le4

Acosta, Timothy John S.2007-16445

ES 201

Problem 1Find the inverse Laplace transform of the function:

F (s) =s2 + 3s+ 1

s6 + 4s5 + 4s4 + 4s3 + 3s2

=s2 + 3s+ 1

s2(s4 + 4s3 + 4s2 + 4s1 + 3)

=s2 + 3s+ 1

s2(s+ 1)(s2)(s+ 3)

=s(s+ 3)

s2(s2 + 1)(s+ 1)(s+ 3)+

1

s2(s2 + 1)(s+ 1)(s+ 3)

=1

s2(s2 + 1)(s+ 1)+

1

s2(s2 + 1)(s+ 1)(s+ 3)

Double Convolution of first term:

F (s) =1

sH(s) =

1

s2 + 1G(s) =

1

s+ 1(1)

f(t) = 1 H(t) = sint G(t) = et (2)

=

t0

(1)(sin(t )det+d2

=

t0

(1 cost)et+d2

= et+ t0coset+d

u = et+ dv = cosddu = et+ v = sin

= et+ sinet+ [et+ +et+ cosd ]

= et+ +1

2(sin + cos)et+

t0

= [1 +1

2(sint+ cost)] [et + 1

2et]

Double Convolution of second term:

F (s) =1

s2(s2 + 1)H(s) =

1

s+ 1G(s) =

1

s+ 3

f(t) = t sint H(t) = et G(t) = e3t

=

( sin)et+d1e3t+3d2et+ sinet+d

u = dv = et+ u = sin dv = et+ddu = d v = et+ du = cosd v = et+

et+ et+ sinet+ +coset+d

1


ES 201

u = cos dv = et+

du = sind v = et+

et+ et+ sinet+ + [coset+ + sinet+d ]et+ et+ sin cos

2et+

t0

= [t 1 (sint cost2

] [0 + 12

]

= t0 [t

3

2 sint cost

2]e3t+3d

= t0 e

3t+3 32e3t+3 1

2sine3t+3 +

1

2cose3t+3d

= A+B + C +D

A =

e3t+3d

u = dv = e3t+3

du = d v =e3t+3

3

A =1

3e3t+3 1

9e3t+3

B =

3

2e3t+3d

B =e3t+3

2

C =

1

2sine3t+3d

u = sin dv = e3t+3

du = cosd v =e3t+3

3

1

3sine3t+3 1

3

cose3t+3d

u = cos dv = e3t+3

du = sind v = e3t+3

3

1

3sine3t+3 1

3[1

3cose3t+3 +

1

3

sine3t+3d ]

simplifying: sine3t+3d =

9

8[e3t+3 (

sin

3 cos

9)]

C =1

2(9

8[e3t+3 (

sin

3 cos

9)])

C =9

16[e3t+3 (

sin

3 cos

9)]

2


ES 201

D =

cos

2e3t+3d

u = cos dv = e3t+3

du = sind v = e3t+3

3

cos

3e3t+3 +

1

3

sine3t+3d

u = sin dv = e3t+3

du = cosd v =e3t+3

3

simplifying: cose3t+3d =

9

10[e3t+3 (

sin

9+cos

3)]

D =1

2[

9

10(cos

3+sin

9)e3t+3 ]

Adding all parts:A + B + C + D

= e3t+3 e3t+3

9 e

3t+3

2 9e

3t+3

16(sin

3 cos

9) +

9e3t+3

20(sin

9+cos

3)

t0

=t

3 1

9 1

2 9

16(sin

3 cos

9) +

9

20(cos

3+sin

9) [0 e

3t

9 e

3t

2 9e

3t

16(19

)] +9

20(e3t)

1

3]

=t

3 11

8+

17

80cost 11

80sint+

287

720e3t

3


ES 201

Problem 2Find the solution to the differential equation given below. The initial conditions are x(0) = 0 ,x(0) = 0

d2x

dt2+ 4x = (t 2)e84t

d2x

dt2+ 4x = e8(t 2)e4t

s2X sx(0) x(0) + 4x = e8[ 1(s+ 4)2

2s+ 4

]

X(s2 + 4) = e8[1 2(s+ 4)

(s+ 4)2]

X = e8 2s+ 7(s+ 4)2(s2 + 4)

Using Partial Fractions:

2s+ 7

(s+ 4)2(s2 + 4)=

A

s+ 4+

B

s+ 4)2+Cs+D

s2 + 4

A+ C = 0

4A+B + 8C +D = 0

4A+ 16C + 8D = 2

16A+ 4B + 16D = 7

A =2

25, B =

120, C =

220, D =

37

100

Substituting...

X = e8[ 225

1

(s+ 4) 1

20

1

(s+ 4)2 2

25

s

(s2 + 4)+

37

100

1

(s2 + 4)]

x(t) = e8[ 225e4t 1

20e4t 2

25cos2t+

37

200sin2t]

4


ES 201

Problem 3Derive the temperature profile for the 1D Conduction along the length of the rod using Laplacetransformation.

u

t= 4

2u

x2

sU u(x, 0) = 42u

x2

sU 20 = 42u

x2

dl2U

dx2 s

4U = 5

We have a 2nd order non-homogenous differential equation. We will have a complementary andparticular soln:

U(x, s) = Uc(x, s) + Up(x, s)

Solving for the particular solution using Method of Undetermined Coefficients:

Up(x, s) = A

U p (x, s) = 0

0 s4A = 5

A =20

s

Now we solve for the complementary solution:

dl2U

dx2 s

4U = 0

This gives us the characteristic equation:

m2 s4

= 0

m = s

4

Uc(x, s) = C1e

s

4x

+ C2es

4x

U(x, s) = C1e

s

4x

+ C2es

4x

+20

s

Boundary Conditions:

u(0, t) = 20

xu(0, t) = 0

u(1, t) = 100

xU(0, s) = 0

U(1, s) =100

s

After transforming the boundary conditions, we use U(0, s) = 0

xU(0, s) = C1

s

4e

s4(0) + C2

s

4e

s4(0) + 0

0 = C1

s

4 C2

s

4

C1 = C2

5


ES 201

Next we use U(1, s) =100

sto solve for the constant

U(x, s) = C1(e

s

4x

+ es

4x

) +20

s

U(1, s) = C1(e

s

4(1)

+ es

4(1)

) +20

s

100

s= C1(e

s

4(1)

+ es

4(1)

) +20

s

C1 =80

s

1

e

s4 + e

s4

U(x, s) =80

s[e

s4x + e

s4x

e

s4 + e

s4

] +20

s

note that we can write this in power series form:

1

1 + u=n=0

(1)nun for|u| < 1

[e

s4(1x) + e

s4(1+x)]

n=0

(1)ne2n

s4

n=0

(1)n[e

s4(2n+1)(1x) + e

s4(2n+1)(1+x)]

Substituting...

U(x, s) =20

s+

80

s

n=0

(1)n[e

s4(2n+1)(1x) + e

s4(2n+1)(1+x)]

note to perform the inverse Laplace transformation we need the error function given below:

erf(t) =2pi

t0eu

2du

erfc(t) =2pi

t

eu2du

erfc(t) = 1 erf(t)L [erfc(

k

2t)](s) =

1

seks

Therefore the solution is:

u(x, t) = 20 + 80n=0

(1)n(erfc

[(2n+ 1) x

2

4t

]+ erfc

[(2n+ 1) + x

2

4t

])

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