Need quick help with Series, it will only take a few seconds

  • Thread starter Thread starter twisted079
  • Start date Start date
  • Tags Tags
    Seconds Series
twisted079
Messages
25
Reaction score
1
So the differential equation I have to solve using power series is
y''-xy=0 when x0 = -1

So i set it up
Ʃ(n+2)(n+1)an+2(x+1)n - x Ʃ an(x+1)n

I know how to generally solve equations like this, but I never solved one like this, where I have to distribute the x ... x(x+1)n ... I just need to figure this part out (I know I left out the n=0 to ∞)
 
Physics news on Phys.org
Ok I figured it out using a cheap (but valid) math trick... in case anyone is wondering...

xƩan(x+1)n = (x+1-1)Ʃan(x+1)n

Now the x+1 can be distributed to give -Ʃan(x+1)n+1

...anyone care to double check me on this?
 
twisted079 said:
Ok I figured it out using a cheap (but valid) math trick... in case anyone is wondering...

xƩan(x+1)n = (x+1-1)Ʃan(x+1)n

Now the x+1 can be distributed to give -Ʃan(x+1)n+1

...anyone care to double check me on this?

Yep, that's the trick you want to use.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top