How Do You Solve for Coefficients in a Power Series for a 2nd Order DE?

[bhajee]

After setting out in the sums and collecting the terms in $$x^j$$ I'm left with a series of expressions in
$$a_2, a_3$$ etc as I believe I'm supposed to. However my first expression reads
$$2a_{2}+2a_{1}+a_-_{1}=0$$

Now I'm told that
$$y(o) = 1$$ and
$$y'(o) = 0$$
I think this means that
$$a_0 = 1$$
and
$$a_1 = 0$$

does this mean that
$$a_-_1 = x$$?

(My other expressions are
$$6a_3+6a_2+a_0=0$$
and
$$24a_4+12a_3+a_1=0$$)

i'm always left with an
$$a_-_1$$
when finding the other
$$a_{j}$$'s

 

[Dr Transport]

usually, negative subscript coefficients are set equal to zero initially in series solutions for differential equations.

 

[M98Ranger]

It is correct that after setting up the power series and collecting terms in x^j, you will be left with a series of expressions in a_j. In order to solve for the coefficients a_j, you will need to use the initial conditions y(0) = 1 and y'(0) = 0. These conditions will give you the values for a_0 and a_1, as you have correctly stated.

However, it is not correct to say that a_-1 = x. The subscript j in a_j represents the power of x, so a_-1 would mean a coefficient for x^-1, which is not present in the power series. Instead, you will need to use the equations you have listed (2a_2+2a_1+a_-1=0, 6a_3+6a_2+a_0=0, and 24a_4+12a_3+a_1=0) to solve for the remaining coefficients.

For example, to solve for a_2, you can substitute in the values a_0 = 1 and a_1 = 0 into the first equation, and then solve for a_2. Similarly, you can use the second equation to solve for a_3 and the third equation to solve for a_4. This process can be continued for higher values of j to find the remaining coefficients.

So while a_-1 is not equal to x, it is still an important coefficient that needs to be solved for in order to find the solution to the second order differential equation using the power series method.