What is the solution of the follwoing differential equation

$\displaystyle \frac{\partial^{2}y}{\partial x^{2}}-ay^{-1}\frac{dy}{dx}=0$

where $\displaystyle a$ is a constant.

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- Dec 22nd 2012, 08:00 AMJulieKNonlinear second order differential equation
What is the solution of the follwoing differential equation

$\displaystyle \frac{\partial^{2}y}{\partial x^{2}}-ay^{-1}\frac{dy}{dx}=0$

where $\displaystyle a$ is a constant. - Dec 22nd 2012, 08:41 AMjakncokeRe: Nonlinear second order differential equation
Ok, This method works for all equations of the form $\displaystyle y'' = f(y', y) $.

Now Assume a new variable, v = y'. So $\displaystyle \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{\mathrm{d}v}{\mathrm{d}y} \frac{\mathrm{d}y}{\mathrm{d}x} $

Since $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}} = v$ This now becomes $\displaystyle = \frac{\mathrm{d}v}{\mathrm{d}x} = \frac{\mathrm{d}v}{\mathrm{d}y} v $

Now you have a Differential equation of the first order, $\displaystyle v \frac{\mathrm{d}v}{\mathrm{d}y} = g(v, y) $. For your specific equations turns into a seperable ODE,

$\displaystyle v \frac{\mathrm{d}v}{\mathrm{d}y} = \frac{a}{y} * v $ which turns into

$\displaystyle \mathrm{d}v} = \frac{a}{y} \mathrm{d}y $

So v = $\displaystyle v = ln(y) + k_0 $

Now again Since $\displaystyle v = \frac{\mathrm{d}y}{\mathrm{d}x}$ which means $\displaystyle \frac{1}{ln(y)+k_0} \mathrm{d}y = \mathrm{d}x $

and you get some super ugly integral. - Dec 22nd 2012, 11:00 AMJulieKRe: Nonlinear second order differential equation
Thank you

- Dec 23rd 2012, 12:06 AMJJacquelinRe: Nonlinear second order differential equation
Hi !

The obvious solutions are y(x)=constant.

The other solutions cannot be expressed as a combination of a finit number of elementary functions.

They involves some special functions such as li(x) or Ei(x) :

Logarithmic Integral -- from Wolfram MathWorld

Exponential Integral -- from Wolfram MathWorld

About special functions, pp.18-36 of the paper "Safari on the contry of special Functions" :

JJacquelin's Documents | Scribd - Dec 26th 2012, 12:47 AMtomjacksonRe: Nonlinear second order differential equation
Can someone please confirm that the above solution is OK? I tried a lot but I could not find the result. Is there any other way to solve this problem? I could not find any divisor for R[x].

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