We're looking for a solution of the form

with derivatives


Substituting these into the ODE gives

Shifting indices to get each term in the summand to start at the same power of
and pulling the first few terms of the resulting shifted series as needed gives

Then the coefficients in the series solution are given according to the recurrence

Given the complexity of this recursive definition, it's unlikely that you'll be able to find an exact solution to this recurrence. (You're welcome to try. I've learned this the hard way on scratch paper.) So instead of trying to do that, you can compute the first few coefficients to find an approximate solution. I got, assuming initial values of
, a degree-8 approximation of

Attached are plots of the exact (blue) and series (orange) solutions with increasing degree (3, 4, 5, and 65) and the aforementioned initial values to demonstrate that the series solution converges to the exact one (over whichever interval the series converges, that is).
A quadrilateral does have 4 side


So, yes! Both fractions are equivalent.
The answer is a.
When you are solving problems like this you would use the equation y^2=4ax. The x and Y would be the coordinates. Based on the equation you have given the coordinates would be (6,12) and (6,-12), (24 is split up evenly between the y-axis).
Then we go on to solve the equation by plugging in the numbers 12^2=4a(6)
144=24a
144/24=24a/24
6=a
Answer:
I belive the answer would be 4ajjdbnmsfnksfnbdnsdbcs,mdv bn,zdb xzvndzsmbznx,vsdanvbf. : ) ahaha