tutorial3.pdf

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


Problem (2.4: 4)


Solution:M(x, y) = sin y − y sin x, My = ∂M

∂y = cos y − sin x

3


Problem (2.4: 4)


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


Problem (2.4: 4)


Solution:




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


Problem (2.4: 4)


Solution:




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


Problem (2.4: 4)


Solution:




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


Problem (2.4: 8)

If the differential equation (1 + ln x + yx)dx = (1 − ln x)dy is exact, solve it.

Solution:

8


Problem (2.4: 8)


Solution:

(1 + ln x + yx)dx + (ln x − 1)dy = 0

9


Problem (2.4: 8)


Solution:

(1 + ln x + yx)dx + (ln x − 1)dy = 0

M(x, y) = 1 + ln x + yx, My = ∂M

∂y = 1x

10


Problem (2.4: 8)


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


Problem (2.4: 22)

Solve the initial value problem (ex + y)dx + (2 + x + yey)dy = 0, y(1) = 1.

Solution:

12


Problem (2.4: 22)


Solution:

M(x, y) = ex + y, My = 1, N(x, y) = 2 + x + yey, Nx = 1My = Nx (the equation is exact)

13


Problem (2.4: 22)


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


Problem (2.4: 22)


Solution:


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


Problem (2.4: 22)


Solution:



g′(y) = 2 + yey, g(y) = 2y + yey − ey

f (x, y) = c, ex + yx + 2y + yey − ey = c

16


Problem (2.4: 22)


Solution:



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


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



Solution:

M(x, y) = 6xy3 + cos y, My = 18xy2 − sin y

19



Solution:

M(x, y) = 6xy3 + cos y, My = 18xy2 − sin yN(x, y) = 2kx2y2 − x sin y, Nx = 4kxy2 − sin y

20



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


Problem (2.4: 32)Find the integrating factor and solve the equationy(x + y + 1)dx + (x + 2y)dy = 0.

Solution:

22



Solution:

M(x, y) = yx + y2 + y, My = x + 2y + 1, N(x, y) = x + 2y, Nx = 1

23



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



Solution:


N = x+2yx+2y = 1

Integrating factor µ(x) = e∫

1dx = ex

25



Solution:


N = x+2yx+2y = 1


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



Solution:


N = x+2yx+2y = 1


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


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



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



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



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



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



Solution:

Substitution is y = ux, dy = udx + xdu

34



Solution:

Substitution is y = ux, dy = udx + xduuxdx = 2(x + ux)(udx + xdu), (u + u2)dx + x(2 + u)du = 0

35



Solution:


1

xdx =

(1

u + 1− 2

u

)du

36



Solution:


1

xdx =

(1

u + 1− 2

u

)du

ln |x| + ln c = ln |u + 1| − 2 ln |u|

37



Solution:


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


Problem (2.5: 10)

Solve xdydx = y +

√x2 − y2, x > 0 by substitution.

Solution:

39


Problem (2.5: 10)

Solve xdydx = y +


Solution:

Substitution is y = ux

x(udx + xdu) = uxdx +√

x2 − u2x2dx

40


Problem (2.5: 10)

Solve xdydx = y +


Solution:


x(udx + xdu) = uxdx +√

x2 − u2x2dx

xdu =√

1 − u2dx, sin−1 y

x= ln x + c

41


Problem (2.5: 12)Solve initial value problem (x2 + 2y2)dx

dy = xy, y(−1) = 1.

Solution:

42



dy = xy, y(−1) = 1.

Solution:


(1 + u2)dx = uxdu

43



dy = xy, y(−1) = 1.

Solution:


(1 + u2)dx = uxdu

2 ln |x| = ln 1 + u2 + c

44



dy = xy, y(−1) = 1.

Solution:


(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


Problem (2.5: 4)Solve xdy

dx − (1 + x)y = xy2 (Bernoulli’s equation) by substitution.

Solution:

46




Solution:

y′ − (1 + 1/x)y = y2, u = y−1

47




Solution:

y′ − (1 + 1/x)y = y2, u = y−1

du

dx+

(1 +

1

x

)u = −1

48




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




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


Problem (2.5: 22)Solve the initial value problemy1/2dy

dx + y3/2 = 1, y(0) = 4.

Solution:

51



dx + y3/2 = 1, y(0) = 4.

Solution:Substitution is u = y3/2

2

3

du

dx+ u = 1

52



dx + y3/2 = 1, y(0) = 4.


2

3

du

dx+ u = 1

2

3ln |1 − u| = x + c

53



dx + y3/2 = 1, y(0) = 4.


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


Problem (2.5: 24)Solve by substitution dy

dx = 1−x−yx+y .

Solution:

55



dx = 1−x−yx+y .

Solution:

Substitution is u = x + y

du

dx− 1 =

1 − u

u, udu = dx,

1

2u2 = x + c

56



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


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


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


Problem (2.5: 32)Find F (y/x) for the equation (5x2 − 2y2)dx = xydy

Solution:

60


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

See you next week :-) !

62

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