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

#### pistache

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
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

$$\displaystyle \chi$$ $$\displaystyle \sigma$$

#### pistache

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 ?