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Aleonysh [2.5K]
3 years ago
9

Find two linearly independent solutions to the equation y"-2xy'+2y=0 in the form of a power series.

Mathematics
1 answer:
ioda3 years ago
6 0

We want a solution in 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 y and its derivatives into the ODE,

y''-2xy'+2y=0

gives

\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n-2\sum_{n\ge0}(n+1)a_{n+1}x^{n+1}+2\sum_{n\ge0}a_nx^n=0

Shift the index on the second sum to have it start at n=1:

\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^{n+1}=\sum_{n\ge1}na_nx^n

and take the first term out of the other two sums. Then we can consolidate the sums into one that starts at n=1:

\displaystyle(2a_2+2a_0)+\sum_{n\ge1}\bigg[(n+2)(n+1)a_{n+2}+(2-2n)a_n\bigg]x^n=0

and so the coefficients in the series solution are given by the recurrence,

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

or more simply, for n\ge2,

a_n=\dfrac{2(n-3)}{n(n-1)}a_{n-2}

Note the dependency between every other coefficient. Consider the two cases,

  • If n=2k, where k\ge0 is an integer, then

k=0\implies n=0\implies a_0=a_0

k=1\implies n=2\implies a_2=-a_0=2^1\dfrac{(-1)}{2!}a_0

k=2\implies n=4\implies a_4=\dfrac{2\cdot1}{4\cdot3}a_2=2^2\dfrac{1\cdot(-1)}{4!}a_0

k=3\implies n=6\implies a_6=\dfrac{2\cdot3}{6\cdot5}a_4=2^3\dfrac{3\cdot1\cdot(-1)}{6!}a_0

k=4\implies n=8\implies a_8=\dfrac{2\cdot5}{8\cdot7}a_6=2^4\dfrac{5\cdot3\cdot1\cdot(-1)}{8!}a_0

and so on, with the general pattern

a_{2k}=\dfrac{2^ka_0}{(2k)!}\displaystyle\prod_{i=1}^k(2i-3)

  • If n=2k+1, then

k=0\implies n=1\implies a_1=a_1

k=1\implies n=3\implies a_3=\dfrac{2\cdot0}{3\cdot2}a_1=0

and we would see that a_{2k+1}=0 for all k\ge1.

So we have

y(x)=\displaystyle\sum_{k\ge0}\bigg[a_{2k}x^{2k}+a_{2k+1}x^{2k+1}\bigg]

so that one solution is

\boxed{y_1(x)=\displaystyle a_0\sum_{k\ge0}\frac{2^k\prod\limits_{i=1}^k(2i-3)}{(2k)!}x^{2k}}

and the other is

\boxed{y_2(x)=a_1x}

I've attached a plot of the exact and series solutions below with a_0=y(0)=1, a_1=y'(0)=1, and 0\le k\le5 to demonstrate that the series solution converges to the exact one.

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