# Thread: "Find the indicated coefficients of the power series solution [...]"

1. ## "Find the indicated coefficients of the power series solution [...]"

I have a vague understanding that I am supposed to find a relationship of a constant C_n (for each n) with a constant C_0 in getting power series solutions to differential equations but I'm not too sure on how to do anything exactly.

Any help would be greatly appreciated!

2. ## Re: "Find the indicated coefficients of the power series solution [...]"

After extensive hard work, this is what I came up with. Assuming that it is correct so far, where do I go from here?

3. ## Re: "Find the indicated coefficients of the power series solution [...]" Originally Posted by s3a I have a vague understanding that I am supposed to find a relationship of a constant C_n (for each n) with a constant C_0 in getting power series solutions to differential equations but I'm not too sure on how to do anything exactly.

Any help would be greatly appreciated!
If we suppose that the solution of the DE...

$\displaystyle (x^{2}+2)\ y^{''} - x\ y^{'} +y=0,\ y(0)=6,\ y^{'}(0)=2$

... is analytic in x=0, then it can be written as...

$\displaystyle y(x)= \sum_{n=0}^{\infty} \frac{y^{(n)}(0)}{n!}\ x^{n}$ (2)

... so that the problem is the computation of the derivatives of y in x=0. The initial conditions give us...

$\displaystyle y(0)=6 \implies a_{0}=6$

$\displaystyle y^{'}(0)=2 \implies a_{1}=2$

The value of $\displaystyle y^{''}(x)$ is obtained from (1)...

$\displaystyle y^{''} (x)= \frac{x\ y^{'} - y}{x^{2}+2}$ (3)

... and that means that...

$\displaystyle y^{''}(0)= -3 \implies a_{2}= -\frac{3}{2}$

Now You can proceed obtaining $\displaystyle y^{'''}(x)$ by deriving (3), then compute it for x=0 and the setting $\displaystyle a_{3}= \frac{y^{'''}(0)}{6}$. An so one...

Kind regards

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