UNIVERSITI TEKNIKAL MALAYSIA MELAKA KERJA KURSUS (ASSIGNMENT)

SESI 2010/2011 FAKULTI KEJURUTERAAN ELEKTRONIK DAN KEJURUTERAAN KOMPUTER

KOD MATAPELAJARAN : BENH 1253 MATAPELAJARAN : PERSAMAAN PEMBEZAAN PENYELARAS : HAMZAH ASYRANI BIN SULAIMAN KURSUS : BENE, BENT, BENC, BENW MASA : HANTAR SEBELUM 1 OGOS 2011 DOWNLOAD DIFFERENTIAL EQUATIONS MATERIALS AT

WWW.ASYRANI.COM/TEACHING

(BENH 1253 ASSIGNMENT SEMESTER III 2010/2011)

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QUESTION 1

Consider an ordinary differential equation below.

(a) Identify the homogeneity and the degree of equations:

(i) 𝑓(𝑥,𝑦) = 𝑥2 + 𝑥𝑦

[2 Marks]

(ii) 𝑓(𝑥,𝑦) = 𝑥2𝑦 + 𝑥𝑦2

[2 Marks]

(iii) 𝑓(𝑥,𝑦) = 609𝑥5

45𝑦2𝑥3+ 𝑥

𝑦

[2 Marks]

(b) Solve the ordinary differential equation by using separation of variable.

(i) 13𝑑𝑦𝑑𝑥

= 2𝑥

[4 Marks]

(ii) 13𝑦2 𝑑𝑦

𝑑𝑥= 2𝑥3

[5 Marks]

(c) Show that this equation

�2𝑥𝑦 −

3𝑦2

𝑥4 �𝑑𝑥 + �

2𝑦𝑥3 −

𝑥2

𝑦2 +1�𝑦

�𝑑𝑦 = 0

is an exact equation.

[5 Marks]

(BENH 1253 ASSIGNMENT SEMESTER III 2010/2011)

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QUESTION 2

Consider the second order differential equation.

𝑦′′ − 10𝑦′ + 25𝑦 = 5

(a) Find the homogeneous solution, 𝑦ℎ(𝑥)

[4 Marks]

(b) Determine the Wronskian, W value.

[6 Marks]

(c) Calculate the particular solution, 𝑦𝑝(𝑥) by using method of variation of parameter. Find

the complete general solution for the problem, 𝑦(𝑥)

[10 Marks]

(BENH 1253 ASSIGNMENT SEMESTER III 2010/2011)

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QUESTION 3

(a) Use the definition of Laplace transforms to determine the Laplace transform of the

function

𝑓(𝑡) = cos 𝑡.

[10 Marks]

(b) The current I(t) in an RLC series circuit is governed by the initial value problem

𝐼′′(𝑡) + 𝐼(𝑡) = 𝑔(𝑡) … … … … … … … . . (𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1)

𝐼(0) = 0, 𝐼′(0) = 0;

Where

𝑔(𝑡) = �10, 0 < 𝑡 < 𝜋

0, 𝜋 < 𝑡 < 2𝜋10, 2𝜋 < 𝑡

�… … … … … … … . . (𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2)

(i) Find the Laplace transform of equation 1

[5 Marks]

(ii) Determine the Laplace transform of g(t) in equation 2.

[5 Marks]

(BENH 1253 ASSIGNMENT SEMESTER III 2010/2011)

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QUESTION 4

(a) Sketch the graph of 𝑓(𝑥) = 𝑥, for −𝜋 < 𝑥 < 𝜋

[2 Marks]

(b) Determine the complete Fourier series of 𝑓(𝑥) = 𝑥, given 0 < 𝑥 < 𝜋

[18 Marks]

QUESTION 5

(a) A metal bar, insulated along its sides is 1m long. It is initially at room temperature

of 15𝑜𝐶 and at time 𝑡 = 0, the ends are placed into ice at 0𝑜𝐶. Find an

expression for the temperature at a point 𝑃 at a distance 𝑥 m from one end at any

time 𝑡 seconds after 𝑡 = 0.

[20 Marks]

(BENH 1253 ASSIGNMENT SEMESTER III 2010/2011)

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LAMPIRAN

A. Table of general solution corresponding to homogeneous part

If m values are Then general solution, hy

1. Real & Distinct xmxmh BeAey 21 +=

2. Real but Repeated xmxmh BxeAey 11 +=

3. Complex conjugates, (a ±b i) )sincos( bxBbxAey axh +=

B. Table of particular solution, py

Form of f(x) Roots )(xyp

011

1 ... αααα ++++ −− xxx n

nn

n m1≠0 and m2≠0 01

11 ... axaxaxa n

nn

n ++++ −−

011

1 ... αααα ++++ −− xxx n

nn

n m1=0 or m2=0

(either one) )...( 01

11 axaxaxax n

nn

n ++++ −−

xkeα m1≠α and m2≠α xCeα

xkeα m1=α or m2=α

(either one)

xCxeα

xkeα m1=m2=α xeCx α2

xk αcos or xk αsin m1≠iα and m2≠iα xqxp αα sincos +

xk αcos or xk αsin m1=iα or m2=iα

(either one)

)sincos( xqxpx αα +

(BENH 1253 ASSIGNMENT SEMESTER III 2010/2011)

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C. Table of Laplace Transform

f(t) is the function F(s) defined as follows :

𝐹(𝑠) = 𝐿{𝑓(𝑡)} = � 𝑒−𝑠𝑡∞

0

𝑓(𝑡)𝑑𝑡

Function Transform Function Transform

𝑓(𝑡) F(s) 1 1𝑠

𝑎𝑓(𝑡) + 𝑏𝑔(𝑡) aF(s)+bG(s) 𝑡 1𝑠2

𝑓′(𝑡) sF(s)-f(0) 𝑡𝑛 𝑛!𝑠𝑛+1

𝑓′′(𝑡) 𝑠2𝐹(𝑠) − 𝑠𝑓(0) − 𝑓′(0) 1√𝜋𝑡

1√𝑠

𝑓𝑛(𝑡) 𝑠𝑛𝐹(𝑠) − 𝑠𝑛−1𝑓(0) −⋯− 𝑓(𝑛−1)(0) 𝑒𝑎𝑡

1𝑠 − 𝑎

� 𝑓(𝜏)𝑑𝜏𝑡

0 𝐹(𝑠)

𝑠 𝑡𝑛𝑒𝑎𝑡

𝑛!(𝑠 − 𝑎)𝑛+1

𝑒𝑎𝑡𝑓(𝑡) 𝐹(𝑠 − 𝑎) cos𝑘𝑡 𝑠

𝑠2 + 𝑘2

𝑢(𝑡 − 𝑎)𝑓(𝑡 − 𝑎) 𝑒−𝑎𝑠𝐹(𝑠) sin 𝑘𝑡 𝑘

𝑠2 + 𝑘2

� 𝑓(𝜏)𝑔(𝑡 − 𝜏)𝑑𝜏𝑡

0 F(s)G(s) cosh 𝑘𝑡

𝑠𝑠2 − 𝑘2

𝑡𝑓(𝑡) −𝐹′(𝑠) sinh𝑘𝑡 𝑘

𝑠2 − 𝑘2

𝑡𝑛𝑓(𝑡) (−1)𝑛𝑑𝑛𝐹(𝑠)𝑑𝑠𝑛

𝑒𝑎𝑡 cos 𝑘𝑡 𝑠 − 𝑎

(𝑠 − 𝑎)2 + 𝑘2

𝑓(𝑡)𝑡

� 𝐹(𝜎)𝑑𝜎∞

𝑠 𝑒𝑎𝑡 sin𝑘𝑡

𝑘(𝑠 − 𝑎)2 + 𝑘2

𝑓(𝑡),𝑝𝑒𝑟𝑖𝑜𝑑 𝑝 1

1 − 𝑒−𝑝𝑠� 𝑒−𝑠𝑡𝑓(𝑡)𝑝

0𝑑𝑡 𝑢(𝑡 − 𝑎) 𝑒−𝑎𝑠

𝑠

𝑡2𝑘

(sin𝑘𝑡) 𝑠

(𝑠2 + 𝑘2)2 𝛿(𝑡 − 𝑎) 𝑒−𝑎𝑠

12𝑘3

(sin𝑘𝑡 − 𝑘𝑡 cos 𝑘𝑡 ) 1

(𝑠2 + 𝑘2)2 1

2𝑘(sin𝑘𝑡 + 𝑘𝑡 cos 𝑘𝑡 )

𝑠2

(𝑠2 + 𝑘2)2

(BENH 1253 ASSIGNMENT SEMESTER III 2010/2011)

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D. Fourier Series

The fourier series of a function f(t) defined on the interval (-L,L) is given by

𝑓(𝑡) =𝑎02

+ ��𝑎𝑛 cos𝑛𝜋𝑡𝐿

+ 𝑏𝑛 sin 𝑛𝜋𝑡𝐿�

∞

𝑛=1

𝑎0 =1𝐿�𝑓(𝑡)𝑑𝑡 𝐿

−𝐿

𝑎𝑛 =1𝐿�𝑓(𝑡) cos

𝑛𝜋𝑡𝐿𝑑𝑡

𝐿

−𝐿

𝑏𝑛 =1𝐿�𝑓(𝑡) sin

𝑛𝜋𝑡𝐿𝑑𝑡

𝐿

−𝐿

E. Identities of Trigonometric Functions

1. sin2 𝑥 + cos2 𝑥 = 1 2. 1 + tan2 𝑥 = sec2 𝑥 3. cot2 𝑥 + 1 = 𝑐𝑜𝑠𝑒𝑐2𝑥 4. sin 2𝑥 = 2 sin 𝑥 cos 𝑥 5. cos 2𝑥 = cos2 𝑥 − sin2 𝑥

= 2 cos2 𝑥 − 1 = 1 − 2 sin2 𝑥

6. tan 2𝑥 = 2 tan𝑥 1−tan2 𝑥

7. sin(𝑥 ± 𝑦) = sin 𝑥 cos 𝑦 ± cos 𝑥 sin 𝑦 8. cos(𝑥 ± 𝑦) = cos 𝑥 cos 𝑦 ∓ sin 𝑥 sin𝑦 9. tan(𝑥 ± 𝑦) = tan𝑥±tan𝑦

1∓tan𝑥 tan𝑦

10. 2 sin 𝑥 cos 𝑦 = sin(𝑥 + 𝑦) + sin(𝑥 − 𝑦) 11. 2 sin 𝑥 sin𝑦 = −cos(𝑥 + 𝑦) − cos(𝑥 − 𝑦) 12. 2 cos 𝑥 cos 𝑦 = cos(𝑥 + 𝑦) + cos(𝑥 − 𝑦) 13. sin(−𝑥) = − sin 𝑥 ; 14. cos(−𝑥) = cos 𝑥 ; 15. tan x = sin𝑥

cos𝑥;

(BENH 1253 ASSIGNMENT SEMESTER III 2010/2011)

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F. Table of Derivatives and Integrals

Differentiation Rules Indefinite Integrals

[ ] 0=kdxd

, k constant ∫ += Ckxkdx

[ ] 1−= nn nxxdxd

∫ ++

=+

Cnxdxx

nn

1

1

, 1−≠n

[ ]x

xdxd 1ln = ∫ += Cx

xdx ln

[ ] xxdxd sincos −= ∫ +−= Cxxdx cossin

[ ] xxdxd cossin = ∫ += Cxxdx sincos

[ ] xxdxd 2sectan = ∫ += Cxxdx tansec2

[ ] xxdxd 2coseccot −= ∫ +−= Cxxdx cotcosec2

[ ] xxxdxd tansecsec = ∫ += Cxxdxx sectansec

[ ] xxxdxd cotcoseccosec −= ∫ +−= Cxxdxx coseccotcosec

[ ] xx eedxd

= ∫ += Cedxe xx

[ ] xxdxd sinhcosh = ∫ += Cxxdx coshsinh

[ ] xxdxd coshsinh = ∫ += Cxxdx sinhcosh