For the differential equation (x+ 2) y\"+xy\' - y = 0 (1): we assume the series
ID: 3087892 • Letter: F
Question
For the differential equation (x+ 2) y"+xy' - y = 0 (1): we assume the series solution at the ordinary point x = 0; of the form y = . The recurrence relation is 4C2 - C0 = 0 12-0 and Cn + 2 = n(c + 1)Cn + 1 + (n - 1)Cn /2(n + 1)(n + 2) = 0 ; n = 1, 2, 3 Use these relation to find 2 linearly independent power series solutions of (1).Explanation / Answer
the differential equation is, y'' - (x - 1)y' - y' - y = 0. here x=0 is an ordinary point since coefficient of y'' is not = 0 at x = 0 Assuming the solution of the given differential equation to be, (x+2)[2c2 + 6c3x + 12c4x^2 + ......+ n(n-1)cnx^(n-2) + ....] + x[c1 + 2c2x + 3c3x^2 + .....+ ncnx^(n-1) + ....] - [c0 + c1x + c2x^2 + .....+ cnx^n +....] = 0 equating to 0 the coefficents of various powers of x, 4c2 - c0 = 0 => c2 = c0/4 2c2 + 12c3 + c1 - c1 = 0 => c2 = -6c3 => c3 = -c0/24 6c3 + 24c4 + 2c2 - c2 = 0 => 6c3 +24c4 + c2 = 0 => 24c4 = -c0/4 +c0/4 = 0 => c4 = 0 12c4 + 40c5 + 3c3 - c3 = 0 => 6c4 + 20c5 + c2 = 0 => c5 = -c0/80 hence, the recurrence relation is, C(n+2) = -[n(n+2)C(n+1) + (n-1)Cn ]/ 2(n+1)(n+2) where n= 1,2,3.... c6 = 3C0/10 / 60 = C0/200 C7 = - [7C0/40 - C0/20 ] / 84 = -C0/8/84 = -c0/672 substituting this values in the eqn, y = C0 + C1x + C2x^2 + C3x^3 + .......+ Cnx^n +..... hence the power series is, y = C0 [ 1 + x^2/4 - x^3/24 - x^5/80 + x^6/200 - x^7/672 ] + C1x +..........