- Information
- AI Chat
Mathematics 2 - fcbcn lbkhugdsjadskv
Btech (kcs-701)
APJ Abdul Kalam Technological University
Recommended for you
Preview text
3
, i __
... ... CONTENTS.
i I KAS 203: MATHEMATICS - II I
I UNIT-I: DIFFERENTIAL EQUATIONS (1-1 F to 1-41 F)
Linear differential equation of nth order with constant coefficients,
I Simultaneous linear differential equations, Second order linear
1 differential equations with variable coefficients, Solution by
l-
changing mdependentvariable, Reduction of order, Normal fonn,
Method of variation of parameters, Cauchy-Euler equation, Series
solutions (Frobenius Method).
i
I UNIT-2 : MULTIVARIABLE CALCULUS-II (2-1 F to 2-25 F)
j hnproper integrals, Beta & Gama function and their properties,
Dirichlet's integral and its applications, Application of definite
I
intE:!gralsto evaluate surface areas and volume of revolutions.
UNIT-3 : SEQUENCES AND SERIES (3-1 F to 3-26 F)
Definition of Sequence .and series with examples, Convergence of
I
,!
sequence and series, Tests for convergence of series, (Ratio test; D'
Alembert's test, Raabe's test). Fourier series, Half range Fourier sine
\ and cosine series.
UNIT-4: COMPLEX VARIABLE-DIFFERENTIATION (4-1 F to 4-27 F)
Limit, Continuity and differentiability, Functions of complex
. variable, Analytic functions, Cauchy- Riemann equations (Cartesian
and Polar form), Harmonic function, Method to find Analytic
functions, Conformal mapping, Mobius transfonnation and their
properties.
UNIT-5 : COMPLEX VARIABLE-INTEGRATION. (5-1 F to 5-32 F)
Complex integrals, Contour integrals, Cauchy- Goursat theorem,
Cauchy integral formula, Taylor's series, Laurent's series,
.1 Liouvilles's theorem, Singularities, Classification of Singularities,
zeros of analytic functions, Residues, Methods of residues,
Cauchy Residue theorem, Evaluation of real integrals of the type
r:"{(cos e, sin e) de and ] {(x)dx.
. SHORT QUESTIONS (SQ-IF to SQ-22F)
No Previous papers are attached because Unit 1 & 3 are from old
Engineering Mathematics-II syllabus. Unit 2 is from old Engineering
Mathematics-I syllabus and Unit 4 & 5 are from old Mathematics-III
syllabus.
Differential Equations. .
I
I 1-1 F (SeIn-2) ftp \I '.' 1-2 F (Sem.-2) Differential Equations CONCEPT OUTLINE Differential Equation: An equation involving derivatives of one or more dependent variables With respect to one or more independent variables is called a differential equation.. For example, = ax + by (l-x 2 ) (l-y)dx = xy(l + y)dy dy = sec (x + y) dx. Order of a Differential Equation : The order of a differential· equation is the order of the highest derivative involved in a differential equation. 4 2 For example, -4- d x + --2d x + (dxJ5_.- =e 5 .IS of 4 th order. dt dt dt Degree of a Differential Equation: The degree of a differential equation is'the power of the highest derivative which occurs in it, after the differential equation has been made free from radicals and fractions as far as the derivatives are concerned. For example, -4- d 4 x + -2-d 2 x + (dx)5-d = e', is offirst degree. dt dt t Linear Differential Equation: A linear differential equation is an equation in which the dependent variable and its derivatives appear only in the first degree. For example, -----'.+ d 2 2_d 9y = 4x 2 - dx 2 dx The above equation is called a LDE (linear differential equation) with constant coefficients. 1,,'
Mathematics - II 1-5 F (Sem.-2) or PI = cosax f(D2) In both cases replace D 2 by - a 2 but f(- a)2 *- O. If after replacing D2 by
- a 2 any term of D exist in denominator then, multiply the operator by its conjugate, again D2 by -a 2. Terms of D in numerator stands for differentiation of function. Case II : When function is sin ax or cos ax and f(-a 2 ) = 0, PI = sinax = x sinax f(D2) f'(-a 2 ) Repeat this step again if('(- a 2 ) = o. Solve --2 d 2 y +4y = sin 2 2x with conditions y(O) = 0, "",,, ""\1,.·,,.'v.!!0i1m\lL, dx y'(O) = o. W_••IllfIB. d 2 dx y 2 + 4y = sin 2 2x ... cos 4x = 1 - 2 sin 2 2X]
- 4y = 1:. _,cos 4x • 2 2 1 - cos 4x. dx 2 2 2 [ SIn. x=----- 2 The auxiliary equation is m' + 4nt = 0 m(m + 4) = 0 m = 0,- CF = C 1 + C 2 e - 4x (1:. _ PI= 2 2 D(D + 4) 1 (1- cos 4x) 2 D(D + 4) II 1 cos 4x ] 2 D(D+4) D(D+4) 1 II {II} cos 4x ] 2 4 D D +4 (-16 + 4D) 1 1 liD
####### ( )
-1] 1. cos 4x
####### '2 l'4 x - '4 + '4 - 2 x 4 CD - 4)
1-6 F (Sem-2) Differential Equations .! _.! DJ _.! (D + 4) cos 4x 2 4 4 4 4 8 (D 2 - 16)
- 1 [x---1J - 1 (D + 4) cos 4x 2 4 4 8(- 16 -16) = - 1 [x- - -IJ + -- 1 [- 4 sm. 4x + 4 cos 4xl 2 4 4 256 Complete solutioIi is Y = C 1 +'C 2 e - 4x + - 12 [x-4. - - 4 IJ + -- 2561 [- 4 SIn. 4x + 4 cos 4xl y = C 1 + C - 4x + .! (x -1) + ...!.. [cos 4x - sin 4xl ... (1.4)
- 8 ··. i;., Now using the conditiony(O) = 0, we have {i, O=C+C--+- 1 1 2 8 64
- .' C 1 + C 2 = 64 .... (1.4) Using another conditiony'(O) = 0, we have :f ( y' = - 4C e -4x + .! + -±.. [_ sin 4x - cos 4xl . 2 8 64 o = - 4C + - 1 + - 1 (- 1) 2 8 16 4C = .! _...! 2 8 16 4C = 2- 2 16 1 C 2 =- 64 From eq. (1.4), C 1 = - 6 64 On putting the value ofC I and C 2 in eq. 0.4), we get y = - 6 + - 1 e - 4", + -1 ( x - 1) + - 1 [ cos 4x - SIn. 4xl 64 64 8 64 A function n(x) satisfies the differential equation d 2 n(x) n(x) dx 2 - = 0, where L is a constant. The boundary conditions are n(O) = x and n(oo) = O. Find the solution to this equation.
' Mathematics - II 1-7 F (SeD1-2) " 1-8 F (Sem.-2) Differential Equations CF = C I cos ax + C 2 sin ax PI = 2 1 dZn(x). n(x)' _ 0 2 sec ax -----;jT - L 2 - D +a The auxiliary equation is ---;:------- 1 sec ax (D 2 - ia) (D + ia) m 2 - -.!- = 0 L 2 1 [ __ 1 l_J sec G m=±- . 1 2ia D - ia D + ia . L I 1 1 i!" 1 [ 1 sec ax _ 1 sec axJ CF = C 2ia (D - ia)' (D + ia) . 1 e-Yx + Cze Yx • ;j g Complete solution, iI -,- 1 [PI - P n(x) = CF+PI 2w 2 ] §" .r x i 1 n(x) = Cle- Y + Cze Y , ('.' PI:= 0) Where, P 1 = ---sec ax D-ia
####### i
Boundary conditions are wrong. So we can't solve it further. eiax fe-iaxsec ax dx Solve --2- d 2 x + 9x = cos 3t. R!1171Illl!,li",_. e iax f(cos ax - i sin ax) sec ax dx dt a e iax f(1- i tan ax) dx .# d 2 x eiax {x + i COg:s ax)} -2- + 9x = cos 3t i dt , 1 -iax { .(log cos ax)} (D2 + 9) x = cos 3t j Similarly, P 2 = ---,-(secax) = e x-£ Auxiliary equation: m 2 + 9 = 0 D+w a m 2 = - 9 =:> m = ± 3i Replacing i by - i) CF = (CI cos 3t + C 2 sin 3t) >
####### 1 PI= [eiax {x + ieOg c:s ax) } - e-iax {x ieOg c:s ax)}J
PI= --- cos 3t % D 2 + --xe 1 [(iax -e -iax) +£,(logcosax)(iax e +e-iax)J PI = t cos 3t =! (sin3t) = t sin 3t 2ia a 2D 2 3 6 -.- 1 [2"£xa SIn ax + -i Iog cos ax 2 cos ax ] Complete solution, x = CF + PI = C I cos 3t + C 2 sin 3t +! sin 3t 2ta a 6 [x sin ax + cos ax log cos ax] Find the particular solution of the differential equation Solve (D2 - 2D + 1) Y =ex sinx d Y 2 t,Jijiki2:i'rii':¥""'" .' ···· .... 2 --dx2 + a. = sec ax .. "".",;, '.-.. -,:', .:,; ,-'.-,', .' ·Btit);."" •• Auxiliary equation is, m 2 + a 2 = 0 (D2 - 2D + l)y = eX sin x m = ±ai Auxiliary equation, m 2 -2m + 1 = 0
Mathematics - II 1-11 F (Sem.-2) Solve (D2 - 2D + 4) y =eX coS x + sin x cos 3 x.
-
Given equation, (D2 - 2D + 4) y = eX cos x + sin x cos 3 x Auxiliary equation, m 2 - 2m + 4 = 0 m == 2 2 ± J=- m= -- 2 m= l±i Complementary function is CF = eX (C 1 cos .J3x + C 2 sin .J3x) Particular integral, PI= PI + P 2 PI == ex cos x 2 1 __-excosx D '- 2D + 4 X 1 cos x e (D + 1)2 - 2(D + 1) of, 4 X 1 e D 2 + 3 cos x eX---cosx 1 -1+ e X cos-- x 2 1 P 2 == D2 _ 2 D + 4 sin x cos 3x
- 1 2 1 2 SIn 0 x cos 3 x 2 (D - 2 D + 4) .! 2 1 {sin x + 3x) + sin (x - 3x)) 2D-2D+. 11 - 2· (sIn. 4x - sIn. 2)x 2D -2D+.
- 1 [ 2 1 sIn 0 4x - 2 1 sIn 0 2"Jx 2 D -2D+4 D -2D+ 1 1-12 F (SeIll-2) Differential Equations ;, ] ::1 - 1[ 1 SIn 04 x- 1 sm02Jx i 2 - (4)2 - 2 D + 4. - (2)2 - 2 D + 4 ;j
- 1[ 1 SIn. 4x - --1. SIn 2 x] 2 -12-2D -2D /J - 1 [.--sm -1 . 4 x+-sm1. 2 x]
1
4 D+6 D
- 1 [- (D - 6) SIn 0 4 x----cos 2x J 4 D 2 -36 2 1 i - 1 [- (D - 6) sm. 4 x----cos 2XJ i 4 -52 2
! [4 cos 4x - 6 sin 4x _ cos 2x J
i
I
4 52 2 .! [4 cos 4x - 6 sin 4XJ _ cos 2x 4 52 8
I
Complete solution, y == CF+PI = CF+P I +P 2 Y = eX (C In 0 In) cos X 1 cos ,,3x + C 2 sm ,,3x + eX -2- i
i
i 1; +! [4 cos 4x - 6 sin 4XJ _ cos 2x 4 52 8
I
: CONCEPT OUTLINE SiInultaneous Differential Equation: If two or more dependent variables are functions ofa single independent variable, the equations which consist of the derivatives of such variables are called simultaneous differential equationso $
Mathematics - II 1-13 F (Selll-2)
_'4"""'/iSolve the silllultaneous equation dxdt + 5x - 2 y= t,
dy + x + Y = 0 being given x = 0, y = 0 when t =O. dt (D + 5) x - 2y = t (1.13) x + (D + 1) y = 0 ( 1.13) On multiplying eq. (1.13) by (D + 5) and subtracting from eq. (1.13), we get (D + 1) (D + 5) y + 2y = - t (Dz + 6D + 5 + 2) y == - t Auxiliary equation, m 2 + 6m + 7 = 0 m == =>m=-3± J 2 CF = e-3t (C 1 cosh J2 t + C 2 sinh J2 t) 1 PI = 2 (- t) D +6D+
- D 2 + 6D -. 1 6D 7 ( 1+-- --) (t) =- 7(1- 7 7 )t
PI = - - %)
y == e-3t (C 1 cosh J2 t+ C z sinh J2 t) - (t - *) ...n.13)
dy == (_ C l J2 sinh J2 t + J2 C 2 cosh J2 t) dt 1
- 3e-3t (C 1 cosh J2 t + C 2 sinh J2 t) - 7 From eq. (1.13), x = - --y dy dt x = - e-3t (- C I J2 sinh J2 t + J2 C 2 cosh J2 t) +! 7
- 3e-3t (Cl cosh J2 t + C zsinh J2 t) 3t (C
- e- lcoshJ2 t+ C 2 sinh J2 t) + i (t - *)
;]• 1-14 F (SeJIl-2) Differential Equations x = _e-3t (- C 1 J2 sinh J2 t + J2 C 2 cosh J2 t) . t 1 t: + 2e-3t (C 1 cosh J2 t + C 2 sinh J2 t) + - 7 + - 49 ... (1.13) Boundary conditions. t x(O) = 0, yCO) = 0 From eq. (1.13) and eq. (1.13), we have 0= C 6 1 + - 7 C =-- 6 1 7 and 1 0=- J2 C 2 + 2C 1 + 49 12 1 J2c 2 = --+-. 7 49 83 >. '$ J2 C z = 49 '-' 83 t C2.= - 49
Now, y= e- 3t[- cosh J2t - sinh .J2tJ -! (t -
7 49 7 7
x= - e-3t (_ sinh .J2t - 83 cosh .J2t)
. 7 49
+ 2e- 3t (t - .J2t - 83 sinh J2tJ +!.- +
7 49,,2 7 49 Solve the following siJIlultaneous equations. d 2 x dt 2 + Y = sin t d 2 y dt 2 +x = cos t
- Let :t == D then the given system of equations become
D2x +y = sin t ... (1.14) x + D2y = cos t ... ( 1.14) Multiplying eq. (1.14) by D2, we get D4x + D2y = - sin t ... (1.14) Subtracting eq. (1.14) from eq. (1.14), we get
:l § Mathematics - II 1-17 F (SeIIJ.-2) To eliminate y, mUltiplying eq. (1.16) by (D - 1) and multiplying eq. (1.16) by 4, then subtracting, we get [(D + 2) (D - 1) - 4]x = (D - 1) (1) + 4(D - l)t - 6t 2 (D Z +2D - D - 2 - 4)x == -1 + 4 - 4t - 6t 2 (D 2 + D - 6)x = 3 - 4t - 6t 2 Auxiliary equation is m 2 +m-6= 0 m 2 + 3m - 2m '- 6 = 0 m (m + 3) - 2(m + 3) = 0 . (m + 3) (m - 2) 0 => m = 2, - 3 CF = C I e 2t + C 2 e-3t 1 PI = (D 2 + D _ 6) (3 - 4t - 6t ) --=----e 3 Ot - 4t 6 t 2 (D 2 + D - 6) (D 2 + D - 6) (D 2 + D 6) = :- + i 1 t + 1 t
6 6 [1 + ( _ J 6 r 1 + ( - - J
- + r 1 + ( - JI t + r 1 + ( _ _ JI t 2
_ + [1 + +[1- (_ _ + ( _ _
3 4t 4 2 2t 2 2 2 6t (-18+4+12+2) = --+-+-+t +-+-+- =t+-+------- 6 6 36 6 6 36 6 PI"';t 2 +t So, x = C 2t + C 3t + t 2 + t I e 2 e- Now dx = 2C I e 2t - 3C 2 e-3t + 2t + 1 dt Substituting the values ofx and dx in eq. (1.16), we get dt 4y = - 2C e 2t + 3C 3t 2t - 3t I 2 e- - 2t - 1 - 2C 1 e 2C 2 e-
- 2t 2 - 2t + 1 + 4t y = _ C e 2t + - 1 C e-3t _ - 1 t 2 I 422 Solve the siDlultaneous differential equations d 2 x. dx d 2 d ---4-+4x=yand 4y =25x+ 16e'. dt 2 dt dt 2 dt ••
I
F (SeIIJ.-2) Differential Equations
I
(D Z - 4D + 4) x -y = 0 (1.17) '- 25x + (Dz + 4D + 4)y = 16e t (1.17) Multiplying eq. (1.17) by DZ + 4D + 4 and adding to eq. (1.17), we get (DZ - 4D + 4) (DZ + 4D + 4) x - 25 y = 16e t (D4 - 8D - 9)x = 16e t
I
Auxiliary equation is, m 4 -8m z -9 = 0 => (m Z - 9) (m Z + 1) = 0 => m = ± i, ± 3 CF = C I e- 3t + C z e-3t + C 3 cos t + C 4 sin t PI = 1 '(16 et ) = - et D 4 -8D 2 - x = C I e 3t + C z e- 3t + C 3 cos t + C 4 sin t _e t ... (1.17) dx dt = 3CI e 3t - 3C z e-3t + C 3 (- sin t) + C 4 cos t _e t d 2 3t -- + 9C e- 3t - x dt 2 - - 9C I e Z C 3 CO"'" t - C 4 sin t - e t
I
From eq. (1.17), y = --2d 2 x - 4-dx + 4x dt dt = 9C I e 3t + 9C e- 3t - C 3 cos t - C sin t _e t I 3t z 4
- 4 (3C I e - - C 3 sin t + C 4 cos t - e t )
- 4 (C I e 3t + C z e-3t + C 3 cos t + C 4 sin t - et ) => y = C l e3t + 25Cze- 3t + (3C 3 - 4C 4) cos t + (4C 3 + 3C 4) sin t - e t ... (1.17) Eq. (1.17) and eq. (1.17) when taken together give the complete solution. a
I
I
CONCEPT OUTLINE Second Order Linear Differential Equation: A differential 2
equation of the form + P: +Qy =R is known linear differen tial
equation of second order, where P, Q and R are functions of x alone. ;
Mathematics - II 1-,19 F (Sem.-2) Method· of Reduction of Order to Solve Second Order Linear Differential Equation: Let y = u be a part of the complementary fmiction of the given . differential equation d 2y dy --2 +P- + Qy =R ... (1) dx dx Where u is a function of x, then, we have d 2u du --2 + P -- + Qu = R ... (2) dx dx Lety = uv, be the complete solution ofeq. (1), where v is a function of x. Differentiatingy w.r, dy dv du
- = u-+-v dx dx dx 2 2 Again d y = U d v + 2 du dv + V d u dx 2 dx 2 dx dx dx 2 Substituting the values ofy, :; and c::::. in eq. (1), we get d 2 v du dv d 2 u (dv dU). u--+2--+v--+P u-+v- +Q(uv)=R dx 2 dx dx dx 2 dx dx 2 -u--+d v (du2-+Pu ) -+l--+P-+Qudv (d u du )v=R dx 2 dx dx '- dx 2 dx 2 u--+d v (du2-+Pu ) - dv =R dx 2 dx dx 2 R d v du + p) dv dx 2 U dx dx u ... (3) dv d 2v dp Put - =p then -- dx ' dx 2 dx Now eq. (3) becomes, dpdx + (2-;; dudx + P ) P =-;;R ... (4) Eq. (4), is a linear differential equation of first order inp and x. IF=e f(2u = ,p)dx =e" fPdxJ =ue 2 fPdx Solution of eq. (4) is given by pu 2 e. f P <Ix - - f R u 2 efP dx dx + C 1 U Where C 1 is an arbitrary constant of integration. = P-2 - 1 e fP dx [f R ue fl' dx d x+ C ] 1 U. f i
- f!
I
1-20F (Sem.-2) Differential Equations .. dvdx _- u2 1 e -fpdx[fR u e fp dx + C] 1 Integration yields, v = -fpd% [f Jpdx ] I u 1 2 e Ru e dx + C 1 dx + C 2 where C 2 is an arbitrary constant ofintegration. Hence the complete solution ofeq. (1) is given by, y = uv = y = u I e-fpdx[f Ru efp dx + C 1 ] dx + c 2 u Solve (3x + 2)2 d----{ 2 - (3x,+ d =6x. dx. dx .. 2 d 2 y dy (3x + 2) --2- (3x + 2)- -12y =6x· dx dx· 2 Using 3x + 2 = eZ , (3x + 2)2 d ; = 9D(D - 1)y and (3x + 2) dy T 3Dy, dx dx we get 9D(D - l)y - 3Dy - 12y = 2(e z - 2) (9D2 - 9D - 3D - 12)y = 2(e z - 2) The auxiliary equation is 9m 2 - 12m - 12 = 0 (m - 2) ( m + = 0 m = 2-- 2 , 3 Therefore, the complementary function is 2z -2z CF = C 1 e + C 2 e"" and PI = 1 2(e Z -2) 9D 2 -12D+ 12 2. 1 eZ-2 e 0 } { 9D 2 -12D -12 9D 2 -12D -
Mathematics - II 1-23 F (SeD1-2) u- dz dx JQ az= I Then eq. (1.20) reduces to x---+P,-+uy d 2 y dy 2 -R dx 2 1 dz - I Which can be solved easily provided PI comes out to be a Solve by changing the'independent variable :' 2 d y +(3sinx-cotx) dy +2ysin2x=e-cosxsin2x dx 2 dx __ 4 i:lf; y" + (3 sinx -cotx)y' + 2y sin 2 x =e-<:O'x sin 2 x Changing independent variable. z = fix) 2 dy = dy dz, d y = (d Y dz) = (dY ) dz + d z , dx dz dx dx 2 dx dz dx dx dz dx dx 2 2 (dy),(dz) (dz) + d z dz dz dx dx ·dx 2 2 d y (dz)2 + dy d z dz 2 dx dz dx 2 Now from given equation, d 2 y (dz)2 d 2 z. dy dz.. -- -- +--+(3slnx-cotx)-- +2'\Jsln2x=e-cosxsln2x dz 2 dx dx 2 dz dx '-" . dy dz d 2 d 2z (3 SIn x - cot x) -d' dx 2 sin 2 x. ---:'. 2 + __ 2 + Z + Y = e--eos x sln 2 x dz dx ( 'C::f This can be written as d 2y dy dz2' + PI dz + QIY = R I d 2 _ y... + (3 sm x - cot x) --dz Where, PI = dz 2. dx (:f = 2sin 2 x R = e-cosxsin2x Q I (dz / dx)2' I (dz / dx)
I
I 1-24 F (SeD1-2) Differential Equations Choose Q 2 sin 2 x (dz) 2. 2 dz . I = 2, i., 2 = (:=) 2 ,=? dx = SIn x =:> dx = SIn x z = -cosx -- d 2 z = COSX III dx 2 P = COS X + (3 sin x - cot x) sin x Now, I sin 2 x COS X + 3 sin 2 x - cos:x:. sin x _______ x = sin 2 x e- COSX sin 2 x :::: e- cosx R I = sin 2 x --'.. d 2 + 3id + 2y = e--eo,x dz 2 dz --d 2y + 3--dy + 2y = e-z dz2 dz Auxiliary equation is m 2 + 3m + 2 = 0 m = -1,- CF = CIe- z + C Zz 2 e- i PI= 1 e ------e 1 Z
- , (D + 2) (D + 1) D 2 + 3D+ 2 Put, D =-
I
___ 1 ez - eZ 1+3+2 6 e Z ... Complete solution = CF + PI = C I + C e-Z + e-Z = C e--eos x I ! 6 2 I
- C 2 e--eosx + e-cosxl I ; I f
- (.
Mathematics - II 1-25 F (Sem.-2) •• How can we solve differential equation. by rem the first derivative or converting in nOrInal form.? ..*,,dQ A part of the complementary function is needed to find the complete solution, it is not always possible to f"md an integral belonging to CF in such cases, we reduce the given equation to the form in which the term containing the first derivative is absent. For this, we shall change the dependent variable in the equation. d 2y dy dx2 + P dx + Qy = R .... (I.22) By putting y = iw, where u is some function of x, so that dy dx dv du udx+v dx d 2y d 2v du dv d 2u and -- = u--+2--+v-- dx 2 dx 2 dx dx dx 2 On substituting dy I dx and d 2y I dx 2 in terms ofu and v in eq. (I.22), we get d 2v ( dU) dv (d 2U u--+ Pu+2- -+ --+P-+Qu du )v = R dx 2 dx dx dx 2 dx
--+d 2v ( P+-- 2 dU) -+dv (1---+--+Q d 2u Pdu )v = Rlu (1222)
dx 2 u dx dx U dx 2 U dx .,... Let us choose u such that, P+ = 0 udx _ =--uP dx 2 ... du uu = = _ P dx 2 e- 1I2 lPdx I Now, from eq. (1.22), we have
d 2v dP - P dU) +P du +QJ u =
dx 2 u 2 dx 2 dx u dx R e lf2lPdx d 2 --+ v [. ----- 1 dP P ( --uP) +-P(-P) --u +Q Jv dx 2 2 dx 2u 2 2 2 = R d 2 __ v + [Q dx 2 1 dP 2 dx 1 P 2Jv=Re 112lPdx 4. , 1 1-26 F (SeDl-2) Differential Equations 2 or --+Xv=Yd v· ) dx 2 Where X=Q-----P 1 dP 1 2 dx. 4 ... (I.22) and Y = R elf2fpdx Eq. (1.22) may easily be integrated and is known as normal form of eq. (1.22). . Solve the following equation by reducing into nOrDlal form.. d 2 + 2x 2 d + (x 2 - 8)y =x2 e- x212 • dx 2 dx
-
OR Solve the following differential equation by reducing into nOrDlal fOrDl:. 1 2 y" + 2xy' + (xl! - 8)y = x 2 e -2'
-
d 2y. dy -- + 2x- + (x 2 - 8)y = x2 e- x dx 2 dx .. d 2y dy On companson With, dx 2 + P dx + Qy = R, we have P = 2x, Q = x 2 - 8, R = x 2 e- x2/ 2 -'!'f2xdr 2 2 v=e =e =e We know that, u is given by d 2 u dx2 + Q 1 U = R 1 ..... (1.23) Where Q = Q - - 1 dP- - - p 2 = x 2 - 8 - - 1 (2) - --4x 2 , 1 2dx 4 2 4 Q 1 =- R 2 -x' R 1 = v e-;J:'1/ 2 On putting the value ofQ 1 and R 1 in eq. (1.23), we get ---9u d 2 u = x 2 dx 2 (D 2 _ 9)u = x 2 Auxiliary equation, m 2 - 9 = 0 m= ± V Ii ci
Mathematics - II 1-29 F (Sem.-2) PI = 1 eez D 2 +3D+ (Using General method to find PI)
1 e Z (1 l)e z
(D+1)(D+2) e = D+l - D+2 e --e 1 eZ ---ele z D+l D+ Z z z
e- JeZ ee dz - e- 2z Je 2zee dz
Let eZ = t => eZ dz = dt
e- z Jet dt - e- 2z Jte' dt = e--Z et - e-2z (te' - e')
e-z ee z _e-2z (eZ ee z _ee z ) = e-2z eez Complete solution, y = CF + PI z y = C 1 e- z + C 2 e-2z + e-2z ee x y = C 1 (;) +C 2 (:2) +(:2)e Using variation of param m, solve X 2 d--+2x--12y 2 y dy = 0 dx 2 dx
-
Same as Q. 1, Page 1-28F, Unit-I. (Ans'Wer:y = C 1 x 3 + C 2 / x 4 ) Apply m of variation of param to find the general solution
--- d 2 x 4dx-+3x _ --e' r.".."iill·"'..•.·u;,·,'_h." -'''.; " "'''.''.
de dt - l+e t ---4-+3x d 2 x dx e t dt 2 dt 1 + et e l (D 2 - 4D + 3)x = ---t 1 + e Auxiliary equation, m 2 - 4m + 3 = 0 m = 1, CF = C , e t + C z e 3t g
I
1-30 F (Sem.-2) '" Differential Equations e t Here, part'ofCF are U =et , v = e3t .Also, R = ---t 1+e Let x = Ae' + Be 3t be the complete solution of the given equation where A and B are suitab\e of t. To determine A have
- Rv f e t e3t A = fUV1-ut dt + C = - dt + Cl. v 1 (1+et)(3e4t_e4t) .
I = - f 2(1 + e t )e 4t dt + C l = - f 2(e- t + 1) dt + C 1
" 1 !n(e _ t + 1) + C l Ru B= dt+C 2 '
I
I
f uV l - ut v t I 2t f e e dt + C = f e dt + C (1 + e t ) (3e 4t _ e 4t ) 2 2(1 + et)e4t 2 M 1 e- 2t 1
- 2 f (1 + e t ). dt + C 2 = - 2 f F -. t -,dt + C 2 F W - 1 (e-t + .1)2 - - 1 In (e-t + 1) + C 4 2. 2
I Hence the complete solution is
i x = (e- t + 1) + C l Jet + [ - (e- t + 1)2 In (e- t + 1) + (e- t + 1) + C 2 Je3t
rJ !J: Solve by m of variation of param for the differential equation:
E d y _6 dy +9y __ "'01.
".' ",,,' " 2
dx 2 dx (::)
Il W
I
w d 2 y dy ---6-+9y = dx 2 dx Auxiliary equation, m 2 -6m + 9 = 0 (m - 3)2 = 0 m = 3, So, CF = (Cl + CzX)e 3x ., Here U = e N 3x and u = x e 3x are two parts of CF e 3x Also, R= - X f "'"
1-32 F (SeID-2) Differential Equations 1-31 F (SeID-2) ',' Mathematics - II Auxiliary equation, m 2 - 1 == 0 =:> m == ± 1 Let the complete solution be 3x z y == A e 3x + Bx e CF == C1e z + C 2 e- == C1x + C 2 ..!.x To determine the values of A and B, we have 1 J Rv Hence parts of CF are x and - A == - dx + C 1 x uV 1 - ,(e Let u==xandv== 1 7 3X ) 3x xe x A == f- 3x ) 3x 3e 3 dx + C 1 Let y == Ax + B be the complete sol uti on, where A and B are some x e3x (e3x + 3x e _ xe >: e 6x I x suitable functions of x. A and B are determined as follows: A == -J--dx+C6x 1 Rv e \ A == - J dx + C 1 \ uV l - A == - J'!- dx + C 1 x 1 X e - A == -logx + C 1 J Ru - J dx+C 1 B == dx+ C z uV 1 - 3x x 1 e-2- e 3x
- f( e -::2)
- dx + C 1 == 2. 1 x ,. dx +C z eX + C 1 B== f e 3x (e 3x + 3x e 3x) _ 3e3 " xeS" Ru J eXx B = and B == f uv _ u v dx + C 2 == -(_y--)(1)'dx + C 2 x 1 1 X - - - 1 , " 'x 2 x B == --+ C z eXx. 1 2 X Hence the complete solution is x J(-:) dx + C 2 2' J e dx + C 2 y = (-logx +C 3 1 )e 3X + xe >:,
- %[x 2 eX - J2x eX dx ] + C 2 2 == - i[x - 2(x - 1) eX I + C 2 lil'iIttllill Use variation of paraIDeters IDethod to solve the differential equation x 2 y" + xy' - y == x 2 ex. - .! x 2 ex+ (x - 1) eX C 2 ' 2 .• ' Hence the complete solution is given by y == Ax + B == (..!. eX + C ) x + [_.! x 2 eX + (x - 1) eX + C x 2 I 2 2 ] 1:.x X Z y" + xy' -y = x 2 e x . ... (1.29) " y' y ... ( 1.29) Y == C·IX + C2 + (1 -:;1) eX Y Here, R = C" Consider the equation y" + L - Y2 == 0 for finding parts of CF x x Put x == eZ so that z == log x So, [D (D - 1) + D - 11 y = 0 (D2 _ 1) y == 0 ... <I.29) F t t
I
Mathematics - II 1-35 F (Sem.-2) Find the series solution of the' following differential equation.
-
2x(l-x) --2 31 + (1-x) -+d31 331=0 1 __ t'I'(I.'IIJ dx dx 1!:".'f',,'"'. >"r
r_
d 2 2x(1-x)y"+(1-x)y'+3y=0 ... (1.31) Dividing eq. (1.31) by 2x(1-x), we get 1 y" + 2 x y' + 2 x (31- x) y = 0. ...(1.31) Comparing eq. (1.31) with y" + P(x) y' + Q(x) y = 0, we get P(x) = - 1 and Q(x) = ---- 3 2x 2x(1- x) Here P(x) and Q(x) both are non-analytic at x = O xP(x) = .!. and 2 x 2Q(x) = are analytic therefore x = 0 is a regular singular point. (1- x) Let the solution of the given differential equation is 1-36F (Sem.-2) Differential Equations .. m = 0,- 1 2 ._ Roots are different and not differing by an integer: The general term is obtained by replacing k by k + 1 in second summation ofeq. (1.31). a k (m + k + 1) (- 2m - 2k + 3) + a k+, (m + k + 1) -t:.D = 0
- (m+ k+l)(-2m-2k+3) .. ak+l = (m + k + 1)(2m + 2k + ak 2m+2k- Thus, a k + 1 = 2m+2k+l ak Putting k = 0, 1, 2. 2m- a 1 = -2-m-+-l- aO (2m-I) a 2 = (2m+3) a 1 (2m+l) a= 3 (2m+5) (2m+3) a 4 = (2m+7)aa (2m+5) a 5 = (2m+9) a 4 1 3 1 At m = 0, a, = - 3a o' a 2 =a o' a 3 = 5" a o' a 4 = 35 a o' a 5 = 21 a o Yl 5 . + asx + .... )
_o( 2 1 3 3 4 1 5 )
- x a o 1 - 3x + x +"5 x + 35 x + 21 x + .... y, = a 1 - 3x + -- 3X2 + - 3 X 3 .+ - 3 X 4 + - 3 x 5 + .)
I
co o ( 1 3 5 7. y = L ak Xm+ k At In = 1/2, a, = -a o' a?, = 0·, a 3 =0, a 4 = a 5 = as = = 0 k=O Y2= (Y)m=1I2 =x 1l2a o(l-x+0 ) y' = ak(m + k)xm + k - , Y 2 = jXa o(l-x) General solution is y = Ay, + BY
y" = a/m + k) (m + k _1)x m + k - 2 Y = A(1- 3x + _3_x2+_3_x3+_3_x4+_3_x5+ ..... ) +B";(1-x)
. 1 3. 5 7.. Putting all these values differential equation and collecting the like terms, we get _ Solve in series : +x (2x+ l)y'-y=O. 00 00 L ak (m + k+ IX -2m - 2k+3)x m+ k + L ak (m + k)(2m+2k _l)x m + k - 2 = 0 _. "''''''''''"""".. -==.. """".=.. =.,=""=·,q."="ili%j k=O k=O ... ( 1.31) Equating the coefficient of lowest degree term x m - 2 to zero. a om(2m - 1) = 0 2x 2 y" + x(2x + 1) y' - y = 0 . .. (1.32) a o '" 0 t x = 0 is a regular singular point. , f:
I
Mathematics - II 1-37 F (Se-2) Let,. y = L m k 00 ak x + k=O 00 y' = Lak(m + k) k=O 00 y" = Lak(m + k) X m k+ - 2 (m + k-1) k=O Putting the value ofy,y' andy" in eq. (1.32), we get 00 00 2 Lak(m + k) (m + k-1) +2 Lak(m + k) X m+k+ 1 k=O k=O 00 00 +L a k (m + k)x(m+k) - L a kx m+ k = 0 kd ;' tlO co m k 1 k k=O k=O / 2 Lak(m + k) X + + + La k [(m + k) (2m + 2k -2+ l)-lJx m+ = 0 00 00 2 Lak(m+k)x m+k+ 1 + = 0 k .. k=O Equating the lowest degree term to zero by putting k = 0 in second sUmmation, a o (m - 1) (2m + 1) = 0 a o "" 0 m= 1 -- 1 , 2 Roots are different and their difference is not an integer. Th us, y.= G1(y)m=1 + G - 2 (y)m = 2 Equating·the general terms, 2ak (m + k) + a k + 1 (m + k) (2m + 2k + 3) = 0
- 2ak a k+1 = (2m + 2k + 3) Putting k == 0, 1, 2, ... -2ao a 1 = 2m+ a 2 = (2m + 5) a -2az 3 = (2m + 7) and so on 3
I 1-38 F (Se-2)
Differential Equations At m = 1, At m = - - 1 , 2 a -2ao -2ao 1 = -- 5 a 1 = _.- 2 =-a o
- 2 (- 2 ao ) 4ao - 2 ao a 2 == 7 -5- = 35 a 2 = 4(-ao)= a 3 = a 3 = a - 16a o a _ -2(-ao) _ ao 4- 5.7.9 4- 8 ·6 - Th us, Y .== C 1xaO [21--x+-x 4 2 ---x. 8 3 + 16- . x 4·.... ]
- 35 5.7 5.7. +C lfZ 1 -2 . 1 3 1 4 ] 2 x- a o [ 1-x+"2 x -6"x + 24 x .... Use Frobenius series .method to find the series solution
of (l-.)y" -xy' + 4y = 0 -_.
0- x 2 )y" -xy' + 4y = 0 Let x+1=t t(2 - t)y" - (t - 1)y' + 4;y = 0 ... (1.33) Dividing eq. (1.33) by t(2 - t), we get "_ (t -1) , + 4 _ 0 ... (1.33) y t(2-t) Y t(2-t) Y - Comparing eq. 0.33) withy" + P (t) y' + Q (t) y == 0
- (t-1). 4 P(t) = and Q(t) =-- t(2-t) t(2-t) t = 0 is a singular point for the given differential equation. 00 Let, y == '"L.." a k t m + k • IS a soI u t'Ion k=O y' = Lak(m+k)t m+k- 1 y" == L ak (m + k)(m + k _1)t m + k - 2 From eq. (1.33), t(2 - t) Lak(m + k) (m + k - 1) F + k - 2 - (t - 1) L ak(m + k) t m + k - 1
- 4 L akt m + k = 0 , 2 L ak(m + k) (m + k - 1) t m + k- 1_ Lak(m + k) (m + k _ l)tm + k ( 0:
11
I
e
Mathematics 2 - fcbcn lbkhugdsjadskv
Course: Btech (kcs-701)
University: APJ Abdul Kalam Technological University
- Discover more from:
