# Power Series Solutions of D.E.s

1. Jun 28, 2012

### ElijahRockers

1. The problem statement, all variables and given/known data

Find two linearly independent power series solutions about the ordinary point x=0 for

y'' + x2y' +y =0

3. The attempt at a solution

Alright so we are supposed to try
y(x) = Ʃn=0 {Cn(x-x0)n} [but x0=0 so i wont include it in the derivatives]

so y'(x) = Ʃn=1 {nCnxn-1}

and y''(x) = Ʃn=2 {n(n-1)Cnxn-2}

Shifting indices and taking out terms to make the exponents of x and the starting points of the series equal (and then finally adding the series):

C0 + 2C2 + Ʃn=1 [(n+2)(n+1)Cn+2 + (n-1)Cn-1 + Cn]xn = 0

To me, this would mean that C0 and C2 both equal 0 right?

Then also That whole jumble inside the final series (except for the xn) also must equal zero...

I guess I am lost at this point. The teacher did it differently. He starts the final series at n=2 instead of n=1. Shouldn't it still be possible to do it my way?

I have attached his solution for reference.

Thanks for the help, final exam is on monday, wish me luck! Diff Eq in 5 weeks has been pretty rough on me...

EDIT: I forgot to mention.... I am not quite sure how to arrive at two different solutions... it seems to me that in my notes the teacher factored something out and somehow got two solutions but I am pretty confused as to what is going on there.

#### Attached Files:

• ###### HW8 Solutions.pdf
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2. Jun 28, 2012

### Dens

The series can only be 0 if the coefficients are 0

Basically the recurrence relation is often difficult to find (if even possible). The solution sets the two free variables (also call parameters in other context hehe) to two easy values like 0 and 1. He basically set the parameters to 0 and 1 and "unset" that later. The two solutions form one solution. So the actual solution is

$$y = c_1 y_1 + c_2y_2$$