J4H09-67H08-7H0

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
|