# Differential Equation w/ Power Series Solution

1. Dec 12, 2004

### DivGradCurl

Problem

$$y^{\prime} = x^2 y$$

General Comments

There must be some kind of flaw in my solution as I don't get to the same result as the one my book provides:

$$y = c_0 \sum _{n=0} ^{\infty} \frac{x^{3n}}{3^n n!} = c_0 e^{x^3 / 3} \qquad \fbox{CORRECT ANSWER}$$

Any help is highly appreciated.

My Solution

If

$$y = \sum _{n=0} ^{\infty} c_n x^n \Longrightarrow x^2 y = \sum _{n=0} ^{\infty} c_n x^{n+2} = \sum _{n=3} ^{\infty} c_{n-3} x^{n-1}$$

and

$$y^{\prime} = \sum _{n=1} ^{\infty} n c_n x^{n-1}$$

Then

$$\sum _{n=1} ^{\infty} n c_n x^{n-1} = \sum _{n=3} ^{\infty} c_{n-3} x^{n-1} \Longrightarrow c_{n-3} = nc_n \quad n=3,4,5,\ldots$$

Hence, I ultimately get

$$y = c_0 + c_1 x + c_2 x^2 + c_0 \sum _{n=1} ^{\infty} \left[ \frac{x^{3n}}{3\cdot 6\cdot 9\cdot \cdots \cdot \left( 3n \right) } \right] + c_1 \sum _{n=2} ^{\infty} \left[ \frac{x^{3n-2}}{4\cdot 7\cdot 10\cdot \cdots \cdot \left( 3n-2 \right) } \right] + c_2 \sum _{n=2} ^{\infty} \left[ \frac{x^{3n-1}}{5\cdot 8\cdot 11\cdot \cdots \cdot \left( 3n-1 \right) } \right] \qquad \fbox{MY ANSWER}$$

Thank you very much!!

2. Dec 12, 2004

### Tide

If you do your recursion relation carefully you will find that $c_1 = c_2 = 0$ since $c_n = 0$ for n < 0.

3. Dec 12, 2004

### DivGradCurl

Oh... I've got it. Thanks

4. Dec 12, 2004

### dextercioby

Cheating method:Why bother with stinky recursion relations,when u can use a series already known??

$$\frac{dy}{dx} =y x^2$$ from which,by separating variables and integrating,one gets:
$$y=C\exp(\frac{x^3}{3})$$.To find the series the problem asks you about,plug $\frac{x^3}{3}$ in the power series of $\exp{x}$ and you'll have your answer.

PS.This cheating trick works,as the differential eq.is easily solvable exactly.In general,it does not.

5. Dec 12, 2004

### DivGradCurl

Just a final check...

------------------------------

Method #1

How about...

$$y = c_0 + c_1 x + c_2 x^2 + c_0 \sum _{n=1} ^{\infty} \left[ \frac{x^{3n}}{3\cdot 6\cdot 9\cdot \cdots \cdot \left( 3n \right) } \right] + c_1 \sum _{n=2} ^{\infty} \left[ \frac{x^{3n-2}}{4\cdot 7\cdot 10\cdot \cdots \cdot \left( 3n-2 \right) } \right] + c_2 \sum _{n=2} ^{\infty} \left[ \frac{x^{3n-1}}{5\cdot 8\cdot 11\cdot \cdots \cdot \left( 3n-1 \right) } \right]$$

So, $$c_1 = c_2 = 0$$ ( I don't see clearly why ) . Then, we find

$$y = c_0 + c_0 \sum _{n=1} ^{\infty} \left[ \frac{x^{3n}}{3\cdot 6\cdot 9\cdot \cdots \cdot \left( 3n \right) } \right] = c_0 \sum _{n=0} ^{\infty} \frac{x^{3n}}{3^{n}n!} = c_0 e^{x^3 /3}$$

------------------------------

Method #2

$$\frac{dy}{dx} = x^2 y \Longrightarrow \int \frac{1}{y}dy = \int x^2 dx$$

Then

$$\ln y = \frac{x^3}{3} \Longrightarrow y = Ce^{x^3/3}$$

where $$C = c_0$$

Last edited: Dec 12, 2004
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