w trójkącie równoramiennym abc podstawa AB ma długość 10cm. Odcinek AD jest wysokością tego trójkąta. Wiedząć że |DB|=6cm,oblicz
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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).
