Respuesta :

Answer:

Step-by-step explanation:

Given is a differential equation

[tex]y"+y=0.[/tex]

SInce it is a power series solution let us assume

[tex]y =[tex]}[/tex][/tex]

Find I and II derivative

[tex]y=a_0+a_1x+...+a_nx^n+...[/tex]

[tex]y'=a_1+2a_2x+3a_3x^2+....na_n x^n +...\\y"= 2a_2 +3(2)a_3x^2+...+n(n-1)x^{n-2} +...[/tex]

Now substitute in the given DE

2a_2 +3(2)a_3x^2+...+a_n n(n-1)x^{n-2} +... a_0+a_1x+....+a_nx^n +....=0\\\\a_{n-2} +n(n-1) a_n =0\\a_n = \frac{-a_{n-2}}{n(n-1)

Thus the solution is a power series with recurring formula

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

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