LECTURE NOTES ON LINEAR ALGEBRA AND ORDINARY DIFFERENTIAL EQUATIONS I B. Tech I semester
UNIT-I THEORY OF MATRICES
Solution for liner systems Mtri : A system of mn numbers rel (or) comple rrnged in the form of n ordered set of m rows, ech row consisting of n ordered set of n numbers between [ ] (or) ( ) (or) is clled mtri of order m n. Eg:............ m m... n n mn m n = [ ij ] mn where i m, j n. some types of mtrics :. squre mtri : A squre mtri A of order nn is sometimes clled s n- rowed mtri A (or) simply squre mtri of order n eg : is nd order mtri. Rectngulr mtri : A mtri which is not squre mtri is clled rectngulr mtri, is mtri. Row mtri : A mtri of order m is clled row mtri eg:. Column mtri : A mtri of order n is clled column mtri Eg: 5. Unit mtri : if A= [ ij ] nn such tht ij = for i = j nd ij = for i j, then A is clled unit mtr.
Eg:I = I = 6. Zero mtri : it A = [ ij ] mn such tht ij = I nd j then A is clled zero mtri (or) null mtri Eg: O = 7. Digonl elements in mtri A= [ ij ] nn, the elements ij of A for which i = j. i.e. (,. nn ) re clled the digonl elements of A Eg: A= digonl elements re,5,9 5 6 7 8 9 Note : the line long which the digonl elements lie is clled the principle digonl of A 8. Digonl mtri : A squre mtri ll of whose elements ecept those in leding digonl re zero is clled digonl mtri. If d, d.. d n re digonl elements of digonl mtri A, then A is written s A = dig (d,d.d n ) Eg : A = dig (,,-)= 9. Sclr mtri : A digonl mtri whose leding digonl elements re equl is clled sclr mtri. Eg : A=. Equl mtrices : Two mtrices A = [ ij ] nd b= [b ij ] re sid to be equl if nd only if (i) A nd B re of the sme type(order) (ii) ij = b ij for every i&j. The trnspose of mtri : The mtri obtined from ny given mtri A, by inter chnging its rows nd columns is clled the trnspose of A. It is denoted by A (or) A T. If A = [ ij ] mn then the trnspose of A is A = [b ji ] nm, where b ji = ij Also (A ) = A
Note : A nd B be the trnsposes of A nd B repectively, then (i) (A ) = A (ii) (A+B) = A +B (iii) (KA) = KA, K is sclr (iv) (AB) = B A. The conjugte of mtri : The mtri obtined from ny given mtri A, on replcing its elements by corresponding conjugte comple numbers is clled the conjugte of A nd is denoted by A Note : if A nd B be the conjugtes of A nd B respectively then, (i) A = A (ii) (A+B) = A+B (iii) (KA) = KA, K is ny comple number (iv) (AB)= B A Eg ; if A= i i 5i i then A i 5i i i. The conjugte Trnspose of mtri The conjugte of the trnspose of the mtri A is clled the conjugte trnspose of A nd is denoted by A θ Thus A θ = ( A ) where A is the trnspose of A. Now A = [ ij ] m n A θ =[b ij ] n m, where bij = ij i.e. the (i,j) th element of A θ conjugte comple of the (j, i) th element of A Eg: if A = 5 i i i i X 5 then A θ = i i i i
Note: A θ = A ( A) nd( A ) =A. (i) Upper Tringulr mtri : A squre mtri ll of whose elements below the leding digonl re zero is clled n Upper tringulr mtri. i.e, ij= for i> j Eg; 8 5 is n Upper tringulr mtri (ii) Lower tringulr mtri ; A squre mtri ll of whose elements bove the leding digonl re zero is clled lower tringulr mtri. i.e, ij= for i< j Eg: 5 7 6 is n Lower tringulr mtri (iii) Tringulr mtri: A mtri is sid to be tringulr mtri it is either n upper tringulr mtri or lower tringulr mtri 5. Symmetric mtri : A squre mtri A =[ ij ] is sid to be symmetric if ij = ji for every i nd j Thus A is symmetric mtri iff = A Eg: h g h b f g f c is symmetric mtri 6. Skew Symmetric : A squre mtri A = [ ij ] is sid to be skew symmetric if ij = ji for every i nd j.
Eg : b c b c is skew symmetric mtri Thus A is skew symmetric iff A= -A (or) -A= A Note: Every digonl element of skew symmetric mtri is necessrily zero. Since ij = - ij ij = 7. Multipliction of mtri by sclr. Let A be mtri. The mtri obtin by multiplying every element of A by sclr K, is clled the product of A by K nd is denoted by KA (or) AK Thus : A + [ ij ] then KA = [k ij ] = k[ ij ] m n m n m n 8. Sum of mtrices : Let A = [ ij ] m n,b = [b ij ] m n be two mtrices. The mtri C = [c ij ] m n where c ij = ij +b ij is clled the sum of the mtrices A nd B. The sum of A nd B is denoted by A+B. Thus [ ij ] m n + [b ij ] m n = [ ij +b ij ] m n nd [ ij +b ij ] m n = [ ij ] m n + [b ij ] m n 9. The difference of two mtrices : If A, B re two mtrices of the sme type then A+(-B) is tken s A B Theorem : Every squre mtri cn be epressed s the sum of symmetric nd skew symmetric mtrices in one nd only wy Proof : let A be ny squre mtri. We cn write A= ½ (A+A )+ ½ (A-A )=P+Q (sy). Where P = ½ (A+A ) Q = ½ (A-A )
We hve P = {½ (A+A )} = ½ (A+A ) since [(KA) = KA ] P is symmetric mtri. = ½ [A +(A ) ]= ½ [A+A ]=P Now, Q = [ ½ (A-A )] = ½ (A-A ) = ½ [A -(A ) ] = ½ (A -A) = - ½ (A-A )= -Q Q is skew symmetric mtri. Thus squre mtri = symmetric + skew symmetric. Then to prove the sum is unique. It possible, let A = R+S be nother such representtion of A where R is symmetric nd S is skew symmetric mtri. R = R nd S = -S Now A = (R+S) = R +S = R-S nd ½ (A+A ) = ½ (R+S+R-S) = R ½ (A-A ) = ½ (R+S-R+S) = S R = P nd S=Q Thus, the representtion is unique. Theorem: Prove tht inverse of non singulr symmetric mtri A is symmetric. Proof : since A is non singulr symmetric mtri A - eists nd A T = A () Now, we hve to prove tht A - is symmetric we hve (A - ) T = (A T ) - = A - (by ()) Since (A - ) T = A - therefore, A - is symmetric.
Theorem : If A is symmetric mtri, then prove tht dj A is lso symmetric Proof : Since A is symmetric, we hve A T = A () Now, we hve (dja) T = dj A T [ since dj A = (AdjA) ] = dj A [by () ] (dja) T = dja therefore, dja is symmetric mtri. [ ik ]m n. Mtri multipliction: Let A =, B = [b kj ] np then the mtri C = [c ij ] mp where c ij is clled the product of the mtrices A nd B in tht order nd we write C = AB. The mtri A is clled the pre-fctor & B is clled the post fctor Note : If the number of columns of A is equl to the number of rows in B then the mtrices re sid to be conformble for multipliction in tht order. Theorem : Mtri multipliction is ssocitive i.e. If A,B,C re mtrices then (AB) C= A(BC) m n n p [ C kl ] p q Proof : Let A= [ ij ] B = [b jk ] nd C= Then AB = [u ik ] m p where u ik = b ij jk ------() n j Also BC = [v jl ] n q where v jl = b c jk kl ------() p k Now, A(BC) is n mq mtri nd (AB)C is lso n mq mtri. let A(BC) = [w il ] m q where w il is the (i,j) th element of A(BC) Then w il = n j ij v jl n p = ij bjkckl by eqution () j k
= p n ij bjk ckl j k (Since Finite summtions cn be interchnged) p = u c ik kl (from ()) k = The (i,j) th element of (AB)C A(BC) = (AB)C. Positive integrl powers of squre mtri: Let A be squre mtri. Then A is defined A.A Now, by ssocitive lw A = A.A = (AA)A = A(AA) = AA Similrly we hve A m- A = A A m- = A m where m is positive integer Note : I n = I O n = Note : Multipliction of mtrices is distributiue w.r.t. ddition of mtrices. i.e, A(B+C) = AB + AC (B+C)A = BA+CA Note : If A is mtri of order mn then AI n = I n A = A. Trce of A squre mtri : Let A = [ ij ] n n the trce of the squre mtri A is defined s. And is denoted by tr A n ii i
Thus tra = n ii i = + +. nn Eg : A = h g h b f g f c then tra = +b+c Properties : If A nd B re squre mtrices of order n nd λ is ny sclr, then (i) tr (λ A) = λ tr A (ii) tr (A+B) = tra + tr B (iii) tr(ab) = tr(ba). Idempotent mtri : If A is squre mtri such tht A = A then A is clled idempotent mtri. Nilpotent Mtri : If A is squre mtri such tht A m = where m is +ve integer then A is clled nilpotent mtri. Note : If m is lest positive integer such tht A m = then A is clled nilpotent of inde m 5. Involutry : If A is squre mtri such tht A = I then A is clled involuntry mtri. 6. Orthogonl Mtri : A squre mtri A is sid to be orthogonl if AA = A A = I Theorem 5: If A, B re orthogonl mtrices, ech of order n then AB nd BA re orthogonl mtrices. Proof : Since A nd B re both orthogonl mtrices. AA T = A T A =I -------- BB T = B T B = I -------- Now (AB) T = B T A T Consider (AB) T (AB) = (B T A T ) (AB) = B T (A T A)B = B T IB (by ) = B T B
= I (by ) AB is orthogonl Similrly we cn prove tht BA is lso orthogonl Theorem 6 : Prove tht the inverse of n orthogonl mtri is orthogonl nd its trnspose is lso orthogonl. Proof : Let A be n orthogonl mtri Then A T.A = AA T = I Consider A T A = I Tking inverse on both sides (A T.A) - = I - A - (A T ) - = I A - (A - ) T = I A - is orthogonl Agin A T.A = I Tking trnspose on both sides (A T.A) T = I T A T (A T ) T = I Hence A T is orthogonl Emples: cos sin. Show tht A = is orthogonl. sin cos Sol: Given A = cos sin sin cos
A T = Consider A.A T = = A is orthogonl mtri.. Prove tht the mtri is orthogonl. Sol: Given A = Then A T = Consider A.A T = cos sin sin cos cos sin sin cos cos sin sin cos sin cos sin cos cos sin sin cos sin cos sin cos I 9
= = A.A T = I Similrly A T.A = I Hence A is orthogonl Mtri. Determine the vlues of,b,c when is orthogonl. Sol: - For orthogonl mtri AA T =I So AA T = = I = Solving b -c =, -b -c = We get c = =b +b =b = From the digonl elements of I b +c = b +b = (c =b ) 9 9 9 9 c b c b c b c b c b c b I c c c b b b c b c b c b c b c b c b c b c b c b b b
b = 6 = b = b= 6 c = b = 7. Determinnt of squre mtri: If A =......... i i... n n... n n in nn nn then A i n i... n...... n...... n in nn 8. Minors nd cofctors of squre mtri Let A =[ ij ] n n be squre mtri when form A the elements of i th row nd j th column re deleted the determinnt of (n-) rowed mtri [Mij] is clled the minor of ij of A nd is denoted by M ij The signed minor (-) i+j M ij is clled the cofctor of ij nd is denoted by A ij.. If A = then
A = M + M + M (or) = A + A + A Eg: Find Determinnt of by using minors nd co-fctors. Sol: A = det A = =(--)-(--6)+(-+6) = -++6 = -8 Similrly we find det A by using co-fctors lso. Note : If A is squre mtri of order n then n KA K A, where k is sclr. Note : If A is squre mtri of order n, then A T A Note : If A nd B be two squre mtrices of the sme order, then AB A B 9. Inverse of Mtri: Let A be ny squre mtri, then mtri B, if eists such tht AB = BA =I then B is clled inverse of A nd is denoted by A -. Theorem 7: The inverse of Mtri if eists is Unique Proof: Let if possible B nd C be the inverses of A. Then AB = BA =I AC = CA= I
consider B = BI =B(AC) =(BA)C =IC B=C Hence inverse of Mtri is Unique Note: (A - ) - = A Note : I - = I. Adjoint of mtri: Let A be squre mtri of order n. The trnspose of the mtri got from A By replcing the elements of A by the corresponding co-fctors is clled the djoint of A nd is denoted by dj A. Note: For ny sclr k, dj(ka) = k n- dj A Note : The necessry nd sufficient condition for squre mtri to posses inverse is tht A Note: if A then A ( dj A) A. Singulr nd Non-singulr Mtrices: A squre mtri A is sid to be singulr if A. If A then A is sid to be non-singulr. Note:. Only non-singulr mtrices posses inverses.. The product of non-singulr mtrices is lso non-singulr. Theorem 9: If A, B re invertible mtrices of the sme order, then
(i). (AB) - = B - A - (ii). (A ) - = (A - ) Proof: (i). we hve (B - A - ) (AB) = B - (A - A)B = B - (I B) = B - B = I (AB) - = B - A - (ii). A - A = AA - = I Consider A - A =I (A - A) = I A. (A - ) = I (A ) - = (A - ) Problems ). Epress the mtri A s sum of symmetric nd skew symmetric mtrices. Where A = 5 7 6 Sol: Given A = 5 7 6
Then A T = Mtri A cn be written s A = ½ (A+A T )+ ½ (A-A T ) P = ½ (A+A T ) = Q= ½ (A-A T ) = s A = P+Q where P is symmetric mtri Q is skew-symmetric mtri. 6 7 5 6 7 5 5 7 6 / / / 7 / 6 6 7 5 5 7 6 5 5 5/ / 5/ /
. Find the djoint nd inverse of mtri A = Soln: Adjoint of A = A A A A A A A A A Where Aij re the cofctors of the elements of ij. Thus minors of ij re M M M M M M 6 M 8 M 8 M 8 Cofctors A ij = (-) i+j M ij
Adjoint of A = --(-) +() = MATRIX INVERSE METHOD ). Solve the equtions +y+5z = 8, -y+8z = nd 5-y+7z = Soln: The given equtions in mtri form is AX = B T 8 8 8 6 8 6 8 8 A ) ( A dj A A 8 6 8 8 8 7 5 8 5 z y
det A = (-7+6)-(-)+5(-+5) = 6 co-fctor mtri is D = ( 7 6) (8 ) ( 5) ( ) ( 5) ( ) ( 5) ( 6 ) ( 8) D = 9 8 7 6 6 Adj A = D T = 9 6 8 6 7 A - = /det A dj A = A = B => = A - B 6 9 6 8 6 7 6 9 6 8 6 7 8 6 6 9 7 68 5 8 8 8
Soln is =, y=, z= Sub Mtri: Any mtri obtined by deleting some rows or columns or both of given mtri is clled is submtri. E.g: Let A =. Then is sub mtri of A obtined by deleting first row nd th column of A. Minor of Mtri: Let A be n mn mtri. The determinnt of squre sub mtri of A is clled minor of the mtri. Note: If the order of the squre sub mtri is t then its determinnt is clled minor of order t. Eg: A = X be mtri is sub-mtri of order = - = - is minor of order 6 6 8 6 z y 5 5 9 8 7 6 5 5 9 8 7 6 5 B B
C 5 6 is sub-mtri of order 7 detc= (7-)-(-)+(8-5) = (-5)-()+() = --+ = -8 is minor of order *Rnk of Mtri: Let A be mn mtri. If A is null mtri, we define its rnk to be. If A is non-zero mtri, we sy tht r is the rnk of A if (i) Every (r+) th order minor of A is (zero) & (ii) At lest one r th order minor of A which is not zero. Note:. It is denoted by ρ(a). Rnk of mtri is unique.. Every mtri will hve rnk.. If A is mtri of order mn, Rnk of A min(m,n) 5. If ρ(a) = r then every minor of A of order r+, or more is zero. 6. Rnk of the Identity mtri I n is n. 7. If A is mtri of order n nd A is non-singulr then ρ(a) = n Importnt Note:. The rnk of mtri is r if ll minors of (r+) th order re zero.
. The rnk of mtri is r, if there is t lest one minor of order r which is not equl to zero. PROBLEMS. Find the rnk of the given mtri Soln: Given mtri A = 7 7 det A = (8-)-(6-8)+(-8) = 8-6+6 = - We hve minor of order ρ(a) =. Find the rnk of the mtri 5 8 6 7 7 8 5 Sol: Given the mtri is of order Its Rnk min(,) = Highest order of the minor will be. Let us consider the minor 5 8 7 6 7 Determint of minor is (-9)-(-56)+(5-8) = -9+-9 =.
Hence rnk of the given mtri is. * Elementry Trnsformtions on Mtri: i). Interchnge of i th row nd j th row is denoted by R i R j (ii). If i th row is multiplied with k then it is denoted by R i K R i (iii). If ll the elements of i th row re multiplied with k nd dded to the corresponding elements of j th row then it is denoted by R j R j +KR i Note:. The corresponding column trnsformtions will be denoted by writing c. i.e c i c j, c i k c j c j c j + kc i. The elementry opertions on mtri do not chnge its rnk. Equivlence of Mtrices: If B is obtined from A fter finite number of elementry trnsformtions on A, then B is sid to be equivlent to A. It is denoted s B~A. Note :. If A nd B re two equivlent mtrices, then rnk A = rnk B.. If A nd B hve the sme size nd the sme rnk, then the two mtrices re equivlent. Echelon form of mtri: A mtri is sid to be in Echelon form, if (i). Zero rows, if ny eists, they should be below the non-zero row. (ii). The first non-zero entry in ech non-zero row is equl to. (iii). The number of zeros before the first non-zero element in row is less thn the number of such zeros in the net row. Note:. The number of non-zero rows in echelon form of A is the rnk of A.. The rnk of the trnspose of mtri is the sme s tht of originl mtri.. The condition (ii) is optionl.
Eg:. is row echelon form.. is row echelon form. PROBLEMS. Find the rnk of the mtri A = by reducing it to Echelon form. sol: Given A = Applying row trnsformtions on A. A ~ R R ~ R R R R R -R 7 7 7 9 9 7 7
~ R R /7, R R /9 ~ R R R This is the Echelon form of mtri A. The rnk of mtri A. = Number of non zero rows =. For wht vlues of k the mtri hs rnk. Sol: The given mtri is of the order If its rnk is det A = A = Applying R R -R, R R kr, R R 9R 9 9 k k 9 9 k k
We get A ~ Since Rnk A = det A = [(8-k)]-(8-k)(k+7)] = (8-k) (-k-7) = (8-k)(--k) = (-k)(6+k)= k = or k = -6 Norml Form: Every mn mtri of rnk r cn be reduced to the form (or) (I r ) (or) (or) by finite number of elementry trnsformtions, where I r is the r rowed unit mtri. Note:. If A is n mn mtri of rnk r, there eists non-singulr mtrices P nd Q such tht PAQ = 7 8 8 8 k k k k 7 8 8 8 k k k k I r I r I r I r
.Norml form nother nme is cnonicl form e.g: By reducing the mtri into norml form, find its rnk. Sol: Given A = A ~ R R R R R R A ~ R R /- A ~ R R +R A ~ c c - c, c c -c, c c -c 5 5 6 5 5 6 5 6 5
A ~ c c -c, c c -5c A ~ c c /-, c c /8 A~ c c This is in norml form [I ] Hence Rnk of A is. Guss Jordn method The inverse of mtri by elementry Trnsformtions: (Guss Jordn method). suppose A is non-singulr mtri of order n then we write A = I n A. Now we pply elementry row-opertions only to the mtri A nd the pre-fctor I n of the R.H.S. We will do this till we get I n = BA then obviously B is the inverse of A. 8
*Find the inverse of the mtri A using elementry opertions where A= Sol: Given A = We cn write A = I A = A Applying R R -R, we get = A Applying R R -R, we get = A Applying R R +5R, R R -R, we get = A 6 6 6 6 5 6 8
Applying R R /, we get 8 = A I = BA B is the inverse of A. System of liner equtions Tringulr systems: Consider the system of n liner lgebric equtions in n unknowns + + + n n =b + + + n n =b --------------------------------------- --------------------------------------- n + n + + nn n =bn The given system we cn write A =B i.e... n... n... n n... nn... n b b.. b n Lower Tringulr system: Suppose the co-efficient mtri A is such tht ll the elements bove the leding digonl re zero. Tht is, A is lower tringulr mtri of the form.
... A =...... n n... nn... n b b.. b n In this cse the system will be of the form =b + =b --------------------------------------- --------------------------------------- n + n + + nn =b n from bove equtions, we get = b / = b b b The method of constructing the ect solution is clled method of forwrd substitution. Upper tringulr system: Suppose the co-efficient mtri A is such tht ll the elements below the leding digonl re ll zero. i.e A is n upper tringulr mtri of the form.... n... n...... nn b b..... n b n
Above system cn be of the form + + + n n =b + + n n =b --------------------------------------- --------------------------------------- nn n =b n from the bove equtions, we get n n b b n nn n n, n n, n n n =, n bn bn nd so on. n, n nn The method of constructing the ect solution is clled method of bckwrd substitution Solution of liner systems Direct methods Method of Fctoriztion (Tringulristion): Tringulr Decomposition Method: This method is bsed on the fct tht squre mtri A cn be fctorized into the form LU, where L is the unit lower tringulr mtri nd U is the upper tringulr mtri.
Note:. The principle minors of A must be non-singulr. This fctoriztion Consider the liner system + + =b + + =b + + =b which cn be written in the form A = B --------() A =. = L. U -----() nd X = Where L = l U = l l u u u u u u Then () LUX = B. If we put UX = Y where Y = y y y Then () becomes LY = B The system is equls to y = b l +y =b () l y +l y +y =b
here y,y,y re solved by forwrd substitution using () we get Y = y y y u +u +u =y u +u =y u =y from these we cn sole for,, by bckwrd substitution. The method of computing L nd U is outlined below from () * we get l l l u u u u u u =. Equting corresponding elements, we get u = l u = l = / nd u = l u = l = / u = l u + u = l + u = u = - / l u +l u = l = [ -l u ] / u nd l +l +u = from which u cn be clculted.
E : Solve the following system by the method of fctoriztion +y+8z =, +y+z = -, +y+z = Soln: The given system cn write AX = B; A 8 X = y z nd B = Let A = LU Where L = l nd U = l l u u u u u u u lu lu l l u u u u l u u l l u u u l u u = 8 u =, u =, u = 8 l u = l = nd l u = l = from the equtions l u +u = nd l u +u =, we get u = -l u = - = u = -l u = -8 = -5
By using l u +l u = nd l u +l u +u = we get l = -l u -l u = -(8)-()5 = - Thus L = nd U = 5 8 Let UX = Y where Y = y y y then LY = B 5 8 z y = y y y ---------() y y y = -----() From () y =, y = - nd y +y = y = - from () = - 9/, y =7/ nd z= ¾ Solution of Tridigonl system:
Consider the system of equtions defined by b u +c u =d u +b u +c u = d ------------------------- nu n- +bnun =dn The co-efficient mtri is b c b c b c n bn cn n bn This mtri is solved by using fctoriztion method. E : Solve the following tri-digonl system of Equtions + = + + = + + = + = step: soln: The given system of equtions in mtri nottion cn be coriten s A = B
= Let A = LU w w w w = w w w w w w w = Equting the corresponding elements of first row w =, w α = α = ½ Equting the corresponding elements of second row, we get β =, α β +w = w = / α = / w = / α = ¾ equting the corresponding elements of fourth row, we get
β =, α β +w = w = 5/ substituting these vlues L = 5 nd U= Step: LUX = B LY = B where UX = Y 5 y y y y = y = y = y +/y = y = y + / y = y = / y +5/ y = y = step: UX = Y
= + ½ = +/ =/ +/ = = solving the solution is given by =, =, =, =
UNIT-II LINEAR TRANSFORMATION
Eigen Vlues & Eigen Vectors Def: Chrcteristic vector of mtri: Let A= [ ij ] be n n n mtri. A non-zero vector X is sid to be Chrcteristic Vector of A if there eists sclr such tht AX=λX. Note: If AX=λX (X ), then we sy λ is the eigen vlue (or) chrcteristic root of A. Eg: Let A= X = AX = = Here Chrcteristic vector of A is nd Chrcteristic root of A is. Note: We notice tht n eigen vlue of squre mtri A cn be. But zero vector cnnot be n eigen vector of A. Method of finding the Eigen vectors of mtri. Let A = [ ij ] be nn mtri. Let X be n eigen vector of A corresponding to the eigen vlue λ. Then by definition AX = λx. AX = λix AX λix = (A-λI)X = ------- () This is homogeneous system of n equtions in n unknowns. () Will hve non-zero solution X if nd only A-λI = - A-λI is clled chrcteristic mtri of A - A-λI is polynomil in λ of degree n nd is clled the chrcteristic polynomil of A - A-λI = is clled the chrcteristic eqution Solving chrcteristic eqution of A, we get the roots, chrcteristic roots or eigen vlues of the mtri. These re clled the
- Corresponding to ech one of these n eigen vlues, we cn find the chrcteristic vectors. Procedure to find eigen vlues nd eigen vectors Let A = i. e., A I n n n n nn λ Then the chrcterstic polynomil is A I sy A I... n... n............ nn n n... The chrcteristic eqution is A-λ we solve the λ λ we get n roots, these re clled eigen vlues or ltent vlues or proper vlues. Let ech one of these eigen vlues sy λ their eigen vector X corresponding the given vlue λ is obtined by solving Homogeneous system n n n n nn n And determining the non-trivil solution.
PROBLEMS. Find the eigen vlues nd the corresponding eigen vectors of Chrcteristic eqution of A is A I λ λ λ Eigen vector corresponding to λ λ λ λ λ λ λ λ λ λ λ λ λ λ Put λ in the bove system, we get Let = Eigen vector is Eigen Vector corresponding to λ λ
from () nd () we hve = Sy Eigen vector. Find the eigen vlues nd the corresponding eigen vectors of mtri Sol: Let A = The chrcteristic eqution is A-λI = i.e. A-λI = λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ=,, The eigen vlues of A is,,. λ λ λ λ λ
λ λ
Properties of Eigen Vlues: Theorem : The sum of the eigen vlues of squre mtri is equl to its trce nd product of the eigen vlues is equl to its determinnt. Proof: Chrcteristic eqution of A is A-λI = n n i.e, n n nn ( polynomil of degree n ) nn + ( polynomil of degree n -) + = ( ) n n n ( ) ( ) n n (... ) polynomil of deg ree ( n ) nn n n ( Trce A) polynomil of deg ree ( n ) in s Theorem : If is n eigen vlue of A corresponding to the eigen vector X, then is eigen vlue A n corresponding to the eigen vector X. Proof: Since is n eigen vlue of A corresponding to the eigen vlue X, we hve
AX= ----------() Pre multiply () by A, A(AX) = A( X) (AA)X = (AX) A X= ( X) A X= X is eigen vlue of A with X itself s the corresponding eigen vector. Thus the theorm is true for n= Let we ssume it is true for n = k i.e,, A K X = K X------------() Premultiplying () by A, we get A(A k X) = A( K X) (AA K )X= K (AX)= K ( X) A K+ X= K+ X K+ is eigen vlue of K+ with X itself s the corresponding eigen vector. Thus, by Mthemticl induction., n is n eigen vlue of A n Theorem : A Squre mtri A nd its trnspose A T hve the sme eigen vlues. Proof: We hve (A- I) T = A T - I T = A T - I (A- I) T = A T - I (or) A- I = A T - I A A T A- I = if nd only if A T - I =
Hence the theorm Theorem : If A nd B re n-rowed squre mtrices nd If A is invertible show tht A - B nd B A - hve sme eigen vlues. Proof: Given A is invertble i.e, A - eist We know tht if A nd P re the squre mtrices of order n such tht P is non-singulr then A nd P - AP hve the sme eigen vlues. Tking A=B A - nd P=A, we hve B A - nd A - (B A - )A hve the sme eigen vlues ie.,b A - nd (A - B)( A - A) hve the sme eigen vlues ie.,b A - nd (A - B)I hve the sme eigen vlues ie.,b A - nd A - B hve the sme eigen vlues Theorem 5: If,.. n re the eigen vlues of mtri A then k, k,.. k n re the eigen vlue of the mtri KA, where K is non-zero sclr. Proof: Let A be squre mtri of order n. Then KA- KI = K(A- I) = K n A- I Since K, therefore KA- KI = if nd only if A I i. e., K is neigen vlueof KAif is neigen vlueof A Thus k, k k n re the eigen svlues of the mtri KA if, n re the eigen vlues of the mtri A Theorem 6: If is n eigen vlues of the mtri A then +K is n eigen vlue of the mtri A+KI Proof: Let be n eigen vlue of A nd X the corresponding eigen vector. Then by definition AX=
Now (A+KI)X AX IKX X KX =( Theorem 7: If, n re the eigen vlues of A, then K, K, n K, re the eigen vlues of the mtri ( A KI), where K is non zerosclr Proof: Since The chrcteristic polynomil of A is A I = ( ) ( ) ( n )----------------------- Thus the chrcteristic polynomil of A-KI is (A KI) I = A (k+ )I = [ Which shows tht the eigen vlues of A-KI re Theorem 8: If re the eigen vlues of A, find the eigen vlues of the mtri Proof: First we will find the eigen vlues of the mtri A- Since re the eigen vlues of A The chrcteristics polynomil is A- ( ( The chrcteristic polynomil of the mtri (A- A- -KI = A-( +K)I
= [ ( +K)] = [ K)] Which shows tht eigen vlues of (A- I) re We know tht if the eigen vlues of A re then the eigen vlues of A re Thus eigen vlues of ( A I) re( ),( ),...( n ) Theorem 9: If is n eigen vlue of non-singulr mtri A corresponding to the eigen vector X, then is n eigen vlue of A nd corresponding eigen vector X itself. Proof: Since A is non-singulr nd product of the eigen vlues is equl to A, it follows tht none of the eigen vlues of A is. If s n eigen vector of the non-singulr mtri A nd X is the corresponding eigen vector nd AX=. Premultiplying this with A, we get A (AX) = A ( X) ( ) A A X A X IX A X X = A X A X X ( ) Hence is n eigen vlue of Theorem : If A tri Adj A Proof: Since is n eigen vlue of non-singulr mtri, therfore Also is n eigen vlue of A implies tht there eists non-zero vector X such tht AX = -----() ( dja) A A A X ( dj A) X or ( dj A) X X AI
Since X is non zero vector, therefore the reltion () it is cler tht A is n eigen vlue of the mtri Adj A Theorem : If Proof: We know tht if is n eigen vlue of mtri A, then is n eigen vlue of A Since A is n orthogonl mtri, therefore A = A is n eigen vlue of But the mtrices A nd A hve the sme eigen vlues, since the determinnts A- I nd A - I re sme. Hence is lso n eigen vlue of A. Theorem : If is eigen vlue of A then prove tht the eigen vlue of B = A + A+ I is + + Proof: If X be the eigen vector corresponding to the eigen vlue, then AX = X --- () Premultiplying by A on both sides This shows tht is n eigen vlue of A A + A+ I A + A+ I)X A X+ AX+ X X+ X+ X ( + + )X ( + + ) is n eigen vlue of B nd the corresponding eigen vector of B is X.
Theorem : Suppose tht A nd P be squre mtrices of order n such tht P is non singulr. Then A nd P - AP hve the sme eigen vlues. Proof: Consider the chrcteristic eqution of P - AP It is ( P - AP)-λI = P - AP-λ P - IP ( I P P) = P - (A-λI)P = P - A-λI P = A-λI since P - P = Thus the chrcteristic polynomils of P - AP nd A re sme. Hence the eigen vlues of P - AP nd A re sme. Corollry : If A nd B re squre mtrices such tht A is non-singulr, then A - B nd BA - hve the sme eigen vlues. Proof: In the previous theorem tke BA - in plce of A nd A in plce of B. We deduce tht A - (BA - )A nd (BA - ) hve the sme eigen vlues. i.e, (A - B) (A - A) nd BA - hve the sme eigen vlues. i.e, (A - B)I nd BA - hve the sme eigen vlues i.e, A - B nd BA - hve the sme eigen vlues Corollry : If A nd B re non-singulr mtrices of the sme order, then AB nd BA hve the sme eigen vlues. Proof: Notice tht AB=IAB = (B - B)(AB) = B - (BA)B Using the theorem bove BA nd B - (BA)B hve the sme eigen vlues. i.e, BA nd AB hve the sme eigen vlues. Theorem 5: The eigen vlues of tringulr mtri re just the digonl elements of the mtri.
Proof: Let A =.................. n n... nn be tringulr mtri of order n The chrcteristic eqution of A is A- I = i.e., i.e, ( - ) ( - ).. ( nn - )=,,. nn Hence the eigen vlues of A re,,. nn which re just the digonl elements of A. Note: Similrly we cn show tht the eigen vlues of digonl mtri re just the digonl elements of the mtri. Theorem 6: The eigen vlues of rel symmetric mtri re lwys rel. Proof: Let be n eigen vlue of rel symmetric mtri A nd Let X be the corresponding eigen vector then AX= Tke the conjugte Tking the trnspose Since Post multiplying by X, we get ------- () Premultiplying () with, we get T X AX X X ------ () T () () gives X X T but
is rel. Hence the result follows Theorem 7: For rel symmetric mtri, the eigen vectors corresponding to two distinct eigen vlues re orthogonl. Proof: Let λ, λ be eigen vlues of symmetric mtri A nd let X, X be the corresponding eigen vectors. Let λ λ. We wnt to show tht X is orthogonl to X (i.e., Sice X, X re eigen vlues of A corresponding to the eigen vlues λ, λ we hve AX = λ X ----- () AX = λ X ------- () Premultiply () by Tking trnspose to bove, we hve T T T T X X X T T T X A -------- () s Premultiplying () by Hence from () nd () we get ( ) Note: If λ is n eigen vlue of A nd f(a) is ny polynomil in A, then the eigen vlue of f(a) is f(λ) PROBLEMS. Find the eigen vlues nd eigen vectors of the mtri A nd its inverse, where A =
Sol: Given A = The chrcteristic eqution of A is given by A-λI = nd X is the solution where is rbritry constnt X Is the eigen vector corresponding to 5
k k k X k k Let is the solution X Is the eigen vector corresponding to 5, becomes For X = 9 X is the eigen vector corresponding to
Eigen vlues of A re Eigen vlues of A re,, We know Eigen vectors of A re sme s eigen vectors of A.. Let f(a) = Then eigen vlues of f(a) re f(), f() nd f(-) f() = () +5() -6()+() = f() = () +5() -6()+() = f(-) = (-) +5(-) -6(-)+() = Eigen vlues of re,, Digonliztion of mtri: Theorem: If squre mtri A of order n hs n linerly independent eigen vectors (X,X X n ) corresponding to the n eigen vlues λ,λ.λ n respectively then mtri P cn be found such tht P - AP is digonl mtri. Proof: Given tht (X,X X n ) be eigen vectors of A corresponding to the eigen vlues λ,λ.λ n respectively nd these eigen vectors re linerly independent Define P = (X,X X n ) Since the n columns of P re linerly independent P
Hence P - eists Consider AP = A[X,X X n ] = [AX, AX..AX n ] = [λx, λ X.λ n X n ] [X,X X n ]..................... n Where D = dig,,,... ) ( n AP=PD P (AP) = P (PD) P AP P PD P AP = (I)D = dig,,,... ) ( n Hence the theorem is proved. Modl nd Spectrl mtrices: The mtri P in the bove result which digonlize the squre mtri A is clled modl mtri of A nd the resulting digonl mtri D is known s spectrl mtri. Note : If X,X X n re not linerly independent this result is not true. : Suppose A is rel symmetric mtri with n pir wise distinct eigen vlues then the corresponding eigen vectors X,X X n re pirwise orthogonl. Hence if P = (e,e e n ) Where e = (X / X ), e = (X / X ).e n = (X n )/ X n then P will be n orthogonl mtri., n
i.e, P T P=PP T =I Hence P = P T P P T AP=D Clcultion of powers of mtri: We cn obtin the power of mtr by using digonliztion Let A be the squre mtri then non-singulr mtri P cn be found such tht D = P - AP D =(P AP) (P AP) = P A(PP )AP = P A P (since PP =I) Simlrly D = P A P In generl D n = P A n P..() To obtin A n, Premultiply () by P nd post multiply by P Then PD n P = P(P A n P)P = (PP )A n (PP ) = A n n n A PD P Hence A n = P n n P n n PROBLEMS. Determine the modl mtri P of =. Verify tht is digonl mtri. Sol: The chrcteristic eqution of A is A-λI = i.e, Thus the eigen vlues re λ=5, λ=- nd λ=-
when λ=5 By solving bove we get X = Similrly, for the given eigen vlue λ=- we cn hve two linerly independent eigen vectors X = = is digonl mtri.. Find mtri P which trnsform the mtri A = Sol: Chrcteristic eqution of A is given by A-λI =
i.e, If,, be the components of n eigen vector corresponding to the eigen vlue λ, we hve [A-λI]X = i.e,. +. +. = nd + + = = nd + + = =, =- =, =-, = Eigen vector is [,-,] T Also every non-zero multiple of this vector is n eigen vector corresponding to λ= For λ=, λ= we cn obtin eigen vector [-,,] T nd [-,,] T P = The Mtri P is clled modl mtri of A
P - = AP P Now 8 6 Cyley - Hmilton Theorem: Every squre mtri stisfies its own chrcterstic eqution PROBLEMS. Show tht the mtri A = stisfies its chrcteristic eqution Hence find A - Sol: Chrcteristic eqution of A is det (A-λI) = C C+C
By Cyley Hmilton theorem, we hve A -A +A-I= A A A I A A A = Multiplying with A we get A A + I =A A. Using Cyley - Hmilton Theorem find the inverse nd A of the mtri A = Sol: Let A = The chrcteristic eqution is given by A-λI = 6 6 7.,. e i
By Cyley Hmilton theorem we hve A -5A +7A-I=..() Multiply with A - we get A - A 5 8 7 8 8 8 7 A 79 78 78 6 5 6 6 6 5 A 6 6 5 5 95 9 9 5 5 75 68 68 56 9 56 56 56 8 69 8 6 6 6 6 8 79 8 8 8 79 Problems. Digonlize the mtri (i) (ii)
. Verify Cyley Hmilton Theorem for A =. Hence find A -. Liner dependence nd independence of Vectors :. Show tht the vectors (,,), (,-,), (,-6,-5) from linerly dependent set. Sol. The Given Vector 5 6 X X X The Vectors X, X, X from squre mtri. Let 5 6 A Then 5 6 A = (+6)-(5-)+(-8+) =6+-8= The given vectors re linerly dependent A =. Show tht the Vector X =(,,), X =(,,-) nd X =(,6,-) re linerly independent. Sol. Given Vectors X =(,-,) X =(,,-) nd X =(,6,-) The Vectors X, X, X form squre mtri. 6 A Then 6 A =(-+6)+(-+)+(6-6) =- The given vectors re linerly independent
A Conjugte of mtri: Rel nd comple mtrices If the elements of mtri A re replced by their conjugtes then the resulting mtri is defined s the conjugte of the given mtri. We denote it with A e.g If A= i 5 6 7i 5 i then A = i 5 6 7i 5 i The trnspose of the conjugte of squre mtri: If A is squre mtri nd its conjugte is A, then the trnspose of A is A T. It cn be esily seen tht A T = A T It is denoted by A A = T i) A A = A T Note: If A nd B be the trnsposed conjugtes of A nd B respectively, then A ii) A B A B KA KA iv) AB B A iii) Hermitin mtri: A squre mtri A such tht A = A T (or) A i e.g A= i 7 then A = i i 7 nd i Aθ = i 7 Here A T =A, Hence A is clled Hermitin Note: T =A is clled hermitin mtri ) The element of the principl digonl of Hermitin mtri must be rel ) A hermitin mtri over the field of rel numbers is nothing but rel symmetric. Skew-Hermitin mtri A squre mtri A such tht T A =- A (or) A T =-A is clled Skew-Hermitin mtri i i e.g. Let A= i i then A = i i i i A T =-A nd A T i i i i
A is skew-hermitin mtri. Note: ) The elements of the leding digonl must be zero (or) ll re purely imginry ) A skew-hermitin mtri over the field of rel numbers is nothing but rel skew-symmetric mtri. Unitry mtri: A squre mtri A such tht A T = i.e A T A=A A T =I A If A A=I then A is clled Unitry mtri Theorem: The Eigen vlues of Hermitin mtri re rel. Proof: Let A be Hermitin mtri. If X be the Eigen vector corresponding to the eigen vlue of A, then AX = X -------------------- () Pre multiplying both sides of () by X,we get X AX X X ----------------------- () Tking conjugte trnspose of both sides of () We get X AX X X i.e X A X X X ABC C B A nd KA (or), KA X A X X X X X A A-------------------- () From () nd (), we hve X X X X i.e X X Hence is rel. X X Note: The Eigen vlues of rel symmetric re ll rel Corollry: The Eigen vlues of skew-hermitin mtri re either purely imginry (or) Zero Proof: Let A be the skew-hermitin mtri If X be the Eigen vector corresponding to the Eigen vlue of A, then AX X or ia X i X ( ) From this it follows tht i is n Eigen vlue of ia
Which is Hermitin (since A is skew-hermitin) A A Now ia ia ia i A ia Hence i is rel. Therefore must be either Zero or purely imginry. Hence the Eigen vlues of skew-hermitin mtri re purely imginry or zero Theorem : The Eigen vlues of n unitry mtri hve bsolute vlue l. Proof: Let A be squre unitry mtri whose Eigen vlue is with corresponding eigen vector X AX X AX X T T T X A X T Since A is unitry, we hve A A I () nd () given T T X A AX X X T i.e T T X X X X T X X From () T Since X X,we must hve Since = We must hve = Note : From the bove theorem, we hve The chrcteristic root of n orthogonl mtri is unit modulus.. The only rel eigen vlues of unitry mtri nd orthogonl mtri cn be Theorem : Prove tht trnspose of unitry mtri is unitry. Proof: Let A be unitry mtri Then A. A A. A I Where A is the trnsposed conjugte of A.
Hence T T T AA A A I T T T AA A A I T T T T A A A A I T T T T A A A A I T A is unitry mtri. PROBLEMS i i i i i i i i i i So A = i i nd i i A T i i T A = A ) Find the eigen vlues of A= Sol: we hve A= Thus A is skew-hermitin mtri. The chrcteristic eqution of A is AI i i A T i i i 8 i, i re the Eigen vlues of A i ) Find the eigen vlues of A i Now A nd i i A T i i T We cn see tht A. A I Thus A is unitry mtri The chrcterstic eqution is AI
i i Which gives i nd i nd / / i Hence bove vlues re Eigen vlues of A. 7 i 5i ) If A= 7 i i then show tht 5i i A is Hermitin nd ia is skew-hermitin. 7 i 5i Sol: Given A= 7 i i then 5i i 7 i 5i 7 i 5i T A 7 i i And A 7 i i 5i i 5i i T A A Hence A is Hermitin mtri. Let B= ia i 7i 5 i i.e B= 7i i i then 5 i i i i 7i 5 i B 7i i i 5 i i i i 7i 5 i T B 7i i i 5 i i i i ( ) 7i 5 i 7i i i 5 i i B i
B T =-B B= ia is skew Hermitin mtri. ) If A nd B re Hermitin mtrices, prove tht AB-BA is skew-hermitin mtri. Sol: Given A nd B re Hermitn mtrices A T A And B T Now AB BA T AB BA T B------------- () AB BA T T T T T T T AB BA B A A B BA AB (By ()) AB BA Hence AB-BA is skew-hemitin mtri. ic b id 5) Show tht A= b id ic is unitry if nd only if +b +c +d = ic b id Sol: Given A= b id ic Then ic b id A b id ic ic b id b id T Hence A A ic AA b id b c d = AA I if nd only if ic b id ic ic b id b c d b id ic b c d 6) Show tht every squre mtri is uniquely epressible s the sum of Hermitin mtri nd skew- Hermitin mtri. Sol. Let A be ny squre mtri
Now A A A A A A A A A A A A is Hermitin mtri. Uniqueness: A A is lso Hermitin mtri Now A A A A A A A A Hence A A is skew-hermitin mtri A A is lso skew Hermitin mtri. Let A =R+S be nother such representtion of A Where R is Hermitin nd S is skew-hermitin Then A R S R S R S R R, S S A A P nd S A A Q R Hence P=R nd Q=S Thus the representtion is unique. i 7) Given tht A= i, show tht I A I A is unitry mtri. Sol: we hve I A i i
i i And i i A I = i i ) ( i i i A I 6 i i Let A I A I B ) )( ( ) )( ( 6 6 i i i i i i i i i i i i B 6 i i B Now 6 i B i nd 6 T i B i 6 T i B B i i i 6 6 6 I T B B i.e., B is unitry mtri. A I A I is unitry mtri. 8) Show tht the inverse of unitry mtri is unitry.
Sol: Let A be unitry mtri. Then AA AA I i.e A A I A A I I Thus A is unitry.
UNIT-III DIFFERENTIAL EQUATIONS OF FIRST ORDER AND THEIR APPLICATIONS
ORDINARY DIFFERENTIAL EQUATIONS OF FIRST ORDER & FIRST DEGREE Definition: An eqution which involves differentils is clled Differentil eqution. Ordinry differentil eqution: An eqution is sid to be ordinry if the derivtives hve reference to only one independent vrible. E. () dy 7y d () d y d dy y e d () Prtil Differentil eqution: A Differentil eqution is sid to be prtil if the derivtives in the eqution hve reference to two or more independent vribles. E.g:. z z z y z z y z y. Order of Differentil eqution: A Differentil eqution is sid to be of order n if the derivtive is the highest derivtive in tht eqution. E.g : (). ( +). + y = Order of this Differentil eqution is. d y dy y e d d () Order of this Differentil eqution is. (). + 5 dy +y =. d
Order=, degree=. (). + =. Order is. Degree of Differentil eqution: Degree of differentil Eqution is the highest degree of the highest derivtive in the eqution, fter the eqution is mde free from rdicls nd frctions in its derivtions. E.g : ) y =. + on solving we get (- ) y. =. Degree = ). = on solving. we get. Degree = Formtion of Differentil Eqution : In generl n O.D Eqution is Obtined by eliminting the rbitrry constnts c,c,c --------c n from reltion like y, c, c,... c. ------()., n Where c,c,c,---------c n re rbitrry constnts. Differentiting () successively w.r.t, n- times nd eliminting the n-rbitrry constnts c,c,----c n from the bove (n+) equtions, we obtin the differentil eqution F(, y, =. PROBLEMS.Obtin the Differentil Equtiony= A + B by Eliminting the rbitrry Constnts: Sol. y= A + B --------------------(). + B(5) ----------(). = A (). + B(5) ----------(). Eliminting A nd B from (), () & ().
e. e e e 5 5e 5 5e 5 y y y 5 y 5 y y -y =. The required D. Eqution obtined by eliminting A & B is y - y -y = ). Log y c Sol: Log y c -------------(). => log y log = c => =c ---------------(). () in () => Log y ) = c. Sol: Given eqution ) =[ ]. = c + = dy d y ) y = [Acos +B sin] Sol: Given eqution is y = [Acos +B sin] = [Acos +B sin] + [-Asin +B cos]
= y + -Asin +B cos). d y d dy d dy dy y y d d = d y 5) y= + b. e Asin B cos e Acos Bsin dy y is required eqution d d Sol: = => ( ). +. = => ( ). +. = is the required eqution. 6) y= + b Sol: -y = 7) Find the differentil eqution of ll the circle of rdius Sol. The eqution of circles of rdius is where (h,k) re the co-ordintes of the centre of circle nd h,k re rbitrry constnts. Sol: = 8) Find the differentil eqution of the fmily of circle pssing through the origin nd hving their centre on -is. Ans: Let the generl eqution of the circle is +y +g+fy+c=. Since the circle psses through origin, so c= lso the centre (-g,-f) lies on -is. So the y- coordinte of the centre i.e, f=. Hence the system of circle pssing through the origin nd hving their centres on -is is +y +g=.
Ans. y. + =. 9) =c. Ans:. +y+. = ) y= Sol: ( + y -=. ) r=(+cos Sol: r=(+cos = - sin Put vlue from () in (). = -r tn Hence. Differentil Equtions of first order nd first degree: The generl form of first order,first degree differentil eqution is f(,y) or [Md + Ndy = Where M nd N re functions of nd y]. There is no generl method to solve ny first order differentil eqution The eqution which belong to one of the following types cn be esily solved.
In generl the first order differentil eqution cn be clssified s: (). Vrible seprble type (). () Homogeneous eqution nd (b)non-homogeneous equtions which to ect equtions. () () ect equtions nd (b)equtions reducible to ect equtions. ) () Liner eqution & (b) Bernoulli s eqution. Type I : VARIABLE SEPARABLE: If the differentil eqution =f(,y) cn be epressed of the form or f() d g(y)dy = where f nd g re continuous functions of single vrible, then it is sid to be of the form vrible seprble. Generl solution of vrible seprble is = c Where c is ny rbitrry constnt. PROBLEMS: ) tn y = sin(+y) + sin(-y). Sol: Given tht sin(+y) + sin(-y) = tn y sin.cos = tn y [Note: sinc+sind =sin(.cos( ] sin = tny secy Generl solution is => -cos = secy +c
=> sec y + cos +c =.// ) Solve ( ). + ( ) =, y() =. Sol: Given ( ). + ( ) = + = On Integrtions d y dy => + =c ---------------() Given y()= => At =,y= ---------() () in () => + =c. => + =c => c= Hence the required solution is + = Ect Differentil Equtions: Def: Let M(,y)d +N(,y) dy = be first order nd first degree Differentil Eqution where M & N re rel vlued functions of,y. Then the eqution Md + Ndy = is sid to be n ect Differentil eqution if function f. d[f (,y)] = f d f dy y Condition for Ectness: If M(,y) & N (,y) re two rel functions which hve continuous prtil derivtives then the necessry nd sufficient condition for the Differentil eqution Md+ Ndy = is to be ect is = Hence solution of the ect eqution M(,y)d +N(,y) dy =. Is + = c. (y constnt) (terms free from ). -------------------********---------------
) Solve y y e d e dy y y y Sol: Hence M = e & N = e ( ) y = = y e ( & = y e ( & = y e y e ( y PROBLEMS y + ) e ( ) y eqution is ect Generl solution is + = c. (y constnt) (terms free from ) => y e y + = c. = c = > = C. Solve ( +).cos d + =. Ans: ( +). sin =c = cos. Solve (r+sin cs + r (sin d Ans: r rsin cos c M r N = sin.
. Solve [y( ) +cos y] d+ [ +log siny]dy =. Sol: hence M = y( ) +cos y, N = +log siny. = + -siny = + -siny so the eqution is ect Generl sol + = c. (y constnt) (terms free from ) + = c. y(+ log) + cosy = c. 5. Solve ysind ( +cos).dy =. 6. Solve (cos-cosy)dy (siny+(ysin))d = Sol: N = cos- cosy & M = -siny-ysin = -sin - cosy = -cosy - sin the eqution is ect. Generl sol + = c. (y constnt) (terms free from ) => + = c => -siny+ ycos =c => ycos siny =c. 7. Solve ( sin. siny - ) dy = ( cos-cosy) d Ans: sin.cosy =c. 8. Solve ( +y - ) d +( -y -b ). y.dy = Ans: + y - -b y =c.
REDUCTION OF NON-EXACT DIFFERENTIAL EQUATIONS TO EXACT USING INTEGRATING FACTORS Definition: If the Differentil Eqution M(,y) d + N (,y ) dy = be not n ect differentil eqution. It Md+Ndy= cn be mde ect by multiplying with suitble function u (,y). Then this function is clled n Integrting fctor(i.f). Note: There my eits severl integrting fctors. Some methods to find n I.F to non-ect Differentil Eqution Md+N dy = Cse -: Integrting fctor by inspection/ (Grouping of terms). Some useful ect differentils. d (y) = dy +y d. d ( =. d ( =. d( ) = d + y dy 5. d(log( ) = 6. d(log( ) = 7. d( ( ) = 8. d ( ( ) = 9. d(log(y)) =. d(log( )) =. d( = PROBLEMS:. Solve d +y dy + =.
Sol: Given eqution d + y dy + = d( ) + d( ( on Integrting + ( = c.. Solve y(. ) d + (y +. dy =. Sol: Given eqution is on Regrouping We get y - d+ y dy + dy =. (yd+ dy)+ y ( dy yd ) = Dividing by (yd + dy) +(. ( ) = d ( ) +(.d +( = on Integrting y y e ½ C is required G.S.. Solve (+y) dy + (- y ) y d = Sol: Given eqution is (+y) dy +(-y ) y d =. (dy + y d ) + y ( dy y d ) =. Divided by y => ( ) + ( = ( ) + dy - d =. On integrting => + log y log =log c y - - log +log y =log c.. Solve yd dy = ( d Sol: Given eqution is yd dy = ( d = d
dtn y Integrting on d tn y = +c where c is n rbitrry constnt. Method -: If M +Ny then M(,y) d + N (,y) dy = is homogeneous differentil eqution nd M Ny is n integrting fctor of Md+ Ndy =.. Solve y d ( + y ) dy = Sol : Given eqution is y d ( + y ) dy = -----------------() Where M = y & N = (- - y ) Consider = & = - eqution is not ect. But given eqution() is homogeneous differentil eqution then So M+ Ny = ( y) y ( + y ) = - y. I.F = M Ny y Multiplying eqution () by y = > y d - y y dy = ----------------------() = >- d - This is of the form M d + N dy = y For M = & N = y = > = & = y y = > = eqution () is n ect D.eqution.
Generl sol M d N dy c (y constnt) (terms free from in N ) => + = c..solve d + ( => + log y = c Ans: (-y). = (+y ).. Solve y( + ( dy = Sol:Given eqution is y( + ( dy = -----------() It is the form Md +Ndy = Where M = y(, N= ( Consider = y - & = y - eqution is not ect. Since eqution() is Homogeneous differentil eqution then Consider M+ N y= [y( ] +y [ ( = y (. => I.F. = y y y Multiplying eqution () by y Now it is ect y y y d y y y y d y y dy dy we get
d dy y + =. ydy d y y log +log y + log ( - log (y - )=logc y = c. Solve r ( + ) d ( + ) dr = Ans: + log - log =c. Method- : If the eqution Md + N dy = is of the form y. f(, y).d +. g (, y) dy = & M- Ny then is n integrting fctor of Md+ Ndy =. Problems:. Solve (y siny +cosy) yd + ( y siny cosy ) dy =. Sol: Given eqution (y siny +cosy) yd + ( y siny cosy ) dy = -------(). Eqution () is of the form y. f(y).d +. g ( y) dy =. Where M =(y siny + cos y ) y N= (y siny- cos y) M y N eqution () is not n ect Now consider M-Ny Here M =(y siny + cos y ) y N= (y siny- cos y) Consider M-Ny =ycosy Integrting fctor =
So eqution () I.F y sin y cos yy y sin y cos y y cos y d y cos y ( y tn y + ) d + ( y tn y - ) dy = M d + N d = Now the eqution is ect. Generl sol M d + N dy = c. dy (y constnt) (terms free from in N ) => + - dy y =c. => +log + (-logy) =log c => log sec(y) +log =log c. =>. secy =c.. Solve (+y) y d + (-y) dy = Sol : I.F = => + =c => + log - log y =c. => +log( ) = where c = c.. Solve ( y+) y d + ( + y- y ) dy = Ans: log y + + =c.. solve ( y +y + ) yd +( y - y+ ) dy = Ans: y - + log( ) =c.
--------------------------------------------------------------------------------------- Method -: If there eists continuous single vrible function f such tht =f(),then I.F. of Md + N dy = is f d e PROBLEMS. Solve ( y ) d + ( dy = Sol: Given eqution is ( y ) d + ( dy = This is of the form Md+ Ndy = => M = y & N = = -y & = -y eqution not ect. Now consider y y y => = = f(). => e d is n Integrting fctor of () eqution () Multiplying with I.F then => d + dy = => ( y -y ) d + ( - y) dy = It is the form M d + N dy =
M y y, N y M y M y y, N N y eqution is n ect Generl sol Md + Ndy = c. (y constnt) (terms free from in N ) y y d dy c = > y y =c.. Solve yd-dy+(+ Sol : Given eqution is (y++ ( dy =. M= y++ & N = = = sin y - M y N = > the eqution is not ect. So consider M N y N sin y sin y sin y sin y sin y sin y d ( f ) d log I.F = e e e Eqution () X I.F => dy =
It is the form of M d+ N dy =. Gen soln => + = => - +- cosy =c. => -------------------------------------------------------------------------. Solve y dy ( +y +)d = Ans: - + +. Solve ( +y ) d -y dy = Ans: -y =c. Method -5: For the eqution Md + N dy = if = g(y) (is function of y lone) then is n integrting fctor of M d + N dy =. Problems:.Solve ( y +y)d +( y - ) dy = Sol: Given eqution ( y +y)d +( y - ) dy = -----------------(). Eqution of the form M d + N dy =. Where M = y +y & N = y - M y y N, 6 y eqution () not ect. So consider = = g(y) I.F = = = =.
y y y Eqution () I.F => d dy y y y d y y y dy It is the form M d + N dy = Generl sol M d Ndy c (y constnt) (terms free from in N ) => + =c. => + =c. => =c.. Solve (y +y) d + ( y ++y ) dy = Sol: = = = g(y). I.F = = =y. 5 Gen sol: y y d y dy c + + =c.. solve (y +y)d + ( y + y ) dy = Sol: = = = g(y). I.F = = = Gen soln : y d ydy c. y
+. Solve (y+ y )d + y dy = Ans: + y =c. 5. Solve (y +y) d + ( y ++y )dy =. Ans: ( +y -) =c. LINEAR DIFFERENTIAL EQUATIONS OF FIRST ORDER: dy Def: An eqution of the form P( ). y Q( ) is clled liner differentil eqution of d first order in y. dy Working Rule: To solve the liner eqution P( ). y Q( ) d first find the integrting fctor I.F = Generl solution is y I.F = Q( ) I.F.d c Note: An eqution of the form Q(y) is clled liner Differentil eqution of first order in. Then integrting fctor = PROBLEMS: Generl solutionis = X I.F = Q( y ) I.F.dy c. Solve (+ y ) d=( y ) dy Sol: Given eqution is (+ y ) = ( y ) tn y + ). = y It is the form of + p(y). = Q(y) I.F = e p( y) dy =