Solving y' -x^2 -e^y = 0 using series

**pistache** · May 18th 2010, 08:08 AM

Hello,

I need to solve the differential equation $y'-x^2-e^y=0$ using series, with initial conditions x=0 and y=0.

My problem is $e^y$ . I know what I could have done with $e^x$ (replace $e^x$ , $y$ and $y'$ par their series in the differential equation) ... but what can I do with $e^y$ ??

Thanks to everybody,

**chisigma** · May 18th 2010, 12:48 PM

Lets expand y in MacLaurin series...

$y(x) = y(0) + y^{'} (0)\frac{x}{1!} + y^{''} (0)\frac{x^{2}}{2!} + y^{'''} (0) \frac{x^{3}}{3!} + \dots$ (1)

The 'initial condition' supplies...

$y(0) = 0$ (2)

The DE supplies $y^{'} (0)$ because is...

$y^{'} = e^{y} + x^{2} \rightarrow y^{'} (0) = 1$ (3)

Now we compute from (3)...

$y^{''} = e^{y} y^{'} + 2 x \rightarrow y^{''} (0) = 1$ (4)

... and from (4)...

$y^{'''} = e^{y} y^{' 2} + e^{y} y^{''} + 2 \rightarrow y^{'''} (0) = 4$ (5)

If necessary we can perform further steps. The series expansion of y is obtained from (1) and is...

$y(x) = x + \frac{1}{2} x^{2} + \frac{2}{3} x^{3} + \dots$ (6)

Kind regards

$\chi$ $\sigma$

**pistache** · May 19th 2010, 09:17 AM

Thanks a lot Chi Sigma!

I failed to see a pattern emerging so I computed the next two terms:

$ y^{(4)} = 3e^{y}y^{'}y^{''} + e^{y} y^{' 3} + e^{y} y^{'''} \rightarrow y^{(4)} (0) = 8 $

and

$ y^{(5)} = e^{y}y^{' 4} +6e^{y}y^{' 2}y^{''} +3e^{y}y^{'' 2} +4e^{y}y^{'}y^{'''} +e^{y}y^{(4)} \rightarrow y^{(5)} (0) = 34 $

(I used Maple for the latest one)

Therefore we have:
$ y(x) = \frac{x}{1!} + \frac{x^{2}}{2!}$ $ + 4 \frac{x^{3}}{3!} + 8 \frac{x^{4}}{4!} + 34 \frac{x^{5}}{5!} + \dots $

or

$ y(x) = x + \frac{1}{2} x^{2}$ $ + \frac{2}{3} x^{3} + \frac{1}{3} x^{4} + \frac{17}{60} x^{5} + \dots $

Is there any pattern in that? Because if I don't have the general term how can I say I solved the DE ?

Thanks for your help,

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