DÜZCE ÜN· IVERS· ITES· I · FEN-EDEBIYAT FAKÜLTES· I MATEMAT· IK BÖLÜMÜ 2014-2015 BAHAR YARIYILI · DIFERANS· IYEL DENKLEMLER II ARA SINAV 14 Nisan 2015 Süre: 90 dakika CEVAP ANAHTARI 1. (20p) Belirsiz katsay¬lar yöntemini kullanarak d2 y dx2 dy dx 6y = 8e2x 5e3x ; y (0) = 3; y 0 (0) = 5 başlang¬ç-de¼ ger problemini çözünüz. Çözüm: m2 m 6 = 0 ) (m + 2) (m 3) = 0 ) m1 = 2; m2 = 3 ) yc = c1 e 2x S1 = fe2x g S1 = fe2x g ) S = S1 [ S2 = e2x ; xe3x ) S2 = fe3x g S2 = fxe3x g 8 < yp = Ae2x + Bxe3x y 0 = 2Ae2x + 3e3x + 3Bxe3x : p00 yp = 4Ae2x + 6Be3x + 9Bxe3x 4Ae2x + 6Be3x + 9Bxe3x 2Ae2x 3e3x 3Bxe3x e2x : 4A 2A 6A = 8 e3x : 6B B = 5 ) xe3x = 9B 3B 6B = 0 2x y = yc + yp = c1 e y0 = 2c1 e )y=e 2x 2e2x + 3c2 e3x y (0) = 3 ) y 0 (0) = 5 + 4e3x 6Bxe3x = 8e2x A= 2 ) yp = B= 1 + c2 e3x 2x 6Ae2x xe3x 4e2x e3x c1 + c2 = 5 ) 2c1 + 3c2 = 10 2e2x xe3x 2e2x 3xe3x c1 = 1 c2 = 4 2. (20p) Parametrelerin de¼ gişimi yöntemini kullanarak d2 y + y = sec3 x dx2 diferansiyel denkleminin genel çözümünü bulunuz. Çözüm: m2 + 1 = 0 ) m = i ) yc = c1 sin x + c2 cos x yp = v1 (x) sin x + v2 (x) cos x v2 (x) sin x 0 yp0 = v10 | (x) cos x v20 (x) sin x {z } sec3 x 1 v1 (x) sin x xe3x genel çözüm tam çözüm | yp0 = v10 (x) sin x + v20 (x) cos x + v1 (x) cos x | {z } + c2 e3x v2 (x) cos x 5e3x v10 (x) sin x + v20 (x) cos x = 0 v10 (x) cos x v20 (x) sin x = sec3 x 0 sec x sin x cos x cos x sin x cos x sin x 3 v10 (x) = v20 (x) = sin x 0 cos x sec3 x 1 v1 (x) = Z v2 (x) = = = cos x sec3 x) = sec2 x 1 sin x sec3 x = = Z (0 tan x sec2 x sec2 xdx = tan x tan x sec2 xdx = u = tan x du = sec2 xdx Z tan2 x 2 u2 = 2 udu = 1 tan2 x cos x 2 2 2 sin x sin x sin x sin2 x = sin x cos x = cos x 2 cos2 x cos x 2 cos x 2 sin x 1 1 1 = = sin2 x = sin2 x sec x 2 cos x 2 cos x 2 1 2 ) y = yc + yp = c1 sin x + c2 cos x + sin x sec x genel çözüm | 2 yp = v1 (x) sin x + v2 (x) cos x = tan x sin x 3. (20p) Cauchy-Euler yöntemini kullanarak dy d2 y 2x 2 dx dx başlang¬ç-de¼ ger problemini çözünüz. Çözüm: x2 x = et ) t = ln x; x > 0 ) x 2 1 x2 d2 y dt2 m2 3m dy dt 10y = 0; y (1) = 5; y 0 (1) = 4 dy 1 dy d2 y 1 = ) 2 = 2 dx x dt dx x d2 y 10y = 0 ) 2 dt 1 dy 2x x dt d2 y dt2 3 dy dt dy dt 10y = 0 10 = 0 ) (m + 2) (m 5) = 0 ) m1 2; m2 = 5 c1 y = c1 e 2t + c2 e5t = 2 + c2 x5 genel çözüm x 2c1 + 5c2 x4 y0 = 3 x y (1) = 5 c1 + c2 = 5 c1 = 3 ) ) 0 y (1) = 4 2c1 + 5c2 = 4 c2 = 2 )y= 2 + 3x5 2 x genel çözüm | 4. (20p) Kuvvet serileri yöntemini kullanarak 2 d2 y dy +y =0 + (x + 2) 2 dx dx diferansiyel denkleminin genel çözümünü bulunuz. Çözüm: 1 1 1 X dy X d2 y X cn xn ) = ncn xn 1 ) 2 = n (n y= dx dx n=0 n=1 n=2 (x + 3) 1 X n (n 1) cn x n 1 +3 n=2 1 X n (n + n=2 +2 1 X ncn xn n n (n + 1) cn+1 x + 3 n=1 1 X 1 + 1 X +2 n=0 1 X n (n + 1) (n + 2) cn+2 x + 6c2 + 2c1 + c0 + 1 X ncn xn n=1 n (n + 1) cn+1 x + 1 X cn xn = 0 n=0 n=0 1 X ncn xn cn xn = 0 n=0 1 X 2 n=1 n=1 1 X 1) cn x n 2 1) cn xn n (n + 1)2 cn+1 + 3 (n + 1) (n + 2) cn+2 + (n + 1) cn xn = 0 n=1 i) 6c2 + 2c1 + c0 = 0 ) c2 = 1 c0 6 1 c1 3 ii) n (n + 1)2 cn+1 + 3 (n + 1) (n + 2) cn+2 + (n + 1) cn = 0; n n (n + 1)2 cn+1 + (n + 1) cn ;n 3 (n + 1) (n + 2) cn+2 = 4c2 + 2c1 1 = c0 18 27 n = 1 ) c3 = y= 1 X 1 1 1 c1 27 cn xn = c0 + c1 x + c2 x2 + n=0 = c0 + c1 x + 1 c0 6 = c0 1 x2 x3 + + 6 27 ) y = C1 1 x2 x3 + + 6 27 1 c1 x2 + 3 1 c0 27 + c1 x + C2 x x2 3 1 c1 x2 + 27 x2 3 x3 + 27 x3 + 27 genel çözüm | 5. (20p) Frobenius yöntemini kullanarak d2 y dy + 2 + xy = 0 dx2 dx diferansiyel denkleminin genel çözümünü bulunuz. Çözüm: 1 1 X dy X n+r cn x ) = (n + r) cn xn+r y= dx n=0 n=0 x 3 1 d2 y X (n + r = dx2 n=0 1 1 X (n + r 1) (n + r) cn x n+r 1 +2 n=0 1) (n + r) cn xn+r 1 X (n + r) cn x n+r 1 2 + n=0 1 X (n + r + 1) (n + r) cn xn+r 1 cn xn+r+1 = 0 n=0 1 + n=0 r (r + 1) c0 xr 1 X 1 X cn 2 xn+r 1 =0 n=2 + (r + 1) (r + 2) c1 xr + 1 X n=2 i) r (r + 1) c0 xr 1 f(n + r + 1) (n + r) cn + cn 2 g xn+r = 0 () r (r + 1) = 0 () r1 = 0; r2 = 1 1 ii) (r + 1) (r + 2) c1 xr = 0 () c1 = 0 ) c3 = c5 = c7 = =0 cn 2 iii) (n + r + 1) (n + r) cn + cn 2 = 0; n 2 ) cn = ;n (n + r + 1) (n + r) y1 (x) = 2 1 1 c0 ; n = 4 ) c4 = c0 ; 6 120 n = 2 ) c2 = 1 X =0 cn xn = c0 + c1 x + c2 x2 + n=0 c0 2 c0 4 x + x 6 120 = c0 r1 r2 = 0 = c0 1 x4 x2 + 6 120 ( 1) = 1 = N (2. durum) d2 v dv dv ) y200 (x) = 2y10 (x) + vy100 (x) + y1 (x) 2 dx dx dx 2 dv dv dv x y1 (x) 2 + 2y10 (x) + vy100 (x) + 2 y1 (x) + vy10 (x) + xvy1 (x) = 0 dx dx dx y2 (x) = vy1 (x) ) y20 (x) = vy10 (x) + y1 (x) d2 v dv vfxy100 (x) + 2y10 (x) + xy1 (x)g + xy1 (x) 2 + [2xy10 (x) + 2y1 (x)] =0 | {z } dx dx 0 dv dw d2 v ) = 2 dx dx dx Z Z dw dw 0 xy1 (x) + [2xy1 (x) + 2y1 (x)] w = 0 ) + dx w w= Z 2y10 (x) 2 + dx = d (ln c) y1 (x) x c ln jwj + 2 ln jy1 (x)j + 2 ln jxj = ln c ) w [y1 (x)]2 x2 = c ) w = ; c = 1 olsun. [xy1 (x)]2 Z Z dx dx x2 x4 v= ) y2 (x) = c0 1 + 6 120 [xy1 (x)]2 [xy1 (x)]2 = C1 1 Yrd.Doç.Dr. x2 x4 + 6 120 ) y = C1 y1 (x) + C2 y2 (x) Z dx + C2 1 [xy1 (x)]2 Y¬ld¬r¬m ÖZDEM· IR 4 x2 x4 + 6 120 |