B.Sc. COMPUTER SCIENCE, BCA Allied Mathematics ALLIED COURSE I (AC) - ALGEBRA AND CALCULUS

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1 UNIT I B.Sc. COMPUTER SCIENCE, BCA Allied Mthemtics ALLIED COURSE I (AC) - ALGEBRA AND CALCULUS Theory of Equtios: Reltio betwee roots & coefficiets Trsformtios of Equtios Dimiishig, Icresig & multiplyig the roots by costt- Formig equtios with the give roots Rolle s Theorem, Descrte s rule of Sigs(sttemet oly) simple problems. UNIT II Mtrices : Sigulr mtrices Iverse of o-sigulr mtri usig djoit method - Rk of Mtri Cosistecy - Chrcteristic equtio, Eige vlues, Eige vectors Cyley Hmilto s Theorem (proof ot eeded) Simple pplictios oly UNIT III Differetitio: Mim & Miim Cocvity, Coveity Poits of ifleio - Prtil differetitio Euler s Theorem - Totl differetil coefficiets (proof ot eeded ) Simple problems oly. UNIT IV Itegrtio : Evlutio of itegrls of types,, d, b si d b cos Evlutio usig Itegrtio by prts Properties of defiite itegrls Fourier Series i the rge (, π ) Odd & Eve Fuctios Fourier Hlf rge Sie & Cosie Series UNIT V Differetil Equtios: Vribles Seprbles Lier equtios Secod order of types ( D + b D + c ) y F ( ) where,b,c re costts d F ( ) is oe of the followig types (i) e K (ii) si (k) or cos (k) (iii), beig iteger (iv) e K F ().

2 Let us cosider ALGEBRA AND CALCULUS Uit I Theory of Equtios This polyomil i of degree provided. The equtio is obtied by puttig f() is clled lgebric equtio of degree. RELATIONS BETWEEN THE ROOTS AND COEFFICIENTS OF EQUATIONS Let the give equtio be Let,,, be its roots. sum of the roots tke oe t time sum of the product of the roots tke two t time sum of the product of the roots tke three t time filly we get. Problem: If α d β re the roots of, fid α + β, αβ. Solutio: Here α + β αβ. Problem: Solve the equtio, oe root beig + i. Solutio: Give equtio is cubic. Hece we hve roots. Oe root is (+i) α (sy) comple roots occur i pirs. β i is other root. To fid third root (sy)

3 Sum of the roots tke oe t time α + β +. i.e., + i + i + The roots of the give equtio re + i, i, -. Problem: Solve the equtio hvig give tht + is root. Solutio: Give equtio is cubic. Hece we hve three roots. Oe root is + i α Sice comple roots occur i pirs, i β is other root. Sum of the roots is α + β + i.e., + i + i Hece the roots of the give equtio re + i, i,. Problem: Solve the equtio which hs root +. Solutio: Give () This equtio is biqudrtic, i.e., fourth degree equtio. It hs roots. Give + is root which is clerly irrtiol. Sice irrtiol roots occur i pirs, is lso root of the give equtio.

4 [ ( + )] [ ( )] is fctor of () ALGEBRA AND CALCULUS i.e., is fctor. Dividig () by, we get (-) (-) Hece the quotiet is. Solvig this qudrtic equtio, we get. Hece the roots of the give equtio re +,,,. Problem: Form the equtio, with rtiol coefficiets oe root of whose roots is. Solutio: Oe root is i.e., i.e., Squrig o both sides we get Agi squrig, we get

5 , which is the required equtio. Problem: Form the equtio with rtiol coefficiets hvig + d + s two of its roots. Solutio: Give + d + i.e., [ ( + )] [ ( + )] re the fctors of the required equtio. Sice comple d irrtiol roots occur i pirs, we hve, re lso the roots of the required equtio. i.e., ( ) d ( ) re lso fctors of the required equtio. Hece the required equtio is, [ ( + )] [ ( + )] [ ( )] [ ( )] i.e., ( ( Simplifyig we get Problem: which is the required equtio. Solve the equtio whose roots re i A.P. Solutio: Let the roots be α d, α, α + d. Sum of the roots tke oe t time is, α d + α + α + d α

6 α is root of the give equtio. By divisio we hve, The reduced equtio is Solvig this qudrtic equtio we get the remiig two roots,. Hece the roots of the give equtio re,,. Problem: Fid the vlue of k for which the roots of the equtio re i A.P. Solutio: Give () Let the roots be α d, α, α + d. Sum of the roots tke oe t time is, α d + α + α + d α i.e., α - i.e., α - is root of (). put - i (), we get k. Problem: Solve the equtio whose roots re i G.P. 6

7 Solutio: ALGEBRA AND CALCULUS Give () Let the roots be, α, αr Product of the roots tke three t time is. α. αr i.e., i.e., α. i.e., α is root of the give equtio () i.e., is root of the give equtio () i.e., ( - is fctor of (). (-) (-) Hece the quotiet is i.e., Solvig this qudrtic equtio we get Hece the roots of the give equtio re. 7

8 Problem: Fid the coditio tht the roots of the equtio Solutio: my be i G.P. Give () Let the roots be, α, αr Product of the roots tke three t time. α. αr r i.e., r () But α is root of the equtio (). Put α i (), we get, () Substitutig () i () we get i.e., i.e., r Hece the required coditio is p r q. Trsformtio of Equtios: Problem: If the roots of re -,, 9, fid the equtio whose roots re, -, -9.

9 Solutio: Give () The roots re -,, 9. Now we fid equtio whose roots re, -, -9 ie., to fid equtio whose roots re the roots of () but the sigs re chged. Hece i () we hve to chge the sig of odd powers of. Hece the required equtio is i.e., This gives the required equtio. Problem: Multiply the roots of the equtio by. Solutio: Give () To multiply the roots of () by, we hve to multiply the successive coefficiets begiig with the secod by i.e., i.e., which is the required equtio. Problem: Remove the frctiol coefficiets from the equtio. 9

10 Solutio: ALGEBRA AND CALCULUS Give () Multiply by the roots of () by m, we get m m m () If m (L.C.M. of d ), the frctios will be removed. Put m i (), we get i.e.,. Problem: Solve the equtio 6 give tht its roots re i H.P. Solutio: Give () Its roots re i H.P. to i (), we get 6 6 Now the roots of () re i A.P. (Sice H.P. is reciprocl of A.P.). Let the roots of () be d,, d. Sum of the roots d d Product of the roots tke t the time is ( d) ( d) d.

11 Cse(i) : ALGEBRA AND CALCULUS Whe d i. e.,,,. The roots of the give equtio re the reciprocl of the roots of i. e.,,,. d, re roots of the roots of re,, Cse (ii) : Whe d i. e.,,,. The roots of the give equtio re the reciprocl of the roots of i. e.,,, d., re roots of the roots of re,, Problem: Dimiish the roots of 7 by d fid the trsformed equtio. Solutio : Dimiishig the roots by, we get (costt term of the - - trsformed equtio) (coefficiet of ) (coefficiet of ) (coefficiet of ) (coefficiet of i the trsformed equtio) The trsformed equtio whose roots re less by of the give equtio is

12 Problem: Icrese by 7 the roots of the equtio 7 d fid the trsformed equtio. Solutio : Icresig by 7 the roots of the give equtio is the sme s dimiishig the roots by (costt term of the trsformed equtio) (coefficiet of ) (coefficiet of ) (coefficiet of ) (coefficiet of i the trsformed equtio) The trsformed equtio is Problem: Fid the equtio whose roots re the roots of icresed by. Solutio : (costt term of the - - trsformed equtio) (coefficiet of ) (coefficiet of ) (coefficiet of ) (coefficiet of i the trsformed equtio) The trsformed equtio is 9.

13 Problem: If,, re the roots of the equtio 6, fid equtio whose roots re,,. Solutio : The trsformed equtio is. i.e., the roots re,,. i.e.,,, i.e.,,,. Problem: Fid the trsformed equtio with sig chged Solutio: Give tht Give sig Now the trsformed equtio which is the required equtio.

14 Nture of the Roots: Problem: Determie completely the ture of the roots of the equtio. Solutio: Give tht There re times sig chged. There eist positive roots. Put - There is time sig chged. There is oly oe positive root. There re rel roots. The degree of the equtio is. Number of imgiry roots degree of equtio umber of rel roots The umber of imgiry roots.

15 UNIT - MATRICES A mtri is defied to be rectgulr rry of umbers rrged ito rows d colums. It is writte s follows:- m m ` m m Specil Types of Mtrices: (i) A row mtri is mtri with oly oe row. E.g., [ ]. (ii) A colum mtri is mtri with oly oe colum. E.g.,. (iii) Squre mtri is oe i which the umber of rows is equl to the umber of colums. If A is the squre mtri. m m ` m m the the determit is clled the determit of the mtri A d it is deoted by A or deta. (iv) Sclr mtri is digol mtri i which ll the elemets log the mi digol re equl.

16 6 E.g., (v) Uit mtri is sclr mtri i which ll the elemets log the mi digol re uity. I, I (vi) Null or Zero mtri. If ll the elemets i mtri re zeros, it is clled ull or zero mtri d is deoted by. (vii) Trspose mtri. If the rows d colums re iterchged i mtri A, we obti secod mtri tht is clled the trspose of the origil mtri d is deoted by A t. (viii) Additio of mtrices. Mtrices re dded, by ddig together correspodig elemets of the mtrices. Hece oly mtrices of the sme order my be dded together. The result of dditio of two mtrices is mtri of the sme order whose elemets re the sum of the sme elemets of the correspodig positios i the origil mtrices. E.g., b b b b b b b b b b b b Problem: Give ; 6 B A ; compute A-B

17 Solutio : ALGEBRA AND CALCULUS A B Problem: Fid vlues of, y, z d tht stisfy the mtri reltioship y z Solutio : From the equlity of these two mtices we y z Solvig these equtios we get, y, z, 7 Multiplictio of Mtrices. get the equtios If A is m mtri with rows A, A,, A m d B is p mtri with colums B, B,.., B p, the the prodduct AB is m p mtri C whose elemets re give by the formul C ij A i. B j. Hece C AB 7

18 Iverse of Mtri Problem: Fid the iverse of the mtri Solutio: ALGEBRA AND CALCULUS det ( ) ( + ) (-6) (-) Form the mtri of mior determits:. 6 6 Adjust the sigs of every other elemet (strtig with the secod etry): 6 6 Tke the trspose d divide by the determit: 6 So the iverse mtri is Problem: Show tht A stisfies the equtio A A I. Hece determie its iverse. Solutio: A 9 9 9

19 9 A I A A I Therefore A A I. Multiplyig by A -, we hve A - A A - A A - I i.e., A I A - Therefore A - A I Therefore A -. Rk of Mtri A sub-mtri of give mtri A is defied to be either A itself or rry remiig fter certi rows d colums re deleted from A. The determits of the squre sub-mtrices re clled the miors of A. The rk of m mtri A is r iff every mior i A of order r + vishes while there is t lest oe mior of order r which does ot vish.

20 Problem: Fid the rk of the mtri 6. Solutio: Mior of third order 6. The miors of order re obtied by deletig y oe row d y oe colum. Oe of the miors of orders is 6 Its vlue is. Hece the rk of the give mtri is. Rk of Mtri by Elemetry Trsformtios: Problem: Fid the rk of the mtri A 7. Solutio: The give mtri is A 6 ~ 6 R R R R R R ~ ) ( 6 R R ~ 6 R R R ~ C C C C C C

21 ~ ~ Hece A R C ALGEBRA AND CALCULUS C 6 R Hece the rk of the give mtri is. C which is uit mtri of order. Procedure for fidig the solutios of system of equtios: Let the give system of lier equtios be b b m + m + + m b m Step : Costruct the coefficiet mtri which is deoted by A m m ` m m Step : Costruct the ugmeted mtri which is deoted by [A, B] [ A, B] m m m Step : Fid the rks of both the coefficiet mtri d ugmeted mtri which re deoted by R(A) d R(A, B). Step : Compre the rks of R (A) d R(A, B) we hve the followig results. () If R(A) R(A, B) (umber of ukows) the the give system of equtios re cosistet d hve uique solutios. b b b b m

22 (b) If R(A) R(A, B) < (umber of ukows) the the give system of equtios re cosistet d hve ifiite umber of solutios. (c) If R(A) R(A, B) the the give system of equtios re icosistet (tht is the give system of equtios hve o solutio). Problem: Test for cosistecy d hece solve.,, z y z z y Solutio: The coefficiet mtri A The ugmeted mtri [A, B] ~ ~ 9 R R R R R R ~ 9 R R ~ 9 R R ~ R R Here rk of coefficiet mtri is. Rk of ugmeted mtri is. Hece the give system of equtios re cosistet d hve uique solutio.

23 Problem: Test the cosistecy of the followig system of equtios d if cosistet solve y z, y z, 7y z. Solutio: The coefficiet mtri A The ugmeted mtri [A, B] ~ 7 7 ~ R ~ R ~ ~ R R R Here rk of coefficiet mtri is R(A). Rk of ugmeted mtri is R(A, B). i.e., R(A) R(A, B) < (the umber of ukows) Hece the give system of equtios re cosistet but hve ifiite umber of solutios. Here the reduced system is y +z + y + z z i.e., y z z ( ) 6 z 6 k z i.e.,, y, z k where z k is the prmeter. R R R R R R

24 Solutio of Simulteous Equtios Problem: Solve the system of equtios Solutio: It c be represeted s: To see whether solutio eists we eed to fid This determit is ALGEBRA AND CALCULUS y z 6 y z 6. y z 6 y 6. z det (-) (-7) + (-) - Therefore we kow tht the equtios do hve uique solutio. To fid the solutio we eed to fid the iverse of the mtri.. Fid the determit: we hve lredy foud tht this is -. Form the mtri of mior determits (which, for prticulr etry i the mtri, is the determit of the by mtri tht is left whe the row d colum cotiig the etry re deleted): 7 Adjust the sigs of every other elemet (strtig with the secod etry): 7 Tke the trspose d divide by the determit: So the iverse mtri is Hece the solutios to the equtios re foud by Therefore., y - d z. y z

25 Cyley Hmilto theorem: Every squre mtri stisfies its ow chrcteristic equtio. Problem: Verify Cyley Hmilto theorem for the mtri iverse of A. Solutio : The chrcteristic equtio of mtri A is λ λ (++6) + λ(--+) [(-)-(-)+(-)] λ -λ - λ+, which is the chrcteristic equtio. By Cyley Hmilto theorem, we hve to prove A -A - A+ 6 d hece fid the A A.A A A A A -A - A+I Hece the theorem is verified. To fid A - We hve A -A - A+I I - A +A + A A - -A -A+ I A -

26 Problem: Fid ll the eige vlues d eige vectors of the mtri A Solutio : Give A The chrcteristic equtio of the mtri is λ λ (++) + λ(-++) [(-)-(-)-(-)] λ - λ - λ+, which is the chrcteristic equtio λ is root. The other roots re λ - λ - (λ -)( λ +) λ, - Hece λ,, -. The eige vectors of the mtri A is give by (A- λi)x i.e. (- λ) (- λ)..() (- λ) Whe λ, equtio () becomes Tke first d secod equtio,

27 Whe λ -, equtio () becomes Whe λ, equtio () becomes Hece Eige vector 7

28 Problem: Fid ll the eige vlues d eige vectors of Solutio : Give A The chrcteristic equtio of the mtri is λ λ (++) + λ(++) [()-()+(-)] λ -7 λ + λ-, which is the chrcteristic equtio λ is root. -6 The other roots re λ -6λ+ (λ -)( λ -) λ, Hece λ,,. The eige vectors of the mtri A is give by (A- λi)x i.e. (- λ) + + +(- λ) +..() + +(- λ) Whe λ, equtio () becomes Here ll the equtios re sme. Put, we get +

29 - For λ, put,we get + - Whe λ, equtio() becomes (tkig first d secod equtio) 6. Hece Eige vector 9

30 Problem: Fid the eige vlues d eige vectors of Solutio : The chrcteristic equtio of mtri A is )] ( () ) [( ) ( ) ( λ is root. The other roots re, ) )( ( Hece λ,, - The eige vectors of mtri A is give by ) ( ) ( ) ( ) ( X I A () Whe λ,equtio () becomes X

31 Whe λ -,Equtio () becomes ALGEBRA AND CALCULUS - X Whe λ, Equtio () becomes (tkig first d secod equtio) X Hece Eige vector

32 Uit III Mim Ad Miim If cotiuous fuctio icreses up to certi vlue d the decreses, tht vlue is clled mimum vlue of the fuctio. If cotiuous fuctio decreses up to certi vlue d the icreses, tht vlue is clled miimum vlue of the fuctio. Theorem: If, the f() hs mimum if d miimum if. Problem: Fid the mim d miim of the fuctio 6. Solutio: Let f() be 6. At the mimum or miimum poit f () Here f () (-) (+) d - give mimum or miimum. To distiguish betwee the mimum d miimum,we eed Whe, Whe -, - gives the mimum d gives the miimum. Hece Mimum vlue f(-) d Miimum vlue f() -7. Problem: Fid the mimum vlue of for positive vlues of. Solutio : Let f() be

33 At mimum or miimum,. log. e., i.e., - ve. e gives mimum. Mimum vlue of the fuctio f(e). Cocvity d Coveity, Poits of ifleio: If the eighbourhood of poit P o curve is bove the tget t P, it is sid to be Cocve upwrds; if the curve is below the tget t P, it is sid to be cocve dowwrds or cove upwrds. If t poit P, curve chges its cocvity from upwrds to dowwrds or vice vers, P is clled poit of ifleio. Problem: For wht vlues of is the curve upwrds? cocve upwrds d whe is it cove Solutio: The, If, is egtive d so cove upwrds. If, is positive d so cocve upwrds. If, d so there is poit of ifleio t. i.e., t the poit

34 Prtil Differetitio ALGEBRA AND CALCULUS Let u f(, y) be fuctio of two idepedet vribles. Differetitig u w.r.t. keepig y costt is kow s the prtil differetil coefficiet of u w.r.t.. It is deoted by. mes differetite u w.r.t. keepig y costt. Similrly if we differetite u w.r.t. y keepig costt is kow s the prtil differetil coefficiet of u w.r.t. y. It is deoted by. mes differetite u w.r.t. y keepig costt. Symboliclly, if u f(, y), the. Problem: If u log ( + y + z ), prove tht. Solutio: Give u log ( + y + z ) () Similrly ()

35 () Addig (), () d () we get. Problem: If u log(t + ty + tz), show tht. Solutio: Give u log(t + ty + tz) si () Similrly, siy () siz () Addig (), () d () we get

36 Euler s Theorem o Homogeeous Fuctio ALGEBRA AND CALCULUS Theorem: If u is homogeeous fuctio of degree i d y, the. Problem: If u, show tht. Solutio: Give u i.e., si u si u f, where f si u is homogeeous fuctio of degree. By Euler s theorem. Problem: Verify Euler s Theorem whe u + y + z + yz. Solutio: + yz. y + z. z + y. ( + yz) + y (y + z) + z (z + y) ( + y + z + yz) u. 6

37 Totl Differetil Coefficiet: ALGEBRA AND CALCULUS Problem: Fid where u, e t, y e t sit d z e t cost. Solutio: + + e t + y (e t sit + e t cost) + z (e t cost - e t sit) e t ( + y sit + y cost + z cost z sit) e t (e t + e t si t + e t sit cost + e t cos t - e t sit cost) e t. e t e t. Problem: If + y + y, fid. Solutio: + y + y, i.e., f(, y). y y. 7

38 UNIT IV Evlutio of Itegrls & Fourier Series Itegrls of the form Problem: Evlute. Solutio: Put + y; d dy. Problem: Evlute. Solutio: Put y; d dy

39 Itegrl of the form Problem: Evlute Solutio: Let + A ( ) + B + A ( ) + B Equtig coefficiet of o both sides we get A A Equtig costt coefficiets we get, - A + B B A ( ) +. log ( + log ( + log ( + log ( +. 9

40 Itegrls of the form Problem: Evlute Solutio: Itegrl of the form Problem: Evlute. Solutio: Let + A ( ) + B + A ( +) + B Equtig coefficiet of o both sides we get -A A - Equtig costt coefficiets we get, A + B B - A +. + ( + )

41 + ALGEBRA AND CALCULUS +. Properties of Defiite Itegrls: where If f() is odd fuctio i.e., f(-) - f() the 7. If f() is eve fuctio i.e., f(-) f() the Problem: Evlute Solutio: Let I () Also I Addig () d () we get I ()

42 I I. Problem: Evlute. Solutio: Let I () Also I () Addig () d () we get I I () To evlute Put y, d dy. Whe, y ;, y

43 I i.e., () Substitutig () i (), we get I I + I i.e.,. FOURIER SERIES Prticulr Cses Cse (i) If f() is defied over the itervl (,l). f() [ cos b si ] l l b l l l l l l f ( ) d f ( )cos f ( )si If f() is defied over the itervl (, ). l l d, d, f() [ cos b si ] f ( ) d,,

44 f ( )cosd,,,.. b f ( )si d,,.. Cse (ii) If f() is defied over the itervl (-l, l). f() [ cos b si ] l l l l l l l l f ( ) d f ( )cos l d,, l b f ( )si d, l l l,, If f() is defied over the itervl (-, ). f() [ cos b si ] f ( ) d f ( ) cosd,,,.. b f ( ) si d,,..

45 Problem: Obti the Fourier epsio of f() i - < < Solutio: f ( ) d ( ) d f ( )cosd ( ) cosd Here we use itegrtio by prts, so tht b ( si ( )si d ) cos cos ( ) Usig the vlues of, d b i the Fourier epsio we get, f ( ) f ( ) ( ) si cos b si ( ) si This is the required Fourier epsio of the give fuctio.

46 Problem: Obti the Fourier epsio of f()e - i the itervl (-, ). Deduce tht cosech Solutio: Here, e e e e ( d ) e sih cos d e cos si ( ) sih b e si d e si cos Thus, ( ) sih f() sih For,, the series reduces to or f() or sih sih sih sih sih sih ( ) ( ( ) ( cos sih si ) ( ) ) 6

47 Thus, cosech This is the desired deductio. ALGEBRA AND CALCULUS ( ) Problem: Obti the Fourier epsio of f() over the itervl (-, ). Deduce tht 6 Solutio: The fuctio f() is eve. Hece f ( ) d d or f ( ) d f ( ) cos d f ( )cosd, sice f()cos is eve cosd Itegrtig by prts, we get ( ) si cos si Also, b f ( )si d sice f()si is odd. 7

48 Thus ALGEBRA AND CALCULUS f ( ) 6 ( ) cos Hece,... 6 Problem: Obti the Fourier epsio of f ( ),, Deduce tht Solutio: Here, f ( ) d f ( ) d d f ( )cos d cosd f ( )cos d sice f()cos is eve. Also, b si ( ) cos f ( )si d, sice f()si is odd

49 Thus the Fourier series of f() is For, we get or Thus, f ( ) ALGEBRA AND CALCULUS ( ) cos f ( ) ( ) cos cos( ) ( ) ( ) or This is the series s required. Problem: Obti the Fourier epsio of, f(), Deduce tht Solutio: Here, d d ( ( Fourier series is b f() ) ) cos d si d cos d si d ( ) ( ) cos si 9

50 Note tht the poit is poit of discotiuity of f(). Here f( + ), f( - )- t. Hece [ f ( ) f ( )] The Fourier epsio of f() t becomes or [( [( Simplifyig we get, ) ] ) ] Problem: Obti the Fourier series of f() - over the itervl (-,). Solutio: The give fuctio is eve, s f(-) f(). Also period of f() is -(-) Here ( Itegrtig by prts, we get f ( ) d f ( ) d ) d f ( )cos( f ( )cos( ) d ) d ( ) cos( ) d ( ) The Fourier series of f() is f() si ( ) s f() cos( ) is eve cos ( ) b f ( )si( ) d, sice f()si( ) is odd. ( ) cos( ) ( ) si ( )

51 Problem: Obti the Fourier epsio of f(),, ALGEBRA AND CALCULUS Deduce tht Solutio: The period of f() is Also f(-) f(). Hece f() is eve Also, Thus puttig, we get b / / f() f() / / / / / / / f ( ) d d f ( )cos ( f ( )cos si ) f ( )si ( ) ( ) / / d d cos / d f ( ) d cos /

52 or Thus, HALF-RANGE FOURIER SERIES The Fourier epsio of the periodic fuctio f() of period l my coti both sie d cosie terms. My time it is required to obti the Fourier epsio of f() i the itervl (,l) which is regrded s hlf itervl. The defiitio c be eteded to the other hlf i such mer tht the fuctio becomes eve or odd. This will result i cosie series or sie series oly. Sie series : Suppose f() () is give i the itervl (,l). The we defie f() - (-) i (-l,). Hece f() becomes odd fuctio i (-l, l). The Fourier series the is where f ( ) b si () l b l l f ( )si l The series () is clled hlf-rge sie series over (,l). d Puttig l i (), we obti the hlf-rge sie series of f() over (, ) give by Cosie series : Let us defie f ( ) b si b f ( )si d f ( ) ( ) ( ) i (,l)... give i (-l,)..i order to mke the fuctio eve. The the Fourier series of f() is give by f ( ) cos () l where, l l l l f ( ) d f ( )cos l d

53 The series () is clled hlf-rge cosie series over (,l) Puttig l i (), we get f ( ) where Problem: Epd f() ( -) s hlf-rge sie series over the itervl (, ). Solutio: We hve, b Itegrtig by prts, we get b The sie series of f() is f ( ) ( f ( )si d ( )si d ) f ( ) d cos f ( )cos d cos,,, ( ) si.. si ( ) cos Problem: Obti the cosie series of Solutio:, f ( ) over (, ), Here d ( cos d ) d ( )cos d

54 Performig itegrtio by prts d simplifyig, we get ( ) cos,,6,,... Thus, the Fourier cosie series is cos cos6 cos f() Problem: Obti the hlf-rge cosie series of f() c- i <<c Solutio: Here c ( c ) d c c c ( c )cos c c Itegrtig by prts d simplifyig we get, d c ( ) The cosie series is give by c f() c ( ) cos c

55 UNIT-V DIFFERENTIAL EQUATIONS Defiitio: A differetil equtio is equtio i which differetil coefficiets occur. Differetil equtios re of two types(i) Ordiry d (ii) Prtil. A ordiry differetil equtio is oe which sigle idepedet vrible eters, either eplicity or implicity. For emple, dy d d d d si, dr y dy y d m y si re ordiry differetil equtios. Vrible seprble. Suppose equtio is of the form f ( ) d F( y) dy. We c directly itegrte this equtio d the solutio is f ( ) d F( y) dy c, where c is rbitrry costt. dy y Problem: Solve d Solutio: dy dy We hve. y Itegrtig, si - y + si - c. dy Problem: Solve ty cot. d Solutio: dy ty cot d ty dy cot d

56 t y dy cot d log secy log si + logc log secy log si logc log sec y si log c sec y c. si Problem: Solve t sec y dy + ty sec d Solutio: t sec y dy - ty sec d sec t y y sec t dy y y dy put t ty sec t sec t d d put u t dt sec y dy du sec (- d) log t - log u + log c log t + log u log c log (tu) log c tu c t y t c. Problem: Solve sec dy + secy d Solutio: dy sec y sec dy - secy d d sec cos y dy cos d si y - si + c si + si y c. 6

57 Lier Equtio: ALGEBRA AND CALCULUS A differetil equtio is sid to be lier whe the depedet vrible d its derivtives occur oly i the first degree d o products of these occur. The lier equtio of the first order is of the form fuctios of oly. dy d Py Q, where P d Q re Problem: Solve ( + dy ) + y. d Solutio: Divided by + (+ ) (+ ) dy d y dy d y This is of the form dy d Py Q. P d Q The solutio is y e Pd Q e Pd d c y e d e d d c () Pd e e d put t + dt d e d dt t e e logt t 7

58 e d +. () Usig () i (), y ( + ) ( ) d c y ( + ) d c y ( + ) c. Problem: Solve dy + y sec t. d Solutio: This is of the form dy d Py Q. The solutio is y e Pd Q e Pd d c P sec & Q t y e sec d t e sec d d c () Nowe sec d e log(sec t) sec t () y (sec + t) t (sec + t) d c t sec d + t d c sec ( sec ) d sec sec d sec t c. d

59 dy Problem: Solve - t y - si. d ALGEBRA AND CALCULUS Solutio: This is of the form dy d Py Q. The solutio is y e Pd Q e Pd d c P - t & Q - si y e t d si e t d d c () Nowe t d e logsec sec - y sec si ( sec ) d c si sec d c si d c cos t d c - y sec log sec + c. Problem: Solve dy cos + y t. d Solutio: Divided by cos. cos dy y t cos d cos cos dy ysec t sec d P sec & Q t sec The solutio is y e Pd Q e Pd d c 9

60 y e sec d t sec e sec d d c () Nowe sec d e t y e t t sec e t d c put t t dt sec d y e t t t e dt c t. e t - e t e t (t ) + c y e t e t ( t ) + c Problem: Solve ( + dy ) + y cos. d Solutio: Divided by + (+ ) (+ ) dy d y dy d y cos cos This is of the form dy d Py Q. P cos d Q The solutio is y e Pd Q e Pd d c y e d cos e d d c () Pd e e d put t + 6

61 dt d ALGEBRA AND CALCULUS e d dt t e e logt t e d +. () Usig () i (), y ( + ) cos ( ) d c y ( + ) y ( + ) cos d si c. c LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS Problem: Solve (D + D + 6) y e. Solutio: To fid the C.F. solve (D + D + 6) y. The uiliry equtio is m + m + 6. Solvig, m - d -. C.F. A e - + B e -. P.I. e D D 6 e o replcig D by. y A e - + B e - + e. 6

62 Problem: Solve (D md + m ) y e m. Solutio: ALGEBRA AND CALCULUS To fid the C.F. solve (D md + m ) y. The uiliry equtio is k - mk + m. i.e., (k m), k m twice. C.F. e m (A + B). P.I. m e (k - m) m e y e m (A + B + ). Problem: Solve (D - D + ) y si. Solutio: To fid the C.F. solve (D + D + 6) y. The uiliry equtio is m - m +. Solvig, m d. C.F. A e + B e. P.I. si D D si 9 D, put D - -9 si 7 D 7 7 D D 7si D(si ) 9 9D 7si (cos ) 9 9( 9) 6

63 7si 9cos 9 7si 9cos ALGEBRA AND CALCULUS 7si 9cos y C.F. + P.I. A e + B e 7si 9cos. Problem: Solve Solutio: d y d dy y d. (D + D + ) y To fid the C.F. solve (D + D + ) y. The uiliry equtio is m + m +. m... i - i α-, β C.F. e - (A cos + B si ) P.I. D D 6

64 6 D D D D D D.... D D D D D D D D D 9 D D D (Neglectig Higher Powers) 9 ) ( ) ( ) ( D D D 9 () ) ( y C.F. + P.I. e - (A cos + B si ) + 9.

65 Problem: Solve (D + ) y e si. Solutio: The uiliry equtio m +. m - m m i. C.F. e (A cos + B si) A cos + B si P.I. e si D e si, replce D by D+ D D e si D e si, replce D by - D e si D D D e [D(si ) si ] 6D 6 e [D(si ) 6( ) si ] 6 e [cos si ] y C.F. + P.I. A cos + B si e [cos si ]. 6