# Series solution to ODE

• May 7th 2010, 05:12 AM
aspotonthewall
Series solution to ODE
I have an equation

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

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

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

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

Any guidance is greatly appreciated :)
• May 7th 2010, 12:27 PM
chisigma
The DE is...

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

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

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

From (2) we derive first...

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

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

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

$\displaystyle a_{0} = y_{0}$

$\displaystyle a_{1} = y_{1}$

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

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

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

... and so on...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• May 7th 2010, 01:02 PM
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?
• May 7th 2010, 10:04 PM
chisigma
Quote:

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...

$\displaystyle a_{0}=6$

$\displaystyle a_{2}=0$

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

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

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

... and so one...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$