# Possible to solve this differential equation?

I was thinking about differential equations, and came across this one: $$\frac{dy}{dx} = x^y.$$ How do I go about solving it?

As far as I can see this ODE doesn't fall into one of the common classes of ODEs. In order to solve it we can pursue two natural paths

a) numerical solution
b) power series around a suitable point $xp$

Numerical study

In order to get an overwiew let's attact the ODE numerically. The tool used here is Mathematica.

The ODE has a singularity at $x = 0$, and the region of interest for real valued solutions is therefore $x > 0$.

First we look at the Streamplot which shows possible trajectories of solutions. Next we show some typical numerical solutions. The initial condition was chosen as $y(x=1) = a$. Notice that numerically the solutions are well behaved in a certain interval $x0 < x < x1$ at the ends of which the solution diverges.

This structure needs further study, preferrably analytical.

Power series

For the time being I leave this as a hint for others.

Hint:

$\dfrac{dy}{dx}=x^y$

$\dfrac{dy}{dx}=(e^{\ln x})^y$

$\dfrac{dy}{dx}=e^{y\ln x}$

Let $u=\ln x$ ,

Then $x=e^u$

$\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=\dfrac{1}{x}\dfrac{dy}{du}=e^{-u}\dfrac{dy}{du}$

$\therefore e^{-u}\dfrac{dy}{du}=e^{uy}$

$\dfrac{dy}{du}=e^{u(y+1)}$

Let $v=y+1$ ,

Then $\dfrac{dv}{du}=\dfrac{dy}{du}$

$\therefore\dfrac{dv}{du}=e^{uv}$

Let $w=uv$ ,

Then $v=\dfrac{w}{u}$

$\dfrac{dv}{du}=\dfrac{1}{u}\dfrac{dw}{du}-\dfrac{w}{u^2}$

$\therefore\dfrac{1}{u}\dfrac{dw}{du}-\dfrac{w}{u^2}=e^w$

$\dfrac{1}{u}\dfrac{dw}{du}=e^w+\dfrac{w}{u^2}$

$(e^wu^2+w)\dfrac{du}{dw}=u$

Let $z=u^2$ ,

Then $\dfrac{dz}{dw}=2u\dfrac{du}{dw}$

$\therefore\dfrac{e^wu^2+w}{2u}\dfrac{dz}{dw}=u$

$(u^2+we^{-w})\dfrac{dz}{dw}=2e^{-w}u^2$

$(z+we^{-w})\dfrac{dz}{dw}=2e^{-w}z$

Let $s=z+we^{-w}$ ,

Then $z=s-we^{-w}$

$\dfrac{dz}{dw}=\dfrac{ds}{dw}+(w-1)e^{-w}$

$\therefore s\dfrac{ds}{dw}+(w-1)e^{-w}s=2e^{-w}(s-we^{-w})$

$s\dfrac{ds}{dw}=(3-w)e^{-w}s-2we^{-2w}$

This belongs to an Abel equation of the second kind.

• Wow I didn't know you would need soooooooooooooooooo many substitutions for this....... – Kw08 Jan 19 '17 at 12:09
• I get it now, what about $\frac{dy}{dx} = y^x$ instead of $x^y$? – Kw08 Jan 21 '17 at 6:41