I shell present 2 questions I came across today related to this subject. I need some explanation about the meaning of the IVP I am given.
We look for a solution of the form $y=\sum a_nx^n$.
So after doing the algebra using these 2: $$y'=\sum_{n=1}^\infty n a_n(x-x_0)^{n-1}$$ $$y''=\sum_{n=2}^\infty n(n-1)a_n(x-x_0)^{n-2}$$ I came up with the following recurrence relations:
- (a) $(x^2-2)y''+6xy'+4y=0$. The recurrence relation to this eq. is $$a_{n+2}=a_n\cdot \frac{n+4}{2(n+2)}$$ The Initial values are $$\left\{\begin{matrix} y'(0)=0\\ y(0)=1 \end{matrix}\right.$$ And my textbook states that this leads to$$\left\{\begin{matrix} a_0=1\\ a_1=0 \end{matrix}\right.$$.
- (b) $(1+x^2)y''+3xy'+y=0$. The recurrence relation to this eq. is $$a_{n+2}=-a_n\cdot \frac{n+1}{n+2}$$ The Initial values are $$\left\{\begin{matrix} y'(0)=1\\ y(0)=2 \end{matrix}\right.$$ And my textbook states that this leads to$$\left\{\begin{matrix} a_0=2\\ a_1=1 \end{matrix}\right.$$.
- (c) $(1+x^2)y''-6y=0$. The recurrence relation to this eq. is $$a_{n+2}=-a_n\cdot \frac{n-3}{n+1}$$ The Initial values are $$\left\{\begin{matrix} y'(0)=0\\ y(0)=1 \end{matrix}\right.$$ And my textbook states that this leads to$$\left\{\begin{matrix} a_0=1\\ a_1=0 \end{matrix}\right.$$.
I simply have no idea how to determine the correct $a_0$ and the correct $a_1$. Who goes to the $y'$ and who goes the $y$?
As Amzoti noted in a comment, $a_0$ is the value of the function at the point of expansion, and $a_1$ is the value of its derivative there. Both facts follows from the Taylor series formula.
Note that $a_0$ and $a_1$ are determined from initial conditions only. From the ODE you get the recurrence relation. Together they determine the rest of coefficients.