solve y'y'''=y''

[lkj-17]

How to use power series to solve this non-linear differential equation?

[HallsofIvy]

First, because this is a third order equation, its general solution will involve 3 undetermined constants. So, assume y(0)= A, y'(0)= B, y"(0)= C.

From y'y"'= y", y"'= y"/y' and so y"'(0)= C/B. Now, differentiating y"'= y"/y' again, yiv= (y'y"'- y"2)/y'2 so yiv(0)= (B(C/B)- C^2)/B2= (C- C2)/B2.

So far, we have y(x)= y(0)+ y'(0)x+ (1/2)y"(0)x2+ (1/3!)y"'(0)x3+ (1/4!)yiv(0)x4+ ...= A+ Bx+ (1/2)Cx2+ (1/6)(C/B)x3+ (1/24){(C- C2)/B2}x^4+ ...

Continue like that to get higher terms.

[Matthew Rodman]

Re-arrange your equation as

y^{\prime \prime \prime} = \frac{y^{\prime \prime}}{y^{\prime}}

Now integrate with respect to x to get

y^{\prime \prime} = \kappa + \ln{y^{\prime}}

where \kappa is a constant of integration. Now re-arrange and integrate to get

\int{\frac{d y^{\prime}}{\kappa + \ln{y^{\prime}}} = x + \epsilon

where \epsilon is another constant.

I checked the Integrator (Wolfram site), and it gave the integral as:

\int{\frac{dw}{a + \ln{w}}} = e^{-a}Ei(a + \ln{w})

where Ei is the Exponential Integral (http://mathworld.wolfram.com/ExponentialIntegral.html).

Hence we can apply this to our integral to get

e^{- \kappa} Ei(\kappa + \ln{y^{\prime}) = x + \epsilon

which we can re-arrange as

y^{\prime} = exp(Ei^{-1}(e^{\kappa}(x + \epsilon)) - \kappa)

We can tidy this up a bit by making

e^{\kappa} = \alpha

and

\epsilon \alpha = \beta

and then integrate to get

y(x) = \frac{1}{\beta}\int{exp(Ei^{-1}(\alpha x + \beta)) dx}

and then you'll have to try some numerical techniques to obtain values for y(x) (I have to idea how to express the inverse of "Ei".)

[Matthew Rodman]

Sorry that should read:

y(x) = \frac{1}{\alpha}\int{exp(Ei^{-1}(\alpha x + \beta)) dx}