depending from the parameter $x_0$




The last formula is accurate only in a small range for $x$ close to $0$. A more extended range would require more terms for the series expansions. This involves arduous calculus of asymptotic expension of special functions.
DETERMINING A PARTICULAR SOLUTION FROM A GIVEN CONDITION :
Often a condition is specified: a given point $(x=x_g , y=y_g)$ determines the particular solution. How to find the corresponding parameter $x_0$ ?


*

*Compute the corresponding value of $t=t_g=exp\left(\frac{y_g}{x_g}\right)$

*Then, $x_0=x_g e^{-\int_1^{t_g} {\frac{d\theta}{\theta\left(\theta-ln(\theta)\right)}}}$

