Exam Proposal GE604

University of Tripoli Faculty of Engineering

Postgraduate department Mustafa Abed Ellsamad - #2215802

1

Advanced Mathematics [ GE 604 ] Midterm exam spring 2015

Q1} Find the solution of the following equation with :

i. Green function.

ii. Variation of parameters.

iii. Laplace transforms.

x3y,xxx + 3x2y,xx 8xy,x = 10x

2

with initial conditions y(1) = 1, y,x(1) = 2 , y,xx(1) = 16

Q2} find the solution of the following equation with :

i. Green function.



x2y,xx 2xy,x + 2y = x2lnx

with initial conditions y(1) = 1, y,x(1) = 1

Q3} Determine Governing Equation by using Calculus of variation.

I = F(x, y, y, y)dx

x2

x1

Q4} Find the solution of the following ODE using power series method :

y + xy = 2 y(0) = 1 , y(0) = 1



2

Q5} Find the solution for x(t) + x(t) = F(t)

With conditions x(0) = x(0) = 0 , by using :

i. Homogeneous and particular solution.

ii. Using superposition.

iii. Using Laplace transform.

1

t

Q6} find the solution of the following :

1

[x2 (1 x )]3

dx 1

0

b cos6 x dx

2

0

c x2dx

2 x

2

0

Q7} Find Fourier series for :

F(x) = {0 5 03 0 5

period =10



3

Advanced Mathematics [ GE 604 ] Midterm exam spring 2015


i. Green function.



x3y,xxx + 3x2y,xx 8xy,x = 10x

2

with initial conditions y(1) = 1, y,x(1) = 2 , y,xx(1) = 16

a0(x)y,xxx a1(x)y,xx + a2(x)y,x + a3(x)y = a4(x)

Check for self ad joint a0(x)=x3 (a0(x)),x=3 x

2 (a0(x)),x=a1(x)

the DE is self ad joint

R(R2 3R + 2) + 3R(R 1) 8R = 0

R3 3R2 + 2R + 3R2 3R 8R = 0

R3 9R = 0

R(R 3)(R + 3) = 0

yh = c1 + c2x3 +

c3x3

i. Green function

G(x, z) = c1(z) + c2(z)x3 +

c3(z)

x3

G(z, z) = c1(z) + c2(z)z3 +

c3(z)

z3= 0

G,x(x, z) = 3c2(z)x2 3

c3(z)

x4

G,x(z, z) = 3c2(z)z2 3

c3(z)

z4= 0

c3(z) = c2(z)z6



4

G,xx(x, z) = 6c2(z)x + 12c3(z)

x5

G,xx(z, z) = 6c2(z)z + 12c3(z)

z5=

1

p(z)=

1

z3

6c2(z)z + 12c2(z)z =1

z3

c2(z) =

1

18z4 c3(z) =

z2

18

c1(z) = c2(z)z3

c3(z)

z3

c1(z) = 1

9z

G(x, z) = 1

9z+

1

18z4x3

+

z2

18x3

yp = G(x, z)

x

0

f(z)dz

yp = (

x

0

1

9z+

1

18z4x3

+

z2

18x3) 10z2 dz

yp = (

x

0

10

9 z +

10

18 z2x3

+

10z4

18x3)dz

yp = 5

9 x2

5

9 x2 +

1

9 x2 = x2

y = yh + yp

y = c1 + c2x3 +

c3x3

x2

1 = c1+c2+c3 1

y,x = 3c2 x2 3

c3

x4 2x



5

2 = 4c2c3

y,xx = 12c2 x2 +

2c3

x3

6

5x3+

12x2

10

12 = 12c2 + 2c3

c3 = 1 c2 = 1 c1 = 2

y = 2 + x3 +1

x3 x2

-----------------------------------------------------------

ii. variation of parameters

From homogeneous solution

yh = c1 + c2x3 +

c3x3

y1 = 1 y2 = 3 y3 =

1

3

w(y1, y2, y3) =

[ 1 x3

1

3

0 3x2 3

x4

0 6x12

x5 ]

=54

3

yp = u1(x) + u2(x)x3 +

u3(x)

3

u1 + u2x

3 + u3

3= 0

3u2x2

3u3x4

= 0

6u2x +12u3

x5= 102



6

[ 1 x3

1

3

0 3x2 3

x4

0 6x12

x5 ]

[

u1u2u3

] = [00

102]

u1 =[ 0 x

3 13

0 3x2 3x4

10 6x

12x5 ]

w(y1, y2, y3)=

602

543

= 10

9

u2 =[ 1 0

13

0 0 3x4

010

12x5 ]

w(y1, y2, y3)=

30 x5

543=

5

9 x2

u3 =

[

1 x3 00 3x2 0

0 6x10

]

w(y1, y2, y3)=

30x

54 3=

5

9 4

u1(x) = u1(x)dx =

10

9 dx =

5

92

u2(x) = u2(x)dx =

5

9 x2dx =

1

9

u3(x) = u3(x)dx =

5

9 4 dx =

1

9 5



7

yp = 5

92

1

9 3 +

19

5

3

yp = 5

9x2

5

9 x2 +

1

9x2 = x2

y = yh + yp y = c1 + c2x3 +

c3x3

x2

1 = c1 + c2 + c3 1

0 = c1 + c2 + c3

y,x = 3c2x2 3

c3x4

2x

2 = 3c2 3c3 2

c2 = c3

y,xx = 6c2x +12c3x5

2

16 = 6c2 + 12c3 2 18 = 18 2 2 = 1

c2 = 1 c3 = 1 c1 = 2

y = 2 + x3 +1

x3 x2

-----------------------------------------------------------

iii. Laplace transform

L[y(t)] = y(s) = y(t)est

0

dt

L[y,x(x)] = sy(s) y(0)

L[y,xx(x)] = s2y(s) sy(0) y,x(0)



8

L[y,xxx(x)] = s3y(s) s2y(0) sy,x(0) y,xx(0)

Let t=lnx t ,x =1

x et = x

y,x = y,t. t ,x xy,x = y,t = R

By diff xy,xx + y,x = y,tt. t ,x x2y,xx + x y,x = y,tt

x2y,xx = y,tt y,t

x2y,xxx + 2xy,xx = y,ttt. t ,x y,t. t ,x

x3y,xxx + 2x2y,xx = y,ttt. y,tt

x3y,xxx = y,ttt. 3y,tt + 2y,t

Form the main equation we found

y,ttt. 3y,tt + 2y,t + 3y,tt 3y,t 8y,t = 10e2t

y,ttt 9y,t = 10e2t

s3y(s) s2y(0) sy,x(0) y,xx(0) 9sy(s) + 9y(0) =10

s 2

s3y(s) s2A sB C 9sy(s) + 9A =10

s 2

(s3 9s)y(s) A(s2 9)Bs C =10

s 2

(s3 9s)y(s) =10

s 2+ A(s2 9) + Bs + C

(s3 9s)y(s) =10 + A(s3 2s2 9s + 18) + B(s2 2s) + C(s 2)

s 2

y(s) =10 + A(s3 2s2 9s + 18) + B(s2 2s) + C(s 2)

s(s 3)(s 2)(s + 3)

p(s) = 10 + A(s3 2s2 9s + 18) + B(s2 2s) + C(s 2)



9

s = 0 s = 3 s = 3 s = 2 p(0) = 10 + 18A 2C p(3) = 10 + 3B + C p(3) = 10 + 15B 5C p(2) = 10 q(s) = s4 2s3 9s2 + 18s q,x(s) = 4 s

3 6s2 18s + 18

q,x(0) = 18 ; q,x(3) = 18 ; q,x(3) = 90 ; q,x(2) = 10

y(t) =p(0)

q,x(0)e0 +

p(3)

q,x(3)e 3t +

p(3)

q,x(3)e3t +

p(2)

q,x(2)e2 t

y(t) =10 + 18A 2C

18 +

10 + 3B + C

18e3 t +

10 + 15B 5C

90e3t +

10

10e2 t

y(x) =10 + 18A 2C

18 +

10 + 3B + C

18x3 +

10 + 15B 5C

90x3 x2

1 =10 + 18A

18 +

10 + 3B + C

18 +

10 + 15B 5C

90 1

A = 1

y,x(x) =10 + 3B + C

6x2 +

10 + 15B 5C

30x4 2x

2 =10 + 3B + C

6 +

10 + 15B 5C

30 2

B = 2

y,xx(x) = 10 + 3B + C

3x +

40 + 45B 20C

30x5 2

16 =10 + 3B + C

3 +

40 + 45B 20C

30 2

C = 14

y = 2 + x3 +1

x3 x2



10


iv. Green function.

v. Variation of parameters.

vi. Laplace transforms.

x2y,xx 2xy,x + 2y = x2lnx

with initial conditions y(1) = 1, y,x(1) = 1

a0(x)y,xx a1(x)y,x + a3(x)y = a4(x)

Check for self-ad joint a0(x)=x2 (a0(x)),x=2 x a1(x) = -2x (a0(x)),xa1(x)

p(x) = e

a1(x)

a0(x)

dx= e

2 xdx = 2

1

2y,xx

2

3y,x +

2

4y =

lnx

x2

the DE now is self ad joint

let x= t= ln x

R(R 1) 2R + 2 = 0

R2 R 2R + 2 = 0

R2 3R + 2 = 0

(R 1)(R 2) = 0

= c1x + c22

i. Green function

G(x, z) = c1(z)x + c2()2

G(z, z) = c1(z)z + c2()2 = 0

G,x(x, z) = c1 + 2c2

G,z(z, z) = c1 + 2c2 = 1



11

1 = 1 22z

2 = 1

1 = 1 , 2 =1

G(x, z) = x +x2

z

yp = G(x, z)

x

0

f(z)dz

yp = (

x

0

x +x2

z) ln z dz

yp = (

x

0

x ln + x2

zln )dz

yp = (

x

0

x ln )dz + (

x

0

x2

zln )dz

yp = x[z ln z]0

+ 2[ 1

2(ln )2]

0

yp = 2 ln + 2 + 2

2(ln )2

y = y + y

y = c1x + c222 ln + 2 +

2

2(ln )2

1 = c1+c2 + 1 c1 = c2

y,x = c1 + 2 c2x + x x ln + (ln )2

y,x(1) = 1 = c1 + 2 c2 + 1 0 + 0



12

c2 = 2 c1 = 2

y = 2x 222 ln + 2 + 2

2(ln )2

ii. variation of parameters

From homogeneous solution

= c1x + c22 y1 = x y2 =

2

w(y1, y2, y3) = [ 2

1 2] = x2

yp = u1(x)x + u2(x)x2

u1x + u2x

2 = 0

u1 + 2u2x = ln

[[ 2

1 2] ] [

u1u2

] = [0

ln ]

u1 =[ 0

2

ln 2]

w(y1, y2)=

2 ln

2 = ln

u2 =[ 01 ln

]

w(y1, y2)=

x ln

2 =

ln

u1(x) = u1(x)dx = ln dx = [x ln x ]

u2(x) = u2(x)dx =

ln

dx =

1

2 (ln )2



13

yp = u1(x)x + u2(x)x2

yp = 2 ln + 2 + 2

2(ln )2

y = + y

y = c1x + c222 ln + 2 +

2

2(ln )2

1 = c1+c2 + 1 c1 = c2

y,x = c1 + 2 c2x + x x ln + (ln )2

y,x(1) = 1 = c1 + 2 c2 + 1 0 + 0

c2 = 2 c1 = 2

y = 2x 222 ln + 2 + 2

2(ln )2

iii. Laplace transform

L[y(t)] = y(s) = y(t)est

0

dt

L[y,x(x)] = sy(s) y(0)

L[y,xx(x)] = s2y(s) sy(0) y,x(0)

Let t= ln x t ,x =1

x et = x

y,x = y,t t,x xy,x = y,t

xy,xx + y,x = y,tt t,x x2y,xx + x y,x = y,tt x

2y,xx = y,tt y,t

by substitution in the main equation :



14

y,tt y,t 2y,t 2y = te2t

y,ttt 3y,t 2y = te2t

s2y(s) sy(0) sy,x(0) 3sy(s) + 3y(0) + 2y(s) =1

(s 2)2

s2y(s) sA sB 3sy(s) + 3A + 2y(s) =1

(s 2)2

()(s2 3s + 2) A(s 3) sB =1

(s 2)2

()(s2 3s + 2) =1

(s 2)2+ A(s 3) + sB

() =1

(s 2)2(s2 3s + 2)+

A(s 3)

(s2 3s + 2)+

sB

(s2 3s + 2)

() =1

(s 2)2(s 2)( 1)+

A(s 3)

(s 2)( 1)+

sB

(s 2)( 1)

() =1 + A(s 3)(s 2)2 + sB(s 2)2

(s 2)3( 1)

p(s) = 1 + A(s 3)(s 2)2 + sB(s 2)2 s = 2 s = 1 p(2) = 1 p(1) = 1 2A + B q(s) = s4 7s3 18s2 + 20s + 8 q,x(s) = 4 s

3 21s2 36s + 20

q,x(2) = 104 ; q,x(1) = 33

y(t) =p(1)

q,x(1)et +

p(2)

q,x(2)e 2t

y(t) =1 2A + B

33e t +

1

104e2 t



15

y(t) =1 2A + B

33x +

1

1042

1 =1 2A + B

33

1

104

1 2A + B = 33(1 +1

104 )

= 33(105

104) 1 + 2

y,x(x) =1 2A + B

33+

2

104x

1 =1 2A + B

33

2

104

1 2A + B = 33(1 +1

104 )

=1

2(1 + + 33 (

103

104))

y = 2x 222 ln + 2 + 2

2(ln )2



16

Q3} Determine Governing Equation by using Calculus of variation.

I = F(x, y, y, y)dx

x2

x1

I=0= F,yy+F,yy+F,yy)dxx2x1

F, y ydx =F, yy|x1

x2 (F, y ),x ydx

F, y ydx = F, y y |x1

x2 (F, y

x2

x1

),x ydx

F, y ),x ydx = (F, y ),x y (F, y ),xx ydx

x2

x1

I = (F, y (F, y ),x

x2

x1

+ (F, y ),xx )y + F, y y| x1x2

+(F, y ) y |x1

x2 (F, y),x y|

x1

x2

Boundary conditions

(F, y ),xx )y + F, y y| x1x2+ (F, y ) y |

x1

x2 (F, y),x y|

x1

x2

Governing equation

F, y (F, y),x+ (F, y),xx = 0



17

Q4} Find the solution of the following ODE using power series method :

y + xy = 2 y(0) = 1 , y(0) = 1

n

n=2

(n 1)anxn2 + x an

n=0

xn = 2

n

n=2

(n 1)anxn2 + an

n=0

xn+1 = 2

(n + 2)

n=0

(n + 1)an+2xn + an1

n=1

xn = 2

(2)(1)a2 + [(n + 2)

n=1

(n + 1)an+2 + an1]xn + 2a2 = 2

a0 = y(0) = 1 , a1 = y(0) = 1

2a2 = 2 , a2 = 1

an+2 =an1

(n + 2)(n + 1) , n 1

a3 = a06

= 1

6 , a4 =

a14.3

= 1

12 , a5 =

a25.4

= 1

20 ,

a6 = a36.5

=1

180

General Solution

y1(x) = 1 + x + x2

x3

6

x4

12

x5

20+

x6

180+



18

Q5} Find the solution for x(t) + x(t) = F(t)

With conditions x(0) = x(0) = 0 by using :

i. Homogeneous and particular solution.

ii. Using superposition.

iii. Using Laplace transform.

i. Homogeneous and particular solution

F(t) = {+1 , 0 t < 1 , t

Homogeneous solution x(t) + x(t) = 0

m2 + 1 = 0 m = i

xh = Acos t + B sin t

particular solution

xp(t) = c0 x(t) = 0

c0 = 1 c0 = 1

xp(t) = 1

x(t) = xh + xp x(t) = A cos t + B sin t + 1

x(0) = A cos 0 + B sin 0 + 1 = 0

A = 1

x(t) = Asin t + B cos t

x(0) = Asin 0 + B cos t

B = 0

x(t) = 1 cos t for 0 t < . . (1)



19

Now repeat last steps for initial condition :

x(t) = 1 cos t

x() = 1 cos

x() = 2

x(t) = sin t

x() = 0

Homogeneous solution x(t) + x(t) = 0

m2 + 1 = 0 m = i

xh = Acos t + B sin t

particular solution

xp(t) = c0 x(t) = 0

c0 = 1

xp(t) = 1

x(t) = xh + xp x(t) = A cos t + B sin t 1

x() = A cos + B sin 1 = 2 A = 3

x(t) = Asin t + B cos t

x() = Asin + Bcos = 0 B = 0

x(t) = 3cos t 1 for t > . . (2)



20

ii. Using superposition :

f(t) = [u(t) u(t )] u(t )

f(t) = u(t) 2u(t )

1

= tt

1

t

+

-1

f(0) = 1

f(1) = 1 2 = 1

f(2) = 1 2 0 = 1

f(3) = 1 2 = 1

From equation (1) we get :

x(t) = 1 cos t for 0 t . . (1)



21

f(t) = u(t) 2u(t )

x(t) = 1 cos t 2(1 cos(t ))

x(t) = 3cos t 1 for t >

iii. Using Laplace transform:

F(t) = {+1 , 0 t < 1 , t

F(s) = est

0

dt est

dt

xeaxdx =eax

a(x

1

a)

F(s) =est

s|0

est

s|

F(s) = es

s+

1

s

es

s=

1

s

2es

s

F(t) = U(t) 2U(t )

s2x(s) sx(0) x(0) + x(s) =1

s

2es

s

x(s)(s2 + 1) =1

s

2es

s

x(s) =1

s(s2 + 1)

2es

s(s2 + 1)

1

s(s2 + 1)=

A

s+

Bs + C

(s2 + 1)=

As2 + A + Bs2 + Cs

s(s2 + 1)

A + B = 0 , A = 1 , B = 1

1

s(s2 + 1)=

1

s+

s

(s2 + 1)



22

x(s) =1

s

s

(s2 + 1)

2es

s+

2ses

(s2 + 1)

x(t) = 1 cos t 2(1 + cos t)U(t )

Q6} find the solution of the following:

a- 1

[x2 (1 x )]3

dx 1

0

I = x23 (1 x )

13 dx

1

0

= B(1

3,2

3) =

(13) (

23)

(1)

= (1

3) (

2

3) = (

1

3) (1

1

3) =

sin (3)

=

3/2=

2

3

b. cos6 x dx

20

2n 1 = 0 n =1

2 , 2m 1 = 6 m =

7

2

2I = B (1

2,7

2) I =

1

2B (

1

2,7

2) =

(12) (

72)

2(4)=

(12)

52

32

12 (

12)

2 3!

=5

32

c. x2dx

2x

2

0

let x = 2t , dx = 2dt



23

at x = 0 t = 0 , at x = 2 t = 1

I = x2dx

2 x

2

0

= 4t2 2dt

2 2t

1

0

=8

2

t2dt

1 t

1

0

= 42 t2(1 t)12 dt

1

0

= 42 B (3,1

2) = 42

(3) (12)

(72)

42 2! (

12)

52

32

12 (

12)

=642

15

Q7} Find Fourier series for :

F(x) = {0 5 03 0 5

period =10

= 1

5 () cos

5

5

5

1

5 { (0) cos

5 + (3)

5

5

0

0

5} =

3

5 cos

5

5

0

For n 0 = 3

5 (

5

) sin(

5)]

50

If n =0 , = 0 = 3

5 cos

0

5

5

0 =

3

5 = 3

5

0

= 1

5 { (0) sin

5 + (3)

5

5

0

0

5

} = 3

5 sin

5

5

0



24

= 3

5 (

5

cos

5)]

50 =

3(1cos )

The corresponding Fourier series is

0

2+ cos

+ sin

=

3

2=1 +

3(1cos)

=1 sin

5

= 3

2+

6

[sin

5+

1

3sin

3

5+

1

5sin

5

5+ ]

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Exam Proposal GE604

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