Mathematics: Use power series to find two linearly independent solutions of y -2xy +4y = 0

we know that in the rule for the translation of a rectangle the term means a translation of units to the left the term means a translation of units up therefore the answer is the option a translation of 2 units to the left and 7 units up...

Use power series to find two linearly independent

Ответ:

jordan2875

10.03.2022

Mathematics

Assume a general solution of the form

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

with derivatives

$y'(x)=\displaystyle\sum_{n\ge0}na_nx^{n-1}=\sum_{n\ge1}na_nx^{n-1}$

$y''(x)=\displaystyle\sum_{n\ge0}n(n-1)a_nx^{n-2}=\sum_{n\ge2}n(n-1)a_nx^{n-2}$

Substituting into the ODE gives

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-2\sum_{n\ge1}na_nx^n+4\sum_{n\ge0}a_nx^n=0$

The first sum starts with degree 0; the second starts with degree 1; and the third starts with degree 0. So remove the first term from the first and third sums, then consolidate everything into one sum by shifting indices as needed.

First sum:

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}=2a_2+\sum_{n\ge3}n(n-1)a_nx^{n-2}$

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}=2a_2+\sum_{n\ge1}(n+2)(n+1)a_{n+2}x^n$

Third sum:

$\displaystyle\sum_{n\ge0}a_nx^n=a_0+\sum_{n\ge1}a_nx^n$

So the ODE can be expressed as

$\displaystyle\left(2a_2+\sum_{n\ge1}(n+2)(n+1)a_{n+2}x^n\right)-2\sum_{n\ge1}na_nx^n+4\left(a_0+\sum_{n\ge1}a_nx^n\right)=0$

$\displaystyle2a_2+4a_0+\sum_{n\ge1}\bigg((n+2)(n+1)a_{n+2}+(4-2n)a_n\bigg)x^n=0$

Then the coefficients are given by the recurrence,

$\begin{cases}a_0=a_0\\a_1=a_1\\2(2-n)a_n+(n+2)(n+1)a_{n+2}=0&\text{for }n\ge0\end{cases}$

Notice that we should have $a_0\neq0$ and $a_1\neq0$ in order to get non-zero solutions.

For n=2 , we would find that a_4=0 , which would imply that a_n=0 for all even $n\ge2$ . Then the even-indexed terms in the series contribute one solution,