# Math Help - Series solution to ODE

1. ## Series solution to ODE

I have an equation

$y''-8y'+16y=0$

And I need to find a power series solution for it. I can write this as

$\sum{a_{n+2}(n+2)(n+1)x^n}-8\sum{a_{n+1}(n+1)x^n}+16\sum{a_{n}x^n}=0$

Now, as I understand it, in order to solve this, I need to get rid of one of the a terms (either $a_{n}$ , $a_{n+1}$ or $a_{n+2}$ ), but I don't know how to do this...

Any guidance is greatly appreciated

2. The DE is...

$y^{''} - 8 y^{'} + 16 y= 0$ (1)

... and we suppose its solutions are expandable is McLaurin series as...

$y(x)= a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + ...$ (2)

From (2) we derive first...

$y^{'} (x) = a_{1} + 2 a_{2} x + 3 a_{3} x^{2} + ...$

$y^{''} (x) = 2 a_{2} + 6 a_{3} x + ...$ (3)

The (1) can be solved only if we know the 'initial conditions' that usuallly are expressed as $y(0)= y_{0}$ , $y^{'} (0) = y_{1}$. If now we impose these 'initial conditions' we obtain...

$a_{0} = y_{0}$

$a_{1} = y_{1}$

Now we can valuate the succesive terms $a_{n}$ inserting (2) and (3) in (1)...

$2 a_{2} - 8 a_{1} + 16 a_{0}= 0 \rightarrow a_{2} = 4 a_{1} - 8 a_{0}$

$6 a_{3} - 16 a_{2} + 16 a_{1} = 0 \rightarrow a_{3} = 8 a_{1} - \frac{64}{3} a_{0}$

... and so on...

Kind regards

$\chi$ $\sigma$

3. The initial conditions for this equation are y(0)=6 and y''(0)=0.

I'm a bit more confused now though. Where am I supposed to plug those in?

4. Originally Posted by aspotonthewall
The initial conditions for this equation are y(0)=6 and y''(0)=0.

I'm a bit more confused now though. Where am I supposed to plug those in?
No problems!... in that case the 'initial conditions' permit us to write...

$a_{0}=6$

$a_{2}=0$

... and the other $a_{n}$ can be found as follows...

$2 a_{2} - 8 a_{1} + 16 a_{0} = 0 \rightarrow a_{1} = 2 a_{0} = 12$

$6 a_{3} - 16 a_{2} + 16 a_{1} = 0 \rightarrow a_{3} = -\frac{8}{3} a_{1} = -32$

... and so one...

Kind regards

$\chi$ $\sigma$