Mathematics: Using power series, solve the lde: (2x^2 + 1) y + 2xy - 4x² y = 0 - -- -

Using power series, solve the lde: (2x^2 + 1) y

Ответ:

dontcareanyonemo

13.03.2022

Mathematics

We're looking for a solution of the form

$y=\displaystyle\sum_{n\ge0}a_nx^n$

with derivatives

$y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n$

$y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n$

Substituting these into the ODE gives

$\displaystyle\sum_{n\ge0}\left(\bigg(2(n+2)(n+1)a_{n+2}-4a_n\bigg)x^{n+2}+2(n+1)a_{n+1}x^{n+1}+(n+2)(n+1)a_{n+2}x^n\right)=0$

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

$2a_2+(2a_1+6a_3)x+\displaystyle\sum_{n\ge2}\bigg((n+2)(n+1)a_{n+2}+2n^2a_n-4a_{n-2}\bigg)x^n=0$

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

$\begin{cases}a_0=y(0)\\\\a_1=y'(0)\\\\a_2=0\\\\2a_1+6a_3=0\implies a_3=-\dfrac{a_1}3\\\\a_n=\dfrac{-2(n-2)^2a_{n-2}+4a_{n-4}}{n(n-1)}&\text{for }n\ge4\end{cases}$

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 y(0)=y'(0)=1 , a degree-8 approximation of

$y(x)\approx1+x-\dfrac{x^3}3+\dfrac{x^4}3+\dfrac{x^5}2-\dfrac{16x^6}{45}-\dfrac{79x^7}{125}+\dfrac{101x^8}{210}$

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).

Using power series, solve the lde: (2x^2 + 1) y + 2xy' - 4x² y = 0 - -- -

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Ответ:

neilaamusic6966

20.02.2020

Mathematics

9514 1404 393

Step-by-step explanation:

The square root of any integer that is not a perfect square will be irrational.

√5 is irrational

Any decimal number that terminates or repeats is a rational number, regardless of sign. All of the decimal numbers shown are rational. (All terminate after a finite number of digits.)

Of course, any ratio of integers is a rational number. That's what the word "rational" means.

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Using power series, solve the lde: (2x^2 + 1) y" + 2xy' - 4x² y = 0 - -- -

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