Power Series Solution for y' = 4xy + 2 with Initial Condition y(0)=1

[Char. Limit]

Homework Statement

I am trying to find the power series solution to y' = 4 x y + 2, with the initial condition of y(0)=1.

Homework Equations

The Attempt at a Solution

Simple enough, I say, as I arrange the equation so I have 0 on one side. I get something like this:

[tex]y' - 4 x y - 2 = 0[/tex]

I then assume that [itex]y = \sum_{n=0}^\infty a_n x^n[/itex]. I also find that [itex]y' = \sum_{n=0}^\infty (n+1) a_{n+1} x^n[/itex] and I pick, for two, a series like [itex]\sum_{n=0}^\infty \frac{1}{2^n}[/itex]. Subbing this all in, I get:

[tex]\sum_{n=0}^\infty \left(a_n - 4 \left(n+1\right) x a_{n+1} - \frac{1}{2^n}\right) x^n = 0[/tex]

Or in other words...

[tex]\left(a_n - 4 (n+1) x a_{n+1} - \frac{1}{2^n}\right) = 0[/tex]

But this doesn't look right. There's an "x" in there that shouldn't be there. What's the best way to remove the x?

 

[tiny-tim]

Hi Char. Limit!

You needed to think ahead …

[itex]y = \sum_{n=0}^\infty a_n x^n[/itex].

… you needed [itex]xy = \sum_{n=0}^\infty a_n x^{n+1}[/itex] ​

(and then change the limits, of course)