How do I solve a cubic ODE using a series solution?

[Winzer]

Homework Statement

Find the solution to the ODE via the power series:
y = \Sigma_{i=0} a_j x^{2j + m}

Homework Equations

y' - y^3 = 0

The Attempt at a Solution

I get
\Sigma_{i=0} a_j (2j+m) x^{2j+m-1} - \Sigma_{i=0} (a_j)^3 x^{3(2j + m)} = 0
I don't know how to deal with the cubic part. No matter where I start my series I can't get the recursion relation without x

 

[Mark44]

Homework Statement

Find the solution to the ODE via the power series:
y = \Sigma_{i=0} a_j x^{2j + m}

Homework Equations

y' - y^3 = 0

The Attempt at a Solution

I get
\Sigma_{i=0} a_j (2j+m) x^{2j+m-1} - \Sigma_{i=0} (a_j)^3 x^{3(2j + m)} = 0
I don't know how to deal with the cubic part. No matter where I start my series I can't get the recursion relation without x

The summation on the right above is incorrect. y^3 is the cube of the summation, not the summation of the cubes of its terms.

It will probably be helpful to work with the a few terms of the expanded series, rather than work with the closed forms of the series. The equation above should be written as follows.

\Sigma_{i=0} a_j (2j+m) x^{2j+m-1} - \left(\Sigma_{i=0} (a_j) x^{2j + m}\right)^3

 

[Winzer]

Thanks for the correction.
So it looks like I would have to expand the cubic part to just a couple terms?
Where do you go from there?

 

[Mark44]

Expand both parts and combine like terms. You'll probably need to expand to more than just two terms, though. I haven't worked this problem, but that's the tack I would take.

The equation is separable, so you can solve it using that technique as a check of your work using series.

 

[Winzer]

I don't see how expanding and combining like terms will set me up for a recursion relation.

 

[Office_Shredder]

Once you do that, you know that the series is supposed to be zero, so each coefficient in the new series has to be zero as well.

 

[jackmell]

I don't see how expanding and combining like terms will set me up for a recursion relation.

I'm not sure this is what you want, but it's the only way I know how to express that DE in terms of a power series. ll do the square via Cauchy product:

\begin{aligned}<br /> \sum_{n=0}^{\infty} a_n x^n \sum_{n=0}^{\infty} a_n x^n &amp;=\sum_{n=0}^{\infty}\sum_{k=0}^n a_k a_{n-k} x^n \\<br /> &amp;=\sum_{n=0}^{\infty} b_n x^n,\quad b_n=\sum_{k=0}^{n} a_k a_{n-k}<br /> \end{aligned}<br />

Now, how would express:

\sum_{n=0}^{\infty} a_n x^n \sum_{n=0}^{\infty} b_n x^n

 

[Winzer]

So I am still having problems
for m = 0 I get:
<br /> a_{i} i x^{i-1} - (a_{i} x^i)^3 =0 <br />
which becomes
<br /> a_{i+1} (i+1) x^{i} - (a_{i} x^i)^3 =0 <br />
so for i = 0
<br /> a_1= a_0<br />
for i =1
<br /> a_2(2)x - (a_{1}^3 x^3) = 0<br />
which is
<br /> a_2(2) - (a_{1}^3 x^2) = 0<br />
Is this right? having an x in the coefficent for a2?

 

[Mark44]

You should not be comparing terms with different powers of x.