Confusing series solution

**rebghb** · May 27th 2010, 10:56 PM

Hello everyone!

Here's a DE: $y' -5x^4y=3x^9$ .
Letting $y=\sum_{n=0}^{\infty} c_n x^n$ and after some algebra, we get:
$c_1+2c_2x+3c_3x^2+4c_4x^3-3x^9$ $+\sum_{k=0}^{\infty}(\left (k+5)c_{k+5}-5c_k \right)\,x^{k+4}=0$ .

does this mean that $c_1=c_2=c_3=c_4=0$ ?
what about $-3x^9$ ? And is the relation $(k+5)c_{k+5}-5c_k$ even valid.

Solved the DE w\out power series, and I checked it, I got $y=-\frac{3}{5}(x^5+1)$

How to do the algebra in the series solution?
Thanks!

**chisigma** · May 28th 2010, 12:29 AM

The DE is...

$y^{'} = 5 x^{4} y + x^{9}$ (1)

... and we are searching for a solution of the form...

$y(x)= c_{0} + c_{1} x + c_{2} x^{2} \dots + c_{n} x^{n} + \dots$ (2)

... so that is...

$y^{'} (x) = c_{1} + 2 c_{2} x + \dots + n c_{n} x^{n-1} + \dots$ (3)

Now in order to proceed we suppose to know $y(0)=c_{0}$ . Inserting (2) and (3) in (1) we obtain first...

$c_{1} = c_{2} = c_{3} = c_{4} = 0$ (4)

... and for $n=5$ ...

$5 c_{5} x^{4}= 5 c_{0} x^{4} \rightarrow c_{5}=c_{0}$ (5)

Increasing n we obtain then...

$c_{6} = c_{7} = c_{8} = c_{9} = 0$ (6)

... and for $n=10$ ...

$10 c_{10} x^{9}= (5 c_{5}+1) x^{9} \rightarrow c_{10}=\frac{5 c_{0}+1}{10}$ (7)

All other $c_{n}$ are 0 so that is...

$y(x)= c_{0} + c_{0} x^{5} + \frac{5 c_{0}+1}{10} x^{10}$ (8)

Kind regards

$\chi$ $\sigma$

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