find a power series solution for (1 + x)y\' +y = 0 please show steps Solution y=

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Explanation / Answer

y= a0+ a1·x + a2·x²+.... = ? an · x^n n>=0 y'= a1 + 2·a2 x + 3·a3 x²+....+ n· an· x^(n-1)+.... y''= 2·a2 + 6·a3·x + ....+ n·(n-1)· an· x^(n-2)+... ((x^2)+2)y" + 4xy' +2y = 0 2·a2 x²+ 6·a3·x³ + ....+ n·(n-1)· an· x^(n)+... 4·a2 + 12·a3·x + ....+ 2·n·(n-1)· an·x^(n-2)+... 4a1·x + 8·a2 x² + 12·a3 x³+....+4 n· an· x^(n)+.... -2a0-2 a1·x -2 a2·x²- .... - 2·an ·x^n -...=0 (4a2-2a0) + (12·a3+4a1-2a1)·x + (2a2+24a4+8a2 -2a2)x² + .. + [ n·(n-1) an+ 2·(n+2)·(n+1)a(n+2) + 4n·an - 2 an] xn +... = 0 2·(2a2-a0) + 2·(6a3+a1)·x + 8(3a4+a2) x² + ... [ n²+3n-2) an + 2·(n+2)·(n+1) a(n+2)] x^n +.... = 0 a2= a0/2 a3= -a1/6 a4= -a2/3 = -a0/6 a5= -2/5·a3= a1/15 a6=-13/30 a4= 13/180 a0 ----- a(n+2)= (n²+3n-2)/[2·(n²+3n+2)] an

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