# Power Series/Differential Equations

• November 10th 2008, 11:41 AM
Oblivionwarrior
Power Series/Differential Equations
Find the first four nonzero terms in each of two linearly independent power series solutions about the origin.

What do you expect the radius of convergence to be for each solution?

$(cosx)y^{''} +xy^{'} -2y=0$

Set the coefficients near zero and first terms to 1

I am not exactly sure what the bolded sentence means. Also, how should I go about starting the problem? Is it best to start writing out the series for each term? like

Cosx = $1-\frac{x^2}{2!} + \frac{x^4}{4!} + ...$

and so on.

Thanks for any help!
• November 10th 2008, 01:13 PM
Mathstud28
Quote:

Originally Posted by Oblivionwarrior
Find the first four nonzero terms in each of two linearly independent power series solutions about the origin.

What do you expect the radius of convergence to be for each solution?

$(cosx)y^{''} +xy^{'} -2y=0$

Set the coefficients near zero and first terms to 1

I am not exactly sure what the bolded sentence means. Also, how should I go about starting the problem? Is it best to start writing out the series for each term? like

Cosx = $1-\frac{x^2}{2!} + \frac{x^4}{4!} + ...$

and so on.

Thanks for any help!

Start by assuming that the solution to this problem is $y=\sum_{n=1}^{\infty}a_nx^n$. Then substitute, differentiate, adjust indicies so they are all the same, and then solve the ensuing recurrence relation.
• November 10th 2008, 01:17 PM
Oblivionwarrior
Ok, ill try that out. What about that bolded sentence? Is that just talking about where I begin once I got the recurrence formula?
• November 10th 2008, 01:19 PM
Mathstud28
Quote:

Originally Posted by Oblivionwarrior
Ok, ill try that out. What about that bolded sentence? Is that just talking about where I begin once I got the recurrence formula?

I'm not entirely sure.
• November 10th 2008, 05:44 PM
shawsend
I think the equation $\cos(x)y''+xy'-2y=0$ would be quite a challenge to solve by substituting $\sum_{n=0}^{\infty}a_nx^n$ because of the $\cos(x)$ term: that's going to involve a Cauchy product of sums for the first term which can be done but is messy. I think in this particular case, just need to differentiate the differential equation to find the first few coefficients: Let $y(0)=a$ and $y'(0)=b$ and write it as:

$y''=-\frac{x}{\cos(x)}y'+\frac{2}{\cos(x)}y$

Now we seek a solution about the origin in which everything is well-behaved until we reach the first singular point at $x=\pi/2$. Therefore we can expect a solution of the form:

$y(x)=\sum_{n=0}^{\infty}y^{(n)}(0)\frac{x^n}{n!};\ quad R=\pi/2$

We already know what the first two terms are. Now differentiate to get the third term:

$\frac{d}{dx}y''\Bigg|_{x=0}=\frac{d}{dx}\left(-\frac{x}{\cos(x)}y'+\frac{2}{\cos(x)}y\right)_{x=0 }$

and substitute the values for $y(0),y'(0),y''(0)$.

Do this again to get the fourth term. However that's only one series but I think the two linearly independent solutions are made up of the separate sums of the even and odd powers of x which would mean you'd need to calculate about 8 derivatives. Might get messy and maybe there is another way.