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

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

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,

2. 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

$\chi$ $\sigma$

3. 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 ?