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Engineering Mathematics II (2M03)Tutorial 3
Marina ChugunovaDepartment of Math. & Stat., office: HH403
e-mail: [email protected]
office hours: Math Help Centre, Thursday 1:30 - 3:30
September 27-28, 2007
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 4)
If the differential equation(sin y − y sin x)dx + (cos x + x cos y − y)dy = 0 is exact, solve it.
Solution:
2
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 4)
If the differential equation(sin y − y sin x)dx + (cos x + x cos y − y)dy = 0 is exact, solve it.
Solution:M(x, y) = sin y − y sin x, My = ∂M
∂y = cos y − sin x
3
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 4)
If the differential equation(sin y − y sin x)dx + (cos x + x cos y − y)dy = 0 is exact, solve it.
Solution:
M(x, y) = sin y − y sin x, My = ∂M∂y = cos y − sin x
N(x, y) = cos x + x cos y − y, Nx = ∂N∂x = − sin x + cos y
My = Nx (the equation is exact)
4
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 4)
If the differential equation(sin y − y sin x)dx + (cos x + x cos y − y)dy = 0 is exact, solve it.
Solution:
M(x, y) = sin y − y sin x, My = ∂M∂y = cos y − sin x
N(x, y) = cos x + x cos y − y, Nx = ∂N∂x = − sin x + cos y
My = Nx (the equation is exact)
To solve the exact equation we should find f (x, y)∂f∂x = M(x, y) = sin y − y sin x, f (x, y) = x sin y + y cos x + g(y)
5
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 4)
If the differential equation(sin y − y sin x)dx + (cos x + x cos y − y)dy = 0 is exact, solve it.
Solution:
M(x, y) = sin y − y sin x, My = ∂M∂y = cos y − sin x
N(x, y) = cos x + x cos y − y, Nx = ∂N∂x = − sin x + cos y
My = Nx (the equation is exact)
To solve the exact equation we should find f (x, y)∂f∂x = M(x, y) = sin y − y sin x, f (x, y) = x sin y + y cos x + g(y)∂f∂y = x cos y + cos x+ g′(y) = N(x, y), x cos y + cos x+ g′(y) = cos x+x cos y− y
6
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 4)
If the differential equation(sin y − y sin x)dx + (cos x + x cos y − y)dy = 0 is exact, solve it.
Solution:
M(x, y) = sin y − y sin x, My = ∂M∂y = cos y − sin x
N(x, y) = cos x + x cos y − y, Nx = ∂N∂x = − sin x + cos y
My = Nx (the equation is exact)
To solve the exact equation we should find f (x, y)
∂f∂x = M(x, y) = sin y − y sin x, f (x, y) = sin yx + y cos x + g(y)∂f∂y = x cos y + cos x+ g′(y) = N(x, y), x cos y + cos x+ g′(y) = cos x+x cos y− y
g′(y) = −y g(y) = −y2
2 ,
f (x, y) = x sin y + y cos x − y2
2 , x sin(y) + y cos x − y2
2 = c
7
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 8)
If the differential equation (1 + ln x + yx)dx = (1 − ln x)dy is exact, solve it.
Solution:
8
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 8)
If the differential equation (1 + ln x + yx)dx = (1 − ln x)dy is exact, solve it.
Solution:
(1 + ln x + yx)dx + (ln x − 1)dy = 0
9
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 8)
If the differential equation (1 + ln x + yx)dx = (1 − ln x)dy is exact, solve it.
Solution:
(1 + ln x + yx)dx + (ln x − 1)dy = 0
M(x, y) = 1 + ln x + yx, My = ∂M
∂y = 1x
10
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 8)
If the differential equation (1 + ln x + yx)dx = (1 − ln x)dy is exact, solve it.
Solution:
(1 + ln x + yx)dx + (ln x − 1)dy = 0
M(x, y) = 1 + ln x + yx, My = ∂M
∂y = 1x
N(x, y) = ln x − 1, Nx = ∂N∂x = 1
xMy = Nx (the equation is exact)
the same method of the solution as in (2.4: 4)
11
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 22)
Solve the initial value problem (ex + y)dx + (2 + x + yey)dy = 0, y(1) = 1.
Solution:
12
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 22)
Solve the initial value problem (ex + y)dx + (2 + x + yey)dy = 0, y(1) = 1.
Solution:
M(x, y) = ex + y, My = 1, N(x, y) = 2 + x + yey, Nx = 1My = Nx (the equation is exact)
13
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 22)
Solve the initial value problem (ex + y)dx + (2 + x + yey)dy = 0, y(1) = 1.
Solution:
M(x, y) = ex + y, My = 1, N(x, y) = 2 + x + yey, Nx = 1My = Nx (the equation is exact)Find f (x, y)
fx = ex + y, f (x, y) = ex + yx + g(y)
14
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 22)
Solve the initial value problem (ex + y)dx + (2 + x + yey)dy = 0, y(1) = 1.
Solution:
M(x, y) = ex + y, My = 1, N(x, y) = 2 + x + yey, Nx = 1My = Nx (the equation is exact)Find f (x, y)
fx = ex + y, f (x, y) = ex + yx + g(y)fy = x + g′(y), x + g′(y) = 2 + x + yey
g′(y) = 2 + yey, g(y) = 2y + yey − ey
15
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 22)
Solve the initial value problem (ex + y)dx + (2 + x + yey)dy = 0, y(1) = 1.
Solution:
M(x, y) = ex + y, My = 1, N(x, y) = 2 + x + yey, Nx = 1My = Nx (the equation is exact)Find f (x, y)
fx = ex + y, f (x, y) = ex + yx + g(y)fy = x + g′(y), x + g′(y) = 2 + x + yey
g′(y) = 2 + yey, g(y) = 2y + yey − ey
f (x, y) = c, ex + yx + 2y + yey − ey = c
16
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 22)
Solve the initial value problem (ex + y)dx + (2 + x + yey)dy = 0, y(1) = 1.
Solution:
M(x, y) = ex + y, My = 1, N(x, y) = 2 + x + yey, Nx = 1My = Nx (the equation is exact)Find f (x, y)
fx = ex + y, f (x, y) = ex + yx + g(y)fy = x + g′(y), x + g′(y) = 2 + x + yey
g′(y) = 2 + yey, g(y) = 2y + yey − ey
f (x, y) = c, ex + yx + 2y + yey − ey = cx = 1, y = 1, e + 3 = c, ex + yx + 2y + yey − ey = e + 3
17
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 28)Find the value of k so that the equation(6xy3 + cos y)dx + (2kx2y2 − x sin y)dy = 0 is exact.
Solution:
18
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 28)Find the value of k so that the equation(6xy3 + cos y)dx + (2kx2y2 − x sin y)dy = 0 is exact.
Solution:
M(x, y) = 6xy3 + cos y, My = 18xy2 − sin y
19
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 28)Find the value of k so that the equation(6xy3 + cos y)dx + (2kx2y2 − x sin y)dy = 0 is exact.
Solution:
M(x, y) = 6xy3 + cos y, My = 18xy2 − sin yN(x, y) = 2kx2y2 − x sin y, Nx = 4kxy2 − sin y
20
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 28)Find the value of k so that the equation(6xy3 + cos y)dx + (2kx2y2 − x sin y)dy = 0 is exact.
Solution:
M(x, y) = 6xy3 + cos y, My = 18xy2 − sin yN(x, y) = 2kx2y2 − x sin y, Nx = 4kxy2 − sin y
My = Nx, 18xy2 = 4kxy2, k =9
2.
21
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 32)Find the integrating factor and solve the equationy(x + y + 1)dx + (x + 2y)dy = 0.
Solution:
22
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 32)Find the integrating factor and solve the equationy(x + y + 1)dx + (x + 2y)dy = 0.
Solution:
M(x, y) = yx + y2 + y, My = x + 2y + 1, N(x, y) = x + 2y, Nx = 1
23
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 32)Find the integrating factor and solve the equationy(x + y + 1)dx + (x + 2y)dy = 0.
Solution:
M(x, y) = yx + y2 + y, My = x + 2y + 1, N(x, y) = x + 2y, Nx = 1My−Nx
N = x+2yx+2y = 1
24
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 32)Find the integrating factor and solve the equationy(x + y + 1)dx + (x + 2y)dy = 0.
Solution:
M(x, y) = yx + y2 + y, My = x + 2y + 1, N(x, y) = x + 2y, Nx = 1My−Nx
N = x+2yx+2y = 1
Integrating factor µ(x) = e∫
1dx = ex
25
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 32)Find the integrating factor and solve the equationy(x + y + 1)dx + (x + 2y)dy = 0.
Solution:
M(x, y) = yx + y2 + y, My = x + 2y + 1, N(x, y) = x + 2y, Nx = 1My−Nx
N = x+2yx+2y = 1
Integrating factor µ(x) = e∫
1dx = ex
Exact equation:exy(x + y + 1)dx + ex(x + 2y)dy = 0fx = exy(x + y + 1), f(x, y) = yxex + y2ex + g(y)
26
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 32)Find the integrating factor and solve the equationy(x + y + 1)dx + (x + 2y)dy = 0.
Solution:
M(x, y) = yx + y2 + y, My = x + 2y + 1, N(x, y) = x + 2y, Nx = 1My−Nx
N = x+2yx+2y = 1
Integrating factor µ(x) = e∫
1dx = ex
Exact equation:exy(x + y + 1)dx + ex(x + 2y)dy = 0fx = exy(x + y + 1), f(x, y) = yxex + y2ex + g(y)fy = xex + 2yex = xex + 2yex + g′(y), g(y) = c, yxex + y2ex = c
27
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 32)Find the functions M(x, y) and N(x, y) so that each differential equationis exact.a) M(x, y)dx + (xexy + 2xy + 1
x)dy = 0,
b)(x−1/2y1/2 + x
x2+y
)dx + N(x, y)dy = 0.
Solution a):
28
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 32)Find the functions M(x, y) and N(x, y) so that each differential equationis exact.a) M(x, y)dx + (xexy + 2xy + 1
x)dy = 0,
b)(x−1/2y1/2 + x
x2+y
)dx + N(x, y)dy = 0.
Solution a):N(x, y) = xexy + 2xy + 1
x, Nx = exy + xyexy + 2y − 1x2 = My
29
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 32)Find the functions M(x, y) and N(x, y) so that each differential equationis exact.a) M(x, y)dx + (xexy + 2xy + 1
x)dy = 0,
b)(x−1/2y1/2 + x
x2+y
)dx + N(x, y)dy = 0.
Solution a):N(x, y) = xexy + 2xy + 1
x, Nx = exy + xyexy + 2y − 1x2 = My
By integration:
M(x, y) = y2 − 1x2y + yexy
30
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 32)Find the functions M(x, y) and N(x, y) so that each differential equationis exact.a) M(x, y)dx + (xexy + 2xy + 1
x)dy = 0,
b)(x−1/2y1/2 + x
x2+y
)dx + N(x, y)dy = 0.
Solution b):
M(x, y) = x−1/2y1/2 + xx2+y
, My = 12
√1xy − x
(y+x2)2= Nx
31
First-Order Differential Equations (2.4 Exact Equations )
Problem (2.4: 32)Find the functions M(x, y) and N(x, y) so that each differential equationis exact.a) M(x, y)dx + (xexy + 2xy + 1
x)dy = 0,
b)(x−1/2y1/2 + x
x2+y
)dx + N(x, y)dy = 0.
Solution b):
M(x, y) = x−1/2y1/2 + xx2+y
, My = 12
√1xy − x
(y+x2)2= Nx
By integration:
N(x, y) =√
xy + 1
2(y+x2)
32
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 4)Solve ydx = 2(x + y)dy by substitution.
Solution:
33
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 4)Solve ydx = 2(x + y)dy by substitution.
Solution:
Substitution is y = ux, dy = udx + xdu
34
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 4)Solve ydx = 2(x + y)dy by substitution.
Solution:
Substitution is y = ux, dy = udx + xduuxdx = 2(x + ux)(udx + xdu), (u + u2)dx + x(2 + u)du = 0
35
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 4)Solve ydx = 2(x + y)dy by substitution.
Solution:
Substitution is y = ux, dy = udx + xduuxdx = 2(x + ux)(udx + xdu), (u + u2)dx + x(2 + u)du = 0
1
xdx =
(1
u + 1− 2
u
)du
36
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 4)Solve ydx = 2(x + y)dy by substitution.
Solution:
Substitution is y = ux, dy = udx + xduuxdx = 2(x + ux)(udx + xdu), (u + u2)dx + x(2 + u)du = 0
1
xdx =
(1
u + 1− 2
u
)du
ln |x| + ln c = ln |u + 1| − 2 ln |u|
37
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 4)Solve ydx = 2(x + y)dy by substitution.
Solution:
Substitution is y = ux, dy = udx + xduuxdx = 2(x + ux)(udx + xdu), (u + u2)dx + x(2 + u)du = 0
1
xdx =
(1
u + 1− 2
u
)du
ln |x| + ln c = ln |u + 1| − 2 ln |u|
ln |x| + ln c = ln |y/x + 1| − 2 ln |y/x|you can simplify it
38
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 10)
Solve xdydx = y +
√x2 − y2, x > 0 by substitution.
Solution:
39
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 10)
Solve xdydx = y +
√x2 − y2, x > 0 by substitution.
Solution:
Substitution is y = ux
x(udx + xdu) = uxdx +√
x2 − u2x2dx
40
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 10)
Solve xdydx = y +
√x2 − y2, x > 0 by substitution.
Solution:
Substitution is y = ux
x(udx + xdu) = uxdx +√
x2 − u2x2dx
xdu =√
1 − u2dx, sin−1 y
x= ln x + c
41
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 12)Solve initial value problem (x2 + 2y2)dx
dy = xy, y(−1) = 1.
Solution:
42
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 12)Solve initial value problem (x2 + 2y2)dx
dy = xy, y(−1) = 1.
Solution:
Substitution is y = ux
(1 + u2)dx = uxdu
43
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 12)Solve initial value problem (x2 + 2y2)dx
dy = xy, y(−1) = 1.
Solution:
Substitution is y = ux
(1 + u2)dx = uxdu
2 ln |x| = ln 1 + u2 + c
44
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 12)Solve initial value problem (x2 + 2y2)dx
dy = xy, y(−1) = 1.
Solution:
Substitution is y = ux
(1 + u2)dx = uxdu
2 ln |x| = ln 1 + y2/x2 + ln c
Find c from x = −1, y = 1
c = 1/2, 2x4 = y2 + x2
45
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 4)Solve xdy
dx − (1 + x)y = xy2 (Bernoulli’s equation) by substitution.
Solution:
46
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 4)Solve xdy
dx − (1 + x)y = xy2 (Bernoulli’s equation) by substitution.
Solution:
y′ − (1 + 1/x)y = y2, u = y−1
47
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 4)Solve xdy
dx − (1 + x)y = xy2 (Bernoulli’s equation) by substitution.
Solution:
y′ − (1 + 1/x)y = y2, u = y−1
du
dx+
(1 +
1
x
)u = −1
48
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 4)Solve xdy
dx − (1 + x)y = xy2 (Bernoulli’s equation) by substitution.
Solution:
y′ − (1 + 1/x)y = y2, u = y−1
du
dx+
(1 +
1
x
)u = −1
Find integrating factor:
e∫
1+1/xdx = exeln x = xex,d
dx[xexu] = xex
49
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 4)Solve xdy
dx − (1 + x)y = xy2 (Bernoulli’s equation) by substitution.
Solution:
y′ − (1 + 1/x)y = y2, u = y−1
du
dx+
(1 +
1
x
)u = −1
Find integrating factor:
e∫
1+1/xdx = exeln x = xex,d
dx[xexu] = xex
xexu = −xex + ex + c, y−1 = −1 +1
x+
c
xe−x
50
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 22)Solve the initial value problemy1/2dy
dx + y3/2 = 1, y(0) = 4.
Solution:
51
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 22)Solve the initial value problemy1/2dy
dx + y3/2 = 1, y(0) = 4.
Solution:Substitution is u = y3/2
2
3
du
dx+ u = 1
52
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 22)Solve the initial value problemy1/2dy
dx + y3/2 = 1, y(0) = 4.
Solution:Substitution is u = y3/2
2
3
du
dx+ u = 1
2
3ln |1 − u| = x + c
53
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 22)Solve the initial value problemy1/2dy
dx + y3/2 = 1, y(0) = 4.
Solution:Substitution is u = y3/2
2
3
du
dx+ u = 1
2
3ln |1 − u| = x + c,
2
3ln |1 − y3/2| = x + c
x = 0, y = 4, c =2
3ln 7
2
3ln |1 − y3/2| = x +
2
3ln 7
54
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 24)Solve by substitution dy
dx = 1−x−yx+y .
Solution:
55
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 24)Solve by substitution dy
dx = 1−x−yx+y .
Solution:
Substitution is u = x + y
du
dx− 1 =
1 − u
u, udu = dx,
1
2u2 = x + c
56
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 24)Solve by substitution dy
dx = 1−x−yx+y .
Solution:
Substitution is u = x + y
du
dx− 1 =
1 − u
u, udu = dx,
1
2u2 = x + c
(x + y)2 = 2x + c
57
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 31)Explain why it is always possible to rewrite HDE M(x, y)dx+N(x, y)dy = 0as dy
dx = F (y/x).
Solution:
58
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 31)Explain why it is always possible to rewrite HDE M(x, y)dx+N(x, y)dy = 0as dy
dx = F (y/x).
Solution:
dy
dx= −M(x, y)
N(x, y)= −xmM(1, y/x)
xmN(1, y/x)= −M(1, y/x)
N(1, y/x)
59
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 32)Find F (y/x) for the equation (5x2 − 2y2)dx = xydy
Solution:
60
First-Order Differential Equations (2.5 Solutions by Substitutions )
Problem (2.5: 32)Find F (y/x) for the equation (5x2 − 2y2)dx = xydy
Solution:
dy
dx=
5x2 − 2y2
xy=
5 − 2(y/x)2
y/x, F (z) =
5 − 2z2
z
61