Question #155636

Using power series method solve the equation

y''+9y=0


1
Expert's answer
2021-01-14T20:10:33-0500

y=n=0+cnxny = \sum\limits_{n=0}^{+\infty} c_n x^n ,

y=n=0+n(n1)cnxn2=n=0+(n+2)(n+1)cn+2xny'' = \sum\limits_{n=0}^{+\infty} n(n-1)c_n x^{n-2} = \sum\limits_{n=0}^{+\infty} (n+2)(n+1)c_{n+2} x^{n}

y+9y=n=0+((n+2)(n+1)cn+2+9cn)xn=0y'' + 9y = \sum\limits_{n=0}^{+\infty} ((n+2)(n+1)c_{n+2}+9c_n) x^{n} = 0

(n+2)(n+1)cn+2+9cn=0(n+2)(n+1)c_{n+2}+9c_n = 0

cn+2=9(n+2)(n+1)cnc_{n+2}=\frac{-9}{(n+2)(n+1)}c_n

c2n=(9)n(2n)!c0c_{2n} = \frac{(-9)^n}{(2n)!}c_0

c2n+1=(9)n(2n+1)!c1c_{2n+1} = \frac{(-9)^n}{(2n+1)!}c_1

y=n=0+(9)n(2n)!c0xn+n=0+(9)n(2n+1)!c1xny=\sum\limits_{n=0}^{+\infty} \frac{(-9)^n}{(2n)!}c_0x^n + \sum\limits_{n=0}^{+\infty} \frac{(-9)^n}{(2n+1)!}c_1 x^n

y=c0cos3x+c1sin3xy= c_0\cos{3x} + c_1\sin{3x}


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