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.


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 $x=x_0$ .

  • $\begingroup$ @Whinter That is a very nice answer ! How would you use this to solve (b) when y becomes infinite ? $\endgroup$ – Nick Jan 30 '15 at 17:07
  • $\begingroup$ @Nick I updated it with a possible way to attack part b. $\endgroup$ – Winther Jan 30 '15 at 17:20
  • $\begingroup$ for the second part I do not directly understand why $-y/x$ would be neglectable since $y\rightarrow\infty$. Wouldn't it be more feasable to use the Taylor expansion again in some way ? I'm trying to figure out how since the demand that $u(x_0)=0$ makes the exponent blow up :(. $\endgroup$ – Nick Jan 30 '15 at 18:39
  • $\begingroup$ Yes $y/x$ does go to $\infty$, but it is still much smaller than the other term ($x(y/x)'$) which goes to $\infty$ much faster. You cannot Taylor expand around a singularity. You can try to get a Laurent expansion $\sum_{n=-\infty}^\infty a_n (x-x_0)^n$. However if the function has an essential singulairy at $x=x_0$ (like here) you will have to include all terms with $n<0$ so that does not seem to work. $\endgroup$ – Winther Jan 30 '15 at 18:51
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    $\begingroup$ @Nick We have $y/x$ in the exponential so it makes sense to try to write the ode in terms of $z=y/x$. $\endgroup$ – Winther Jan 31 '15 at 11:27

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).

enter image description here

  • $\begingroup$ indeed every first order ODE is exactly solvable but I'm not sure to what extent the implicit solution might help. Also your last system of equations says y=ln(t)x, is that correct ? $\endgroup$ – Nick Jan 31 '15 at 11:09
  • $\begingroup$ @ Nick: $y=x\ln(t)$ comes from the second equation at the begining of my main answer. $\endgroup$ – JJacquelin Jan 31 '15 at 11:33
  • $\begingroup$ Ah yes the substitution, my fault ;). $\endgroup$ – Nick Jan 31 '15 at 11:37
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    $\begingroup$ @ Nick : In fact, the parametric form of solution is convenient to find approximates. The result of a first attempt is now in addition to my main answer above. $\endgroup$ – JJacquelin Jan 31 '15 at 19:26
  • $\begingroup$ Note that the approximation I derived (blue) is under the assumption that $x = x_0$ is a singularity of $y$, i.e. that $y$ blows up at $x/x_0 = 1$. And it is only guaranteed to be a good approximation close to $x=x_0$. $\endgroup$ – Winther Feb 1 '15 at 18:14

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 $x_0$

enter image description here

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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.


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

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