
#1
Feb1413, 10:26 AM

P: 3,175

I thought today of the next DE:
[tex] y''(x) = y(x)e^{y'(x)}[/tex] Not sure if it has applications, obviosuly I tried to find a solution via power series around x=0. It seems tough to look for a general recurrence equation for the coefficients. Here's what I have done so far. [tex]y(x)=\sum_{n=0}^{\infty} a_n x^n [/tex] [tex]y'(x)=\sum_{n=0}^{\infty} na_n x^{n1}[/tex] [tex]y''(x)=\sum_{n=0}^{\infty} n(n1)a_n x^{n2} [/tex] equating: [tex] \sum_{n=0}^{\infty} n(n1)a_n x^{n2} =\sum_{n=0}^{\infty} a_n x^n e^{\sum_{n=0}^{\infty} a_n n x^{n1}} [/tex] [tex]e^{a_1} e^{2a_2 x} e^{3a_3 x^2} \cdots = e^{a_1}[1+2a_1 x + \frac{(2a_1 x)^2}{2!}+\cdots]\cdot [1+3a_3 x^2 +\frac{(3a_3 x^2)^2}{2!}+\cdots]\cdot \cdots [/tex] I am not sure if it even converges, is this equation known already, I am quite sure someone already thought of it. Thanks in advance. 



#2
Feb1413, 11:14 AM

P: 745

Hi !
The ODE is solvable on the form of the inverse function x(y) as a special function defined by an integral : 



#3
Feb1413, 11:37 AM

P: 3,175

Thanks.




#4
Feb1513, 08:50 AM

P: 3,175

An ODE I was thinking of.
Well, if I am already in the mood for nonordinary DEs, I'll make this thread a thread with peculiar DEs I have in my mind.
Here's another one: [tex]y^{(n)}+(y^{(n1)})^2+(y^{(n2)})^3+\ldots + (y')^{n+1}+y^{n+2} = 0[/tex] Guessing a solution in the form of power series will be hard work (which I don't have time for right now), so does it have an specail function form solution? P.S n\geq 1 



#5
Feb1513, 09:02 AM

P: 3,175

Maybe some examples if the general case isn't clear enough.
for n=1: [tex] y'+y^2=0[/tex] For n=2: [tex] y''+(y')^2+y^3=0[/tex] For n=3: [tex] y'''+(y'')^2+(y')^3+y^4=0[/tex] Etc. 


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