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

May 2010
2
0
Hello,

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

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

Thanks to everybody,
 

chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
Lets expand y in MacLaurin series...

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

The 'initial condition' supplies...

\(\displaystyle y(0) = 0\) (2)

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

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

Now we compute from (3)...

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

... and from (4)...

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

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

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

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
May 2010
2
0
Thanks a lot Chi Sigma!

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

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

and

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

(I used Maple for the latest one)

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

or

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

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,