Solution of Exam on Ordinary Differential Equations. Trial Aleph ...

<strong>Solution</strong> <strong>of</strong> <strong>Exam</strong> on Ordinary
Di¤erential Equations.
Trial Aleph
19.01.2005, 9:00-12:00
January 26, 2005
1 Problem 1
1.1 <strong>Solution</strong>
3x 2 y 00 + 8xy 0 2y = 0 (1)
x 0 = 0
Here P 1 (x) = 8
3x
) x 2 0 = 0 is a singular point; P 2 (x) = 2
3x
; 2
xP 1 (x) = 8 3 ; x2 P 2 (x) = 2
3 ) x 0 = 0 is a singular regular point;
p 1 (0) = 8 3 ; p 2 (0) = 2
3 ; 2 +(p 1 (0) 1) +p 2 (0) = 2 + 8 3
1
+ 2
3 =
2 + 5 3 2
3 :
The indicial equation is:
3 2 + 5 2 = 0 (2)
1 = 1 3 ; 2 = 2; 1 2 = 5 3 =2 Z: 1

0
X 1
y (x) = x a n x n ; a 0 6= 0; jxj < 1 (3)
n=0
X
1
y 0 (x) = x (n + ) a n x n 1 ;
n=0
X
1
xy 0 (x) = x (n + ) a n x n ; (4)
n=0
X
1
y 00 (x) = x (n + ) (n + 1) a n x n 2 ;
n=0
X
1
x 2 y 00 (x) = x (n + ) (n + 1) a n x n ; (5)
n=0
3x 2 y 00 + 8xy 0 2y =
= 3 (x P 1
n=0 (n + ) (n + 1) a nx n )+8 (x P 1
n=0 (n + ) a nx n ) 2x P 1
n=0 a nx n =
x P 1
n=0 (3 (n + ) (n + 1) + 8 (n + ) 2) a nx n = 0
1X
(3 (n + ) (n + 1) + 8 (n + ) 2) a n x n = 0; jxj < +1 (6)
n=0
(3 ( 1) + 8 2) a 0 = 0; a 0 6= 0 , 3 2 + 8 2 = 0 , 2 + 5 3 2
3 = 0 (7)
(3 (n + ) (n + 1) + 8 (n + ) 2) a n = 0; n = 1; 2; :::;
= 1 ; = 2 : 3 (n + ) (n + 1) + 8 (n + ) 2 6= 0; n = 1; 2; :::; ,
3
, a n = 0; n = 1; 2; :::: (8)
= 1 = 1 3 :
y 1 (x) = x 1
P 1
n=0 a nx n = 3p x a 0 + a 1 x + a 2 x 2 + ::: = a 0
3 p x; 8a 0 2 R; a 0 6=
0 ) a 0 = 1
y 1 (x) = 3p x (9)
= 1 = 2 :
y 2 (x) = x P
2 1
n=0 a nx n = x 2 a 0 + a 1 x + a 2 x 2 + ::: 1
= a 0 x
; 8a 2 0 2 R; a 0 6=
0 ) a 0 = 1
y 1 (x) = 1 x 2 (10)
2

1.2 Answer
y (x) = C 1
3 p x + C 2
1
x 2 ; 8C 1; C 2 2 R; (11)
2 Problem 2
y 00 + y = 1
cos x
(12)
2.1 <strong>Solution</strong>
2.1.1 <strong>Solution</strong> <strong>of</strong> homogeneous equation
, Exact solution is: C 1 cos x + C 2 sin x;
2.1.2 Variation <strong>of</strong> parameters
C 1
def
= C 1 (x) ; C 2
def
= C 2 (x)
y 00 + y = 0 (13)
y 1 (x) = cos x; y 2 (x) = sin x (14)
, <strong>Solution</strong> is: [C 0 1 = tan x; C 0 2 = 1]
C1 0 cos x + C2 0 sin x = 0 (15)
C1 0 sin x + C2 0 cos x =
1
cos x
(16)
Z
C1 0 (x) = tan x , C 1 (x) = tan xdx = ln jcos xj + C 1 (17)
Z
C2 0 (x) = 1 , C 2 (x) = dx = x + C 2 (18)
2.2 Answer
y (x) = (ln jcos xj + C 1 ) cos x + (x + C 2 ) sin x
y (x) = (ln jcos xj) cos x + x sin x + C 1 cos x + C 2 sin x (19)
3 Problem 3-aleph
y 0 =
2xy + y2
x 2 (20)
3

3.1 <strong>Solution</strong>
2xy+y 2
x
= 2xy
2 x
+ y2
2 x
= 2 y 2 x +
y 2
x
, y
0
= 2 y
x +
y 2
x
;so equation (20) is
homogeneous equation.
y
x = z , y = zx; y0 = z 0 x + z; z 0 x + z = 2z + z 2 ; z 0 x = z + z 2 ;
dz
z+z
= dx; R dz
2 z+z 2
1
z+z
= 1 2 z
ln
z
(z+1)x
z = y x )
= R dx
x ;
1
z+1 )R dz
z+z
= R 1
2 z
= C;
z
(z+1)x
z
z+1 = y x
y
x +1 =
dz
dx x = z + z2 (21)
1
z+1
dz = ln jzj
ln jz + 1j = ln
z
z+1
ln
z
z + 1 = ln jxj + C (22)
= e c = jCj
y
x+y ;
z
(z + 1) x = C
z
(z+1)x =
y
(x+y)x ;
y
(x + y) x = C
y = Cx (x + y) = Cxy + Cx 2 ; y Cxy = Cx 2 ; y (1 Cx) = Cx 2
y (1) = 1 : 1 y
2
= ln 1 + C = C;
y =
Cx2
1 Cx
(x+y)x = 1 2 () y =
x2 = 2x2
1 1 2 x 2 x
3.2 Answer
y
(x + y) x = 1 2x2
() y =
2 2 x
(23)
4 Problem 3-bet
4.1 <strong>Solution</strong>.
Equation (24) is an equation <strong>of</strong> the form
where p (x) = 1; q (x) = e x ; m = 3:
y (x) = 0 is a solution (partial or singular?)
Now assume y 6= 0:
y 0 + y = e x y 3 (24)
y 0 + p (x) y = q (x) y m ; (25)
4

4.1.1 General solution. The …rst way
y 0 + y = e x y 3 , y0
y
+ 1 3 y
= e x
2
z = 1
y
) z 0 = 2y0
2 y
, y0
3 y
= 1 3 2 z0
1
2 z0 + z = e x , z 0 2z = 2e x
z 0 2z = 2e x (26)
Ce
z = e 2x C 2 R e 2x e x dx = e 2x C 2 R e x dx = Ce 2x + 2e 2x e x =
2x + 2e x z = Ce 2x + 2e x (27)
z = 1
y 2 , 1
y 2 = Ce 2x + 2e x , y 2 =
1
Ce 2x +2e x
y 2 =
1
Ce 2x + 2e x ; C 2 R (28)
lim y (x) = lim p 1
C!1 C!1 Ce
= 0;so y (x) = 0 is a partial solution.
2x +2e x
1
y (x) = p
Ce
2x
+ 2e ; C 2 R
x
4.1.2 General solution. The second way
y (x) = u (x) v (x) ; y 0 = u 0 v + uv 0 ; u 0 v + uv 0 ; y 0 + y = u 0 v + uv 0 + uv = e x u 3 v 3
u 0 v + u (v 0 + v) = e x u 3 v 3
v 0 + v = 0 ) v = e x
u 0 v = e x u 3 v 3 ; y 6= 0 ) v 6= 0 , u 0 = e x u 3 v 2 = e x u 3 e 2x = u 3 e x
du
dx = u3 e x ; du
u
1
3
C;
e x +C = 2u2 ;
u 0 = u 3 e x
= e x dx; R du
u 3 = R e x dx;
1
2u 2 = e x + C;
1
2u 2 = e x +
u 2 =
1
2(e x +C) ; y2 = u 2 v 2 = u 2 e 2x =
1
2e 2x (e x +C) = 1
2e x +Ce 2x ;
y =
1
p
2ex + Ce 2x
4.1.3 Partial solution satisfying given initial conditions
y (0) = 1 2 : 1 4 = 1
C+2 ; C + 2 = 4; C = 2
y (x) =
1
p
2 (e
2x
+ e x )
5

4.2 Answer
y (x) =
1
p
2 (e
2x
+ e x )
(29)
5 Problem 4
y 00 2y 0 + 2y = e x cos x (30)
5.1 <strong>Solution</strong>
5.1.1 <strong>Solution</strong> <strong>of</strong> homogeneous equation
y 00 2y 0 + 2y = 0 (31)
, <strong>Solution</strong> is: 1 i; 1 + i
2 2 + 2 = 0 (32)
5.1.2 First way(Conjecture Method)
y = e x (C 1 cos x + C 2 sin x) (33)
e x cos x = Re e (1+i)x ; = 1 + i is a root <strong>of</strong> the characteristic equation (32) <strong>of</strong>
multiplicity 1, so the partial solution <strong>of</strong> equation (30) is obtained by conjecture
y p (x) = xe x (A cos x + B sin x) (34)
y 0 p (x) = d
dx (xex (A cos x + B sin x)) =
= A (cos x) e x + B (sin x) e x + Ax (cos x) e x + Bx (cos x) e x Ax (sin x) e x +
Bx (sin x) e x =
= Ae x cos x + Be x sin x + (A + B) xe x cos x + ( A + B) x sin x
y 0 p (x) = e x (A cos x + B sin x + (A + B) x cos x + ( A + B) x sin x) (35)
y 00
p (x) = d2
dx 2 (xe x (A cos x + B sin x)) =
= d
dx (ex (A cos x + B sin x + (A + B) x cos x + ( A + B) x sin x)) =
= 2A (cos x) e x +2B (cos x) e x 2A (sin x) e x +2B (sin x) e x +2Bx (cos x) e x
2Ax (sin x) e x =
= (2A + 2B) e x cos x + (2B 2A) e x sin x + 2Bxe x cos x 2Axe x sin x
y 00
p (x) = 2e x ((A + B) cos x + (B A) sin x + Bx cos x Ax sin x) (36)
y 00
p 2y 0 p + 2y p =
= 2e x ((A + B) cos x + (B A) sin x + Bx cos x Ax sin x)
6

2e x ((A cos x + B sin x + (A + B) x cos x + ( A + B) x sin x)) +
+2e x (Ax cos x + Bx sin x) = e x cos x
2 ((A + B) cos x + (B A) sin x + Bx cos x Ax sin x)
2 ((A cos x + B sin x + (A + B) x cos x + ( A + B) x sin x)) +
+2 (Ax cos x + Bx sin x) = cos x
((A + B) cos x + (B A) sin x + Bx cos x Ax sin x)
((A cos x + B sin x + (A + B) x cos x + ( A + B) x sin x)) +
+ (Ax cos x + Bx sin x) = 1 2 cos x
0
A + B A 1 2
cos x + (B A B) sin x + (B A B + A) x cos x + ( A + A B + B) x sin x
=
cos x A sin x
= 0
A = 0; B = 1 2
B 1 2
(37)
y p (x) = 1 2 ex sin x (38)
y (x) = 1 2 xex sin x + e x (C 1 cos x + C 2 sin x) ; C 1 ; C 2 2 R (39)
5.1.3 Second way(Variation <strong>of</strong> Parameters Method)
C 1
def
def
= C 1 (x) ; C 2 = C 2 (x)
y 1 (x) = e x cos x; y1 0 (x) = d
dx (ex cos x) = e x (cos x sin x)
y 2 (x) = e x sin x; y2 0 (x) = d
dx (ex sin x) = e x (cos x + sin x)
C 0 1e x cos x + C 0 2e x sin x = 0 (40)
C 0 1e x (cos x sin x) + C 0 2e x (cos x + sin x) = e x cos x (41)
C 0 1 cos x + C 0 2 sin x = 0 (42)
C 0 1 (cos x sin x) + C 0 2 (cos x + sin x) = cos x (43)
C1 0 = sin x cos x; C2 0 = cos 2 x
C 1 (x) = R sin x cos xdx = 1 4 cos 2x 1
4 + C 1;
C 2 (x) = R cos 2 xdx = 1 2 x + 1 4 sin 2x + C 2
y (x) = 1 4 cos 2x 1
4 + C 1
e x cos x + 1 2 x + 1 4 sin 2x + C 2
e x sin x =
= 1 4 ex (cos 2x cos x + sin 2x sin x)+e x 1
C 1 4
cos x + C2 sin x + 1 2 xex sin x 1 =
= 1 4 ex cos x+e x 1
C 1 4
cos x + C2 sin x + 1 2 xex sin x = 1 2 xex sin x+e x (C 1 cos x + C 2 sin x)
1 cos 2x cos x + sin 2x sin x = cos (2x x) = cos x
7

5.2 Answer
y (x) = 1 2 xex sin x + e x (C 1 cos x + C 2 sin x) ; C 1 ; C 2 2 R (44)
6 Problem 5
_x =
_y =
cos 2 x
tan y tan x
cos 2 y
tan y tan x
(45)
(46)
6.1 <strong>Solution</strong>
_y
_x = y0 x =
cos2 y
tan y tan x
cos 2 x
tan y tan x
dy
dx = cos2 y
cos 2 x ;
dy
cos 2 y =
= cos2 y
cos 2 x ;
dx
cos 2 x ; R
y 0 = cos2 y
cos 2 x
dy
cos 2 y = R
dx
cos 2 x ; tan y = tan x + C 1;
(47)
cos 2 x
tan y tan x = cos2 x
C 1
;
tan y tan x = C 1
cos 2 y
tan y tan x = cos2 y
C 1
_x = cos2 x
C
_y = cos2 y
C
dx
dt = cos2 x
C ; dx
cos 2 x = dt
C ; R dx
cos 2 x = R dt
C ; tan x = 1 C t+C 1; x = arctan 1 C t + C 1
:
dy
dt = cos2 y
C ; dy
cos 2 y = dt
C ; R dy
cos 2 y = R dt
C ; tan y = 1 C t+C 2; y = arctan 1 C t + C 2
:
tan y tan x = C , 1 C t + C
1
2 C t + C 1
= C2 C 1 = C
6.2 Answer
1
x (t) = arctan t + C 1
C 2 C 1
1
y (t) = arctan t + C 2
C 2 C 1
8C 1 ; C 2 2 R; C 1 6= C 2 :
8

7 Problem 6
y 0 = 1
x
y
7.1 <strong>Solution</strong>. First way
x y = z; 1 y 0 = z 0 ; y 0 = 1 z 0 ; 1 z 0 = 1 z ; z0 1
= 1
dz
dx = z 1
z
; dz
z 1 = dx; zdz
z 1 = dx; R zdz
z 1
z
x y ln jx y 1j = x + C , y + ln jx y 1j = C
z = z 1
z ;
= dx; z + ln jz 1j = x + C;
7.2 Aleph.
7.2.1 Answer
y + ln jx y 1j = C; C 2 R
7.3 Bet
y (3) = 0 :
0 + ln (3 0 1) = ln 2 = C
7.3.1 Answer
ln
jx y 1j
2
= y (48)
7.4 <strong>Solution</strong>. Second way
y 0 = dy
dx = 1
x
y , dx
dy = x 0 = x y (49)
Equation (49) is a linear nonhomogeneous equation <strong>of</strong> the …rst order.
7.4.1 Aleph
x = e y C
Z
e y ydy = e y C e y e y ye y = e y C + y + 1
7.4.2 Answer
7.4.3 Bet
x = e y C + y + 1 (50)
y (3) = 0 , x (0) = 3 :
3 = C + 1; C = 2
9

7.4.4 Answer
x = 2e y + y + 1
8 Problem 7
8.1 <strong>Solution</strong>
x 2 2x + 3 y 00 + (2x + 1) y 0 + y = 0 (51)
y (1) = y 0 (1) = 1
x 0 = 1 ) x 2 0 2x 0 + 3 = 1 2 + 3 = 2 6= 0 ) x 0 = 1 is a regular point.
x 1 = t , x = t + 1; z (t) def
= y (t + 1) , y (x) = z (x 1) ; y 0 = z 0 ; y 00 = z 00
x 2 2x + 3 y 00 +(2x + 1) y 0 +y =
z =
= t 2 + 2 z 00 + (2t + 3) z 0 + z
(t + 1) 2 2 (t + 1) + 3 z 00 +(2 (t + 1) + 1) z 0 +
t 0 = 0 is a regular point.
t 2 + 2 z 00 + (2t + 3) z 0 + z = 0 (52)
z (0) = z 0 (0) = 1
z =
z 0 =
tz 0 =
z 00 =
t 2 z 00 =
1X
a n t n (53)
n=0
1X
1X
na n t n 1 = (n + 1) a n+1 t n (54)
n=0
n=0
1X
na n t n (55)
n=0
1X
1X
n (n 1) a n t n 2 = (n + 2) (n + 1) a n+2 t n (56)
n=0
n=0
1X
n (n 1) a n t n (57)
n=0
t 2 + 2 z 00 + (2t + 3) z 0 + z = t 2 z 00 + 2z 00 + 2tz 0 + 3z 0 + z = 0
P 1
n=0 n (n 1) a nt n + P 1
n=0 2 (n + 2) (n + 1) a n+2t n + P 1
n=0 2na nt n + P 1
P n=0 3 (n + 1) a n+1t n +
1
n=0 a nt n = 0
10

P 1
n=0 (n (n 1) a n + 2 (n + 2) (n + 1) a n+2 + 2na n + 3 (n + 1) a n+1 + a n ) t n =
0
P 1
n=0 (2 (n + 2) (n + 1) a n+2 + 3 (n + 1) a n+1 + (n (n 1) + 2n + 1) a n ) t n =
0
2 (n + 2) (n + 1) a n+2 +3 (n + 1) a n+1 +(n (n 1) + 2n + 1) a n = 0; n = 0; 1; 2; :::
(58)
a n+2 =
3
2 (n + 2) a n+1
n 2 + n + 1
2 (n + 2) (n + 1) a n; n = 0; 1; 2; ::: (59)
z (0) = a 0 = 1; z 0 (0) = a 1 = 1
n = 0 :
a 2 = 3 4 a 1
1 4 a 0 = 1
n = 1 :
a 3 = 3 6 a 2
n = 2 :
a 4 = 3 8 a 3
3
12 a 1 = 1 2 ( 1) 1
4 = 1 4
7
24 a 2 = 3 8 1 4
7
19
24
( 1) =
96
n = 3 :
a 5 =
3
2(3+2) a 4
9+3+1
2(3+2)(3+1) a 3 = 3
10
19
96
13
40 1 4 = 9
64
n = 4 :
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
z (t) = a 0 +a 1 t+a 2 t 2 +a 3 t 3 +a 4 t 4 +a 5 t 5 +::: = 1+t t 2 + 1 4 t3 + 19
y (x) = z (x 1)
8.2 Answer
96 t4 9
64 t5 +:::;
y (x) = 1 + (x 1) (x 1) 2 + 1 4 (x 1)3 + 19
96 (x 1)4 9
64 (x 1)5 + :::
11