In this chapter a student has to learn the Concept of adjoint of a matrix. Inverse of a matrix. Rank of a matrix and methods finding these.

Transcription

1 MATRICES UNIT STRUCTURE.0 Objectives. Introduction. Definitions. Illustrative eamples.4 Rank of matri.5 Canonical form or Normal form.6 Normal form PAQ.7 Let Us Sum Up.8 Unit End Eercise.0 OBJECTIVES In this chapter a student has to learn the Concept of adjoint of a matri. Inverse of a matri. Rank of a matri and methods finding these.. INTRODUCTION At higher secondar level, we have studied the definition of a matri, operations on the matrices, tpes of matrices inverse of a matri etc. In this chapter, we are studing adjoint method of finding the inverse of a square matri and also the rank of a matri.. DEFINITIONS ) Definitions:- A sstem of m n numbers arranged in the form of an ordered set of m horizontal lines called rows & n vertical lines called columns is called an m n matri. The matri of order m n is written as

2 a a a a a j n a a a a a j n a a a a a i i i ij in a a a a a m m m mj mn n n Note: i) Matrices are generall denoted b capital letters. ii) The elements are generall denoted b corresponding small letters. Tpes of Matrices: ) Rectangular matri :- An mn Matri where m n is called rectangular matri. For e.g 4 ) Column Matri : It is a matri in which there is onl one column. 4 ) Row Matri : It is a matri in which there is onl one row ) Square Matri : It is a matri in which number of rows equals the number of columns. i.e its order is n n.

3 e.g. A 4 6 5) Diagonal Matri: It is a square matri in which all non-diagonal elements are zero. e.g. 0 0 A ) Scalar Matri: It is a square diagonal matri in which all diagonal elements are equal. e.g A ) Unit Matri: It is a scalar matri with diagonal elements as unit. e.g. 0 0 A ) Upper Triangular Matri: It is a square matri in which all the elements below the principle diagonal are zero.

4 4 e.g. 0 A ) Lower Triangular Matri: It is a square matri in which all the elements above diagonal are zero. the principle e.g A 4 0 0) Transpose of Matri: It is a matri obtained b interchanging rows into columns or columns into rows. A T A Transpose of A ) Smmetric Matri: T If for a square matri A, A A then A is smmetric 5 A ) Skew Smmeric Matri : T If for a square matri A, A A then it is skew -smmetric matri.

5 5 A Note : For a skew Smmetric matri, diagonal elements are zero. Determinant of a Matri: Let A be a square matri then A = determinant of A i.e det A= A If (i) then A 0 matri A is called as non-singular and If (i) then A 0, matri A is singular. Note : for non-singular matri A- eists. a) Minor of an element : Consider a square matri A of order n Let A = a ij n n The matri is also can be written as A = a a a a n a a a a n a a a a n n n nn Minor of an element a ij is a determinant of order (nd) b deleting the elements of the matri A, which are in 6th row and 5th column of A. E.g. Consider, a a a A = a a a a a a M = Minor of an element a

6 6 A = a a a a II M = a a a a E.g. (ii) Let, 5 8 A = M =, M =, M = M =, M =, M = (b) Cofactor of an element :- If A = element a ij. a ij is a square matri of order n and a ij denotes cofactor of the i j C ij =. M ij Where ij M is minor of a ij. If a b c A = a b c a b c A = The cofactor of B = The cofactor of C = The cofactor of A =(-) b =(-) b =(-) a a b c b c a a c c b b

7 7 E.g. Consider, 4 A = c = M c = 0 6 =. = =. = = 7 = = 7 = = 5 = (C) Cofactor Matri :- A matri C = C ij where C ij denotes cofactor of the element a ij. Of a matri A of order nn, is called a cofactor matri. In above matri A, cofactor matri is 5 6 C = A B C C = A B C A B C Similarl for a matri, A = 9 the cofactor matri is c= 4 (d) Adjoint of Matri :- If A is an square matri then transpose of its cofactor matri is called Adjoint of A.

8 8 Thus in the notations used, Adjoint of A C T A B C Adj A = A B C A B C Adjoint of a matri A is denoted as Adj.A Thus if, 4 A = than Adj. 5 0 A = Note : If a b A = than Adj. c d d -b A = -c a (d) Inverse of a square Matri:- Two non-singular square matrices of order n A and B are said to be inverse of each other if, AB=BA=I, where I is an identit matri of order n. Inverse of A is denoted as A - and read as A inverse. Thus AA - =A - A=I Inverse of a matri can also be calculated b the Formula. A - = A Adj.A where A denotes determinant of A. Note:- From this relation it is clear that A - eist if and onl if A 0 i.e A is non singular matri.. ILLUSTRATIVE EXAMPLES

9 Eample : Find the inverse of the matri b finding its adjoint 9 A Solution: We have, A A A A eists Transpose of matri A=A A We find co-factors of the elements of A (Row-wise) C. F., C. F., C. F. C. F. 7, C. F., C. F. 5 C. F. 5, C. F., C. F. - adj A A A adj (A) Eample : Find the inverse of matri A b Adjoint method, if 0 A =

10 0 Solution: Consider 0 A = = = = Co factor of the elements of A are as follows C =. C =. 8 C =. 5 C =. 0 C =. 6 0 C =. C =. 0 C =. 0 C =. Thus, 8 5 Cofactor of matri C = 6 And Adjoint of A= C

11 = 8 6 A Note:- A Rectangular matri does not process inverse. Properties of Inverse of Matri:- i) The inverse of a matri is unique i.e ii) The inverse of the transpose of a matri is the transpose of inverse T i.e. (A ) T (A ) iii) If A & B are two non-singular matrices of the same order (AB) B A This propert is called reversal law. Definition:-Orthogonal matri.:- If a square matri it satisfies the relation is called an orthogonal matri. & T AA Ithen the matri A T A A Eample : show that A = Cos Sin is orthogonal matri. Solution: To show that A is orthogonal i.e To show that T AA I A = Cos Sin Sin Cos T Cos Sin A = Sin Cos T Cos Sin Cos Sin AA = Sin Cos Sin Cos Cos Sin Cos Sin Sin Cos Sin Cos Cos Sin Sin Cos

12 0 = I 0 A is an orthogonal matri. Check Your Progress: Q. ) Find the inverse of the following matrices using Adjoint method, if the eist. i), ii), 4 iii) cos sin, sin cos iv) 0 5, 5 0 v) cos sin 0 sin cos 0, 0 0 vi) vii) Q.) If A = prove that A= B.C - Q. 4) If cos sin, sin cos 4 A = B = tan, C= tan, prove that Adj. A= A tan,, tan Q. 5) If A = 0, verif if (Adj.A) = (Adj.A ) Q.6) Find the inverse of A = 0, hence find inverse of 6 A =

13 .4 RANK OF A MATRIX a) Minor of a matri Let A be an given matri of order mn. The determinant of an submatri of a square order is called minor of the matri A. order mn then r mif m<n and r n if n<m. e.g. Let 4 A = The determinants , 0 7, 4 7, ,,,, 0,, Are some eamples of minors of A. b) Definition Rank of a matri: almost one minor of the matri which is of order r whose value is non-zero e.g.(i) Let 0 A = 4 5 7

14 4 Consider e.g. Let 0 0 A = 4, A = 8 etc A = Rank of A= (ii) A = 0 Here, A = 0 0 A = 0 Thus minor of order is zero and atleast one minor of order is non-zero Rank of A =. Some results: (i) Rank of null matri is alwas zero. (ii) Rank of an non-zero matri is alwas greater than or equal to. (iii) If A is an mn non-zero matri then Rank of A is alwas equal to rank of A. (iv) Rank of transpose of matri A is alwas equal to rank of A. (v) Rank of product of two matrices cannot eceed the rank of both of the matrices. (vi) Rank of a matri remains unleasted b elementar transformations. Elementar Transformations: Following changes made in the elements of an matri are called elementar transactions. (i) Interchanging an two rows (or columns). (ii) Multipling all the elements of an row (or column) b a non-zero real number.

15 5 (iii) Adding non-zero scalar multitudes of all the elements of an row (or columns) into the corresponding elements of an another row (or column). Definition:- Equivalent Matri: Two matrices A & B are said to be equivalent if one can be obtained from the other b a sequence of elementar transformations. Two equivalent matrices have the same order & the same rank. It can be denoted b [it can be read as A equivalent to B] Eample 4: Determine the rank of the matri. A = 4 Solution: Given 6 5 A = R R - R & R R - R 0 0 Here two column are Identical. hence rd order minor of A vanished Hence nd order minor 0 0 e(a) Hence the rank of the given matri is..5 CANONICAL FORM OR NORMAL FORM

16 6 Ir o If a matri A of order mn is reduced to the form o o using a sequence of elementar transformations then it called canonical or normal Note:- If an given matri of order mn can be reduced to the canonical e.g. () Consider Eample 5: Determine rank of the matri. A if 6 A = Solution: 6 A = R R 6 R R, R R R 7R

17 7 R R, R R R R R, R R I o Rank of A= Eample 6: Determine the rank of matri 7 A = Solution: A = R R, R R R R R R

18 C C C C I 0 Rank of A= Eample 7: Determine the rank of matri A if 4 A = Solution: 4 A = R R, R R, R4 6R, R R

19 R + R, R 4R, R4 9R R 4 R R C- C R + R, R R C4-5C C C

20 0 I Rank of A= Check Your Progress:- Reduce the following to normal form and hence find the ranks of the matrices i) ii) 4 iii) iv) vii) 0 4 v) vi) viii) NORMAL FORM PAQ such that, Ir PAQ We observe that, the matri A can be epressed as A = Im Where Im In are the identit matrices of order m and n respectivel. Appling the elementar transformations on this equation. A in L.H.S. can be reduced to normal form. The equation can be transformal into the equations. Ir PAQ

21 Note that, the row operations can be performed simultaneousl on L.H.S. and prefactor (i.e. Im in equation (i)) and column operations can be performed simultaneousl on L.H.S. and post factor in R.H.S. i.e. [(In in eqn (i)] Eamples 8: Find the non-singular matrices P and Q such that PAQ is in normal and hence find the rank of A. i) A Solution: Consider A= I AI = 0 0 A R R = 0 0 A C 5C, C 4C = 0 0 A R R = 0 A R R, R R, = A C C

22 = A R, = A R R = 0 A Thus 0 0 P = 0 P = 5 Q = 0 Q = 0 0 P and Q are non-singular matrices Also Rank of A = ii) 6 A Solutions: Consider 0 0 A= 0 0 A

23 A 0 0 R R A C C, C C, C4 C A R R, R R A R 6R, A C4 C A C 5C A R, R

24 A R R A I 0 = 0 A P = 0, P Q =, Q 0 0 Also, P&Q are non singular. Rank of A =. Check Your Progress: A) Find the non-singular matrices P and Q such that PAQ is in normal form and hence find rank of matri A. i) ii) iii)

25 5 iv) (v) LET US SUM UP Definition of matri & its tpes. Using Adjoint method to find the A b using formula A adja A Rank of the matri using row & column transformation Using canonical & normal form to find Rank of matri..8 UNIT END EXERCISE ) Find the inverse of matri A if eists ii) Find Adjoint of Matri A 0

26 6 0 iii) Find the inverse of A b adjoint method if A iv) Find Rank of matri A Cos Sin 0 v) Prove that the matri A Sin Cos 0 is orthogonal 0 0 Also find A. vi) find its rank. Reduce the matri 0 0 A to the normal form and 0 0 vii) Find the non singular matri. such that is the normal form when A = Also find the rank of matri B X = 4 & Y viii) Under what condition the rank of the matri will be! A = 4 0

27 7 i) If X = 4 & Y Then show that where denotes Rank. 8 6 ) Find the rank of matri A = ***** LINEAR ALGEBRIC EQUATIONS UNIT STRUCTURE. Objectives. Introduction. Canonical or echelon form of matri.4 Linear Algebraic Equations.5 Let Us Sum Up.6 Unit End Eercise. OBJECTIVES After going through this chapter ou will be able to - Find the rank of Matri. - Find solution for linear equations. - Tpe of linear equations. - Find solution for Homogeneous equations. - Find solution of non-homogeneous equations.. INTRODUCTION In XII th we have solved linear equations b using method of reduction also b rule. Here we are going to find solution of homogeneous

28 8 and non-homogeneous both with different case. Using matri we can discuss consistenc of sstem of equation.. CACONICAL OR ECHOLON FORM OF MATRIX Let A be a given matri. Then the canonical or Echelon form of A is a matri in which (i) One or more elements in each of first r-rows are non-zero and these first r-rows form an upper triangular matri. (ii) The elements in the remaining rows are zero. Note : ) The number of non-zero rows in Echelon form is the rank of the matri. ) To reduce the matri to Echelon form onl row transformations are to be applied. Solved Eamples :- Eample : Reduce the matri to Echelon and find its rank. A Solution: R R R R R R R4 R4 6 R A R R A

29 9 A R R 4 R 5 R R 9 R 5 A R4 R4 R A Rank of A e A No. of non-zero rows Check Your Progress: ) Find the rank of the following matrices b reducing to Echelon form. i) ii) iii) A 4 7 Ans : A A Ans : 4

30 0.4 LINEAR ALGEBRIC EQUATIONS i) Consider a set of equations : a b c z d a b c z d a b c z d The equation can be written in the matri form as : a b c d a b c d a b c z d A X D i. e. AX D Now we join matrices A and D a b c : d A : D a b c : d a b c : d It is called as Augment matri We reduce (A.D.) to Echelon form and thereb find the ranks of A and (A:D) ) If ( A) ( AD) then the sstem is inconsistent i.e. it has no solution. ) If ( AD) ( A) then the sstem is consistent and if (i) ( AD) ( A) Number of unknowns then the sstem is consistent and has unique solution. (ii) ( AD) ( A ) Number of unknowns and has infinitel man solutions. Non- Homogeneous equation:- Sstem of simultaneous equation in the matri form is Pre-multipling both sides of I b A we set

31 A AX A D IX A B X A B which is required solution of the given non-homogeneous equation. Homogeneous linear equation:- Consider the sstem of simultaneous equations in the matri form. AX D If all elements of D are zero i.e then the sstem of equation is known as homogeneous sstem of equations. In this case coefficient matri A and the augmented matri [A,O] are the same. So The rank is same. It follow that the sstem has solution,, , which is called a trivial solution. Eample : Solve the following sstem of equations Solution: The sstem is written as AX Hence the coefficient and augmented matri are the same We consider A A 4 4 R R R

32 A 4 R R - R & R R 4R R R R R +9 R & R R R 0 R R R R + R & R R R Hence Rank of A is ( A), The coefficient matri is non-singular Therefore there eist a trivial solution 0 Eample : Solve the following sstem of equations Solution: The given equations can be written as AX 0

33 Here the coefficient & augmented matri are the same A 4 4 R R - R & R R R R R R Here rank of A is i.e ( A ) So the sstem has infinite non-trivial solutions Let

34 4 Hence and Hence infinite solution as deferred upon value of Eample 4: Discuss the consistenc of 4z z 4 z 5 Solution: In the matri form 4-4 z -5 Consider an Agumental matri 4 : A: D : 4 R R R R R R : 5 4 : A : D : : R R R 4 : A: D : : 7

35 A AD AD The sstem is inconsistent and it has no solution. 5 Eample 5: Discuss the consistenc of z z z 4 Solution: In the matri form, A - z 4 A Now we join matrices A and D Consider : A:D - - : - : 4 X D We reduce to Echelon form R R : 4 A:D - - : - R R R R R R : 4 A:D -7 - : - 5 R R R : -9 :

36 6 : 4 A:D : -...() : 7 7 This is in Echelon form AD A AD A Number of unknowns sstem is consist and has unique solution. Step () : To find the solution we proceed as follows. At the end of the row transformation the value of z is calculated then values of and the value of in the last. The matri in e.g. () in Echelon form can be written as z -8 7 Epanding b R Z z - epanding b R 7 z 7 ( ) ep anding b R z 4 4 4,,z - Eample 6: Eamine for consistenc and solve 5 7z 4 6 z 9

37 7 7 0z 5 Solution: Step () : In the matri form Consider z 6 A X D 5 7 : 4 4 A: D 6 : : 5 z 6 R 5 R : 4 5 A: D 6 : 9 R R R R R 7 R 7 0 : : 4 5 A: D : 5 R R R : 4 5 A: D : : 0 AD A : 5 AD A Number of unknowns The sstem is consistent and has infinitel man solutions. Step () :- To find the solution we proceed as follows: Let

38 -z z put z k k B eapanding 8 z k... k parameter B epanding R 5 5z 5 R z 5 7 6k Check Your Progress: Solve the sstem of equations: i) Ans : consistent ii),,, Ans : Infinitel man solutions, 5 9 iii) k, 4 k, k, 4 k Ans : Inconsistent iv) 0

39 9 0 0 Ans: Trivial Solution. v) Ans : Definitel man solution 0.5 LET US SUM UP In this chapter we have learn Using row echelon from finding Rank of matri. Representing linear equation m n in to argumented matri. Consistenc of matri. Solution of Homogeneous equations. Solution of non homogeneous equations..6 UNIT END EXERCISE ) Reduce the following matri in Echolon form & find its Rank. i) A Ans : Rank = 5 4 ii) 4 A Ans : Rank = 0

40 40 iii) iv) A 0 Ans : Rank = 0 A Ans : Rank = ) Solve the following sstem of equations. i) + + =, + + = 4, =6 Ans:-,, z 0. ii) - - =0, - =0, + - =0 Ans:-.... iii) = = =0 iii) + - =0 iv) = - + = = =7 - + =0. v) = = =. *****

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73 70 Z =+i is called a comple number, where i and, Z i is called a conjugate of the comple number Z R and Let A be a mn matri having comple numbers as its elements, then the matri is called a comple matri. Conjugate of a Matri: The matri of order mn is obtained b replacing the elements b their corresponding conjugate elements, is called conjugate of a matri. It is denoted b A i i For e.g. A i i A i i i i Properties of conjugate of matri: () A A () A B A B () AB A. B Conjugate Transpose: Transpose of the conjugate matri A is called conjugate transpose. It is denoted b A. i i For e.g. A i i i i A then i i i A i i i Properties of Transpose of Conjugate of a matri: () A A () A B A B () AB B. A Hermitian matri:

74 7 A square matri A is called Hermitian matri if A= A i.e. A= A is Hermitian if aij a ji i and j. a ij m n Eample 0: Show that the matri Solution: i i A i i i i is Hermitian Here i i A i i i i i i A i i i i i i A i i i i A A Hence b definition A is Hermitian matri. Skew Hermitian Matri: A Square matri A such that A A is called a Skew Hermitian Matri. i.e. if A aij is Skew Hermitian if aij a ji i and j. m n Here a ij = purel imaginar or re a ij = 0. Eample : Show that the matri Matri. i 5 i 6 i A 5 i 0 i 6 i i 0 is called a Skew Hermitian Solution: i 5 i 6 i Here A 5 i 0 i 6 i i 0

75 7 i 5 i 6 i A 5 i 0 i 6 i i 0 i 5 i 6 i A 5 i 0 i 6 i i 0 i 5 i 6 i A 5 i 0 i 6 i i 0 Hence A = -A The matri A is Skew Hermitian Matri. Note: Let A be a square matri epressed as B+iC where B and C are Hermitian and Skew Hermitian Matrices respectivel. A A A i A A B ic i B A A and Unitar Matri: C A A i A square matri A is said to be unitar matri if A A Eample : Show that the matri A 5 i i i i is Unitar matri. Solution: Here A AA A 5 5 i i 5 i i i i i i i i i i i i i i

76 AA I I Hence A is Unitar Matri. Eample : Epress the matri, i i A i i i i 5 Skew Hermitian Matri. As the Hermitian Matri and Solution: i i Let A i i i...( I ) i 5 i i A i i i i 5 i i A i...( II) i i 5 Adding I and II we get i i i i i A A i i i i i 5 i i 5 4 i 4 4i i i 4 4i i 0 4 i 4 4i B A A i i...( III ) 4 4 i i 0 also i i i i i A A i i i i i 5 i i 5

77 74 i i i i 5 i i 5 i 0 i i i A A i 5 i...( IV ) i 5 i 0 Now, A=B+iC 4 i 4 4i i i i A i i i 5 i 4 4i i 0 i 5 i 0 Eample 4: Prove that the matri, A i Solution: i Let A i i A i i i A A i i i i i i i i 0 0 I 0 0 AA I i Hence A is Unitar. Check Your Progress: () Show that the following matrices are Skew Hermitian. i 4i i i (i) A 4i 6 (ii) A i i 5i 6 0 i 5i i () Show that the following matrices are Unitar matrices.

78 A i i i i 75 (i) A i () If A is Hermitian matri, then show that ia is Skew- Hermitian matri. 4.8 LET US SUM UP In this chapter we have learn Cale Hamilton theorem & it application like Higher power of matri & Inverse of matri. Minimal.polnomial & derogator & non-derogator matri. Comple matri. Hermitian matri. i.e A A Skew Hermitian matri. i.e A Unitar matri= AA I. 4.9 UNIT END EXERCISE. Show that the given matri A satisfies its characteristics equation. i) A 0 0 A i (ii) ii) iii) A A Using Cale Hermitian theorem find inverse of the matri A. i) ii) A A 4 4

79 76. Calculate 5 A b Cale Hamilton Theorem if A 4 4. Let A 6. Find a similarit transformation that 0 diagonalises matri A. 5. Let A 6. Find matri P such that is diagonal matri 6. Diagonalise the matri 0 7. For the matri A Determine a matri P such that is diagonal matri. 8. If show that is Hermitian matri. 9. Show that the following matri are skew Hermitian matri. i) ii) A i 4 i i 0 i i i 0 6i i 6i 0 0. Show that the following matri are unitar matri i) A i i i i

80 MEAN VALUE THEOREMS Unit Structure.0 Objectives. Introduction.. Theorem: Some Important Deductions from the Mean Value Theorem:. Theorem:..... Summar.4 Unit End Eercise.0 OBJECTIVES: After going through this chapter ou will be able to: State and prove. INTRODUCTION: The Mean Value Theorem is one of the most important theoretical tools in Calculus. Let us consider the following real life event to understand the concept of this theorem: If a train travels 0 km in one hour, then its average speed during is 0 km/hr. The car definitel either has to go at a constant speed of 0 km/hr during that whole journe, or, if it goes slower (at a speed less than 0 km/hr) at a moment, it has to go faster (at a speed more than 0 km/hr) at another moment, in order to end up with an average speed of 0 km/hr. Thus, the Mean Value Theorem tells us that at some point during the journe, the train must have been traveling at eactl 0 km/hr. This theorem form one of the most important results in Calculus. Geometricall we can sa that MVT states that given a continuous and differentiable curve in an interval [a, b], there eists a point c [a, b] such that the tangent at c is parallel to the secant joining (a, f(a)) and (b, f(b))... If f is a real valued function such that (i) f is continuous on [a, b], (ii) f is differentiable in (a, b) and (iii)f(a) = f(b) then there eists a point c (a, b) such that

81 : Fig. We know that is the slope of the tangent to the graph of f at = c. Thus the theorem simpl states that between two end points with equal ordinates on the graph of f, there eists at least one point where the tangent is parallel to the X ais, as shown in the Fig.. After the geometrical interpretation, we now give ou the algebraic interpretation of the theorem. Algebraic Interpretation of Rolle's Theorem: We have seen that the third condition of the hpothesis of Rolle's theorem is that f(a) = f(b). If for a function f, both f(a) and f(b) are zero that is a and b are the roots of the equation f() = 0, then b the theorem there is a point c of (a, b), where = 0, which means that c is a root of the equation = 0. Thus Rolle's theorem implies that between two roots a and b of f() = 0 there alwas eists at least one root c of = 0 where a < c < b. This is the algebraic interpretation of the theorem. Eample : () in () in Solution: () Let f, As f is a polnomial in, it is continuous and differentiable everwhere on its domain. Also f f We ma have to find some c such that Now f and lies in ) Let f is polnomial in. f() is continuous and differentiable everwhere on its domain. i.e. (i) f is continuous on [, ] and (ii) f is differentiable in (, ). But we have and f 9 which are not equal. The values of f at the end points are not equal i.e. The function

82 Eample : in Solution: Given in (i) is continuous in since it is a product of continuous functions. (ii) = = eists in (-, 0) (iii) There eists such that c =, - = 0 c = - is the required value. Eample : = in ; a, b > 0 Solution: f is continuous in (a, b) and = eists, since it is not indeterminate or infinite. Also There eists such that (i.e.), which lies in (a, b). Eample 4: = in Solution: Since. e, sin, cos are continuous and differentiable functions, the given functions is also continuous in and differentiable in Also,

83 = = 0 Now, = c /, which lies in Eample 5: f() =,. Solution: We have f() =, Since is continuous and differentiable on [0, ], is also continuous and differentiable in the given domain. Now = = 0 The derivative of f should vanish for at least one point such that Now,. Since c = lies in verified. Eample 6: If,, are differentiable in, show that there eists a value c in (a, b) such that = 0 Solution: Consider the function defined b, F() = Since f,, are differentiable in, is also differentiable in a, b. Further, and since in each case, two rows of the above determinant becomes identical. c a, b such that i.e = 0 Eample 7: If = then show that has three real roots in [, 0].

84 Solution in three intervals,, We observe that (i) is continuous in all the intervals since it is a polnomial in. (ii) is differentiable in all the intervals polnomial in. (iii) has three real roots. Eample 8: Prove that between an two real roots of the equation, there is at least one roots of. Solution: Let a and b be two real roots of the equation (i.e.) of sin e (i.e.) of Let, which is continuous and differentiable. Also,. Since a and b are roots of. c between a and b such that Now, = c is a root of ling between a and b. Eample 9: has a root between 0 and. Solution: Let f = which is obtained b integrating the given equation. Here is continuous in and differentiable in (0, ) and such that Now, and this is zero at = c which means the equation, has a root between 0 and.

85 Eample 0: Show that the equation where has eactl one real root. Solution: Let Since and is a polnomial, it is continuous. Thus, using Intermediate value theorem, we get, there is a number c between 0 and such that Thus the given equation has a root. Now, let if possible have two roots, sa a and b. Then =. Since on represents a polnomial, it is differentiable on (a, b) and continuous c between a and b such that But, Hence for an, which is a contradiction. Thus, the equation The equation = 0 cannot have two real roots. has eactl one root. Check Your Progress. Theorem for the function f defined on the intervals as given below: a) on b) on c) on d) on e) on f) in. Prove that the equation has at least one root between and.. f = in, sin 4. f = in 0, e 5. Show that 4 0 has eactl one real solution.

86 Ans:() () () c (4) c (5) c.. Theorem 6. : If is a real valued function defined on, such that, (i) is continuous on a closed interval, (ii) is differentiable in (a, b) then there eists at least one point such that.. : If (i) is continuous in the closed interval, (ii) is differentiable in the open interval (a, a + h) then there eists at least one number in (0, ) such that,..4 Theorem: Let A a, f a and B b, f b be two points on the curve. The slope m of the line AB is given b, m = Also, is the slope of the tangent at the point C there eists at least one point C c, f c on the graph where the slope of the tangent line is same as the slope of line AB. (i.e.) C is a point on the graph where the tangent is parallel to the chord joining the etremities of the curve. Phsical Significance: We note that is the change in the function as changes from a to b, so that is the change rate of change of the function over. Also is the actual rate of change of the function for c. Thus the theorem states that the average rate of change of a function over an interval is also the actual rate of change of the function at some point of the interval...5 Some Important Deductions from the Mean Value Theorem: Definitions:- (i) Monotonicall increasing function:

87 Let be defined in. Let such that. If (ii) then Monotonicall decreasing function: is said to be a monotonicall increasing function. Let be defined in. Let such that. If Note: then is said to be a monotonicall decreasing function. (i) If is monotonicall increasing in then we can write its maimum value. for all. is its minimum value and is (ii) If is monotonicall decreasing function in then we can write, for all. is its maimum value and is its minimum value. (iii) Let be differentiable in an interval (a, b). Let and, we get or (*) () Let for ever value of in then from equation (*) for and both are positive i.e. We have thus proved: A function whose derivative is positive for ever value of in an interval is a monotonicall increasing function of in that interval. () Let for ever value of in (a, b) from equation we have, for is positive and negative. Hence is a decreasing function of. We have thus proved: A function whose derivative is negative for ever value of in an interval is a monotonicall decreasing function to in that interval. Eample : Verif mean value theorem for on Solution: The given function is on We know that is continuous on and differentiable on (, e). such that

88 Since loge =, log = 0 and we get which lies in the interval (, ) and hence in (, e), since < e <. Eample : Separate the interval in which the polnomial is increasing or decreasing. Solution: We have, = (i) is an increasing function if i.e. i.e. But if ( and ) or ( and ) i.e. if ( and ) or ( and ) i.e. if or < Hence is an increasing function if lies in or (ii) is a decreasing function if But i.e. i.e.if if ( and ) or ( and ) i.e. if ( and ) or ( and ) i.e. if and since and is impossible. i.e. if is decreasing in (, ) Thus is increasing in and and is decreasing in (, ). Eample : Find the interval in which is increasing or decreasing. Solution: We have (i) is an increasing function if i.e. if 0. i.e. if 0. i.e. if or

89 Hence is increasing in the interval and (ii) is a decreasing function if i.e. i.e if <. i.e. if Hence is decreasing in (-, ). Eample 4: Show that if 0, for 0. Solution: Let us assume, for all 0 ecept at and is an increasing function in increasing from 0 and hence., for Consider, for ecept at when it is zero. is an increasing function in increasing from 0 and hence for From (i) and (ii), for

90 Eample 5: Solution: Show that Let f b b f a a f ' c for But, ( c is positive) Eample 6: Show that where and 0 Solution: Let = For we have, putting, a = 0 and h =. = But, But, Eample 7: e, determine in terms of a and h. Hence deduce that, Solution: Let

91 Now, But, Now b substituting h = in the above equation, we get, Eample 8: Show that, Hence show that Solution: Let in where () Since () From () and () b a tan b tan b a For the second part; a b a () Since we put a = and in ()

92 Eample 9: Prove that, for Hence deduce that Solution: Let in Since is (i) continuous in and (ii) differentiable in (a, b) But such that () But a c b, a c From () and () we get, b () For the second part a =, b = 4. Check Your Progress. Eamine the validit of the conditions and the conclusions of LMVT for the functions given below: (i) e on (ii) on (iii) in (iv) on

93 . Appl LMVT to the function Log in and determine in terms of and h. Hence deduce that:.. Appling LMVT show that: (i) for 0 (ii) for (iii) Prove that, Hence deduce that, (i) (ii) 5. Separate the intervals in which the following polnomials are increasing or decreasing. (i) (ii) [ 6. Show that, for. 7. If then show that,. If functions f and g are (i) continuous in a closed interval [a, b], (ii) differentiable in the open interval (a, b) and (iii) (a, b) then for some, ] for an point of the open interval i.e... If two function and are derivable in a closed interval [a, a + h] and for an in (a, a + h) then there eists at least one number such that, Th mean value theorem. Remark:

94 (i) Taking of Cauch mean value theorem. (ii) Usefulness of this theorem depends on the fact that and are f g for some.. : Geometricall, we consider a curve whose paramedic equations are. Then, slope of the curve at an point is, Also the slope of the chord joining the end points f b f a B g b, f b is given b, g b g a Thus under the assumption of Cauch mean value theorem. If that the tangent to the curve at A g a, f a and is parallel to the chord AB. such Eample 0: and in the interval [, ]. Solution: Let and let. As and are polnomials (i) the are continuous on [, ], (ii) the are differentiable on (, ) and (iii) for an value in (, ) If such that, 9c 4 Cauch mean value theorem is verified. Eample : Using CMVT show that Solution: Let and Here, and are continuous on [a, b] and differentiable on (a, b) and for an c in (a, b), thus CMVT can be applied. c a, b such that,

95 Eample : If in CMVT we write and show that c is the arithmetic mean between a and b. Solution: Now and If can be proved that function and are continuous on an closed interval [a, b] and differentiable in (a, b). Also and Then CMVT can be applied. such that, Now and where c e e a b Thus, c is the arithmetic mean between a and b. Eample : Using CMVT prove that there eists a number c such that 0 a c b and. B putting deduce that n n b log b. Solution: Let be a continuous and differentiable function and Then and satisf the condition of continuit and differentiabilit of CMVT. Hence c a, b such that, If f n and then b putting a = we get in the interval (, b) where c b

96 [ as ] Eample 4: If a b, show that there eists c satisfing a c b such that Solution: We have to prove that, This suggests us to take and Now, and are continuous on [a, b] and differentiable on (a, b) and b). CMVT can be applied. such that, for an c in (a, Check Your Progress. Find c (i) f, g, [ a, b], a 0 (ii) on [ 0, ] (iii) f, g on 4. (iv) f e, g e on [0, ] (v) f e, g, [, ] [Ans :- (i) ab (ii) 4 (iii) 5 (iv) (v) 0.]. Summar In this chapter we have learnt about the mean value theorems. The which is the fundamental theorem in analsis has been proved. based on these theorems have been done in order to understand the Mean Value theorems. In the net chapter we are going t its applications..4 Unit End Eercise: i) f ( ) ( )( )( ) in [-,] ii) f ( ) ( ) in [0,] iii) f ( ) tan in [0, ] iv) f ( ) 4 in [-, ]

97 . Verif LMVT for the following functions. i) f ( ) in [-, ] ii) f ( ) ( )( 4)( ) in [0, 7] iii) f ( ) ( ) in [0, ] i) f ( ), g( ) in [,] ii) f ( ) 4, g( ) in [0,] iii) f ( ) ( ) 4, g( ) in [0,]

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126 Gamma & Beta Functions I. Gamma Function Definition Γ(n) = 0 n- e - d ; n > 0 & Γ(n) = Γ(n+) / n ; n Є R - Z 0 Results: () Γ(n+) = n Γ(n) ; n > 0, where Γ() = () Γ(n+) = n! ; n Є N ( convention: 0! = ) () Γ(n) Γ(- n) = π /sin(nπ) ; 0 < n < In Particular; Γ(/) = π

127 Eamples: Eample() Evaluate 0 4 e - d Solution 0 4 e - d = 0 5- e - d = Γ(5) Γ(5) = Γ(4+) = 4! = 4()()() = 4 Eercise Evaluate 0 5 e - d Eample() Evaluate 0 / e - d 0 / e - d = 0 /- e - d = Γ(/) / = ½ + Γ(/) = Γ(½+ ) = ½ Γ(½ ) = ½ π Eercise Evaluate 0 / e - d

128 Eample() Evaluate 0 / e - d 0 / e - d = 0 5/- e - d = Γ(5/) 5/ = / + Γ(5/) = Γ(/+ ) = / Γ(/ ) = /. ½ Γ(½ ) = /. ½. π = ¾ π Eercise Evaluate 0 5/ e - d Eample(4) Find Γ(-½) (-½) + = ½ Γ(-/) = Γ(-½ + ) / (-½) = - Γ(/ ) = - π

129 Eample(5) Find Γ(-/) (-/) + = - ½ Γ(-/) = Γ(-/ + ) / (-/) = Γ(-/ ) / (-/) = ( - π ) / (-/) = 4 π / II. Beta Function Definition B(m,n) = 0 m- ( ) n- d ; m > 0 & n > 0 Results: () B(m,n) = Γ(m) Γ(n) / Γ(m+ n) () B(m,n) = B(n,m) () 0 π/ sin m-. cos n- d = Γ(m) Γ(n) / Γ(m+ n) ; m>0 & n>0 (4) 0 q- / (+). d = Γq) Γ(-q) = Π / sin(qπ) ; 0<q< 4

130 Eamples: Eample() Evaluate 0 4 ( ) d Solution 0 4 ( ) d = 5- ( ) 4- d = B(5,4) = Γ(5) Γ(4) / Γ(9) = 4!.! / 8! =!/( ) = / (8.7.5) = /80 Eercise Evaluate 0 ( ) 6 d Eample() Evaluate I = 0 [ / [ ( )] ] d Solution I = 0 -/ ( ) -/ d = 0 / - ( ) / - d = B(/,/) = Γ(/) Γ(/) / Γ() Γ(/) Γ(/) = Γ(/) Γ(- /) = π /sin(π/) = π / ( /) = π / 5

131 Eample() Evaluate I = 0. ( ) d Solution I = 0 / ( ) d = 0 / - ( ) - d = B(/, ) = Γ(/) Γ() / Γ(7/) Γ(/) = ½ π Γ(5/) = Γ(/+ ) = (/) Γ(/ ) = (/). ½ π = π / 4 Γ(7/) = Γ(5/+ ) = (5/) Γ(5/ ) = (5/). (π / 4) = 5 π / 8 Thus, I = (½ π ).! / (5 π / 8) = 4/5 II. Using Gamma Function to Evaluate Integrals Eample() Evaluate: I = 0 6 e - d Solution: Letting =, we get I = (/8) 0 6 e - d = (/8) Γ(7) = (/8) 6! = 45/8 Eample() 6

132 Evaluate: I = 0 e ^ d Solution: Letting =, we get I = (/) 0 -/ e - d = (/) Γ(/) = π / Eample() Evaluate: I = 0 m e k ^n d Solution: Letting = k n, we get I = [ / ( n. k (m+)/n ) ] 0 [(m+)/n ] e - d = [ / ( n. k (m+)/n ) ] Γ[(m+)/n ] II. Using Beta Function to Evaluate Integrals Formulas () 0 m- ( ) n- d = B(m,n) = Γ(m) Γ(n) / Γ(m+ n) ; m > 0 & n > 0 () 0 π/ sin m-. cos n- d = (/) B(m,n) ; m>0 & n>0 (4) 0 q- / (+). d = Γ(q) Γ(-q) = Π / sin(qπ) ; 0 < q < Using Formula () 7

133 Eample() Evaluate: I = 0 / ( ). d Solution: Letting =, we get I = (8/) 0 ( ) -/ d = (8/). B(, / ) = 64 /5 Eample() Evaluate: I = 0 a 4 (a ). d Solution: Letting = a, we get I = (a 6 / ) 0 / ( ) / d = (a 6 / ). B(5/, / ) = a 6 / Eercise Evaluate: I = 0 (8 ). d Hint Lett = 8 Answer 8

134 I = (8/) 0 -/ ( ) /. d = (8/) B(/, 4/ ) = 6 π / ( 9 ) Using Formula () Eample() Evaluate: I = 0 d / ( + 4 ) Solution: Letting 4 =, we get I = ( / 4) 0 -/4 d / ( + ) = ( / 4). Γ (/4). Γ ( - /4 ) = (/4). [ π / sin ( ¼. π ) ] = π / 4 Using Formula () Eample(4) a. Evaluate: I = 0 π/ sin. cos d b. Evaluate: I = 0 π/ sin 4. cos 5 d Solution: a. Notice that: m - = m = & n - = m = / I = ( / ) B(, / ) = 8/5 b. I = ( / ) B( 5/, ) = 8 /5 Eample(5) 9

135 a. Evaluate: I = 0 π/ sin 6 d b. Evaluate: I = 0 π/ cos 6 d Solution: a. Notice that: m - = 6 m = 7/ & n - = 0 m = / I = ( / ) B( 7/, / ) = 5π / b. I = ( / ) B( /, 7/ ) = 5π / Eample(6) a. Evaluate: I = 0 π cos 4 d b. Evaluate: I = 0 π sin 8 d Solution: a. I = 0 π cos 4 = 0 π/ cos 4 = (/) B (/, 5/ ) = π / 8 b. I = I = 0 π sin 8 = 4 0 π/ sin 8 = 4 (/) B (9/, / ) = 5π / 64 Details 0

136 I. Eample() Evaluate: I = 0 6 e - d = / 6 = 6 /64 d = (/)d 6 e - d = 6 /64 e. (/)d Eample() I = 0 e ^ d = / = /6 d=(/) -/ d e ^ d = /6 e. (/) -/ d Eample() Evaluate: I = 0 m e k ^n d = k n = /n / k /n m = m/n / k m/n d = (/n) (/n-) / k /n d m e k ^n d = ( m/n / k m/n ). e. (/n) (/n-) / k /n d m/n + /n = (m+)/n - -m/n /n = - (m+)/n I = [ / ( n. k (m+)/n ) ] 0 [(m+)/n ] e - d II. Eample()

137 Eample() I = 0 / ( ). d = d=d = 4 ( ) = ( ) = ( ) / ( ). d = 4 / ( ) d =0 when =0 = when = Eample() Evaluate: I = 0 a 4 (a ). d = a, we get 4 = a 4 = a / d= (/)a -/ d (a ) = (a a ) = a ( ) / 4 (a ). d = a 4 a ( ) / (/)a -/ d =0 when =0 = when =a Eample() I = 0 d / ( + 4 ) 4 =

138 = /4 d= (/4) -/4 d d / ( + 4 ) = ( / 4) -/4 d / ( + ) Proofs of formulas () & () Formula () We have, B(m,n) = 0 m- ( ) n- d Let = sin Then d = sin co d & m- ( ) n- d = (sin ) m- ( cos ) n- ( d / sin co ) When =0, we have = 0 When =, we hae = π/ Thus, I = 0 π/ sin m-. cos n- d = sin m-. cos n- d I = 0 π/ sin m-. cos n- d = B(m,n) / Formula ()

139 We have, I = 0 q- / (+) d Let = / (+) Hence, = / -, + = + ( / -) = /(-) & d = - [ (- ) (-)] / (-). d= / (-). d whn = 0, we have = 0 when, we have = lim / (+) = Thus, I = 0 [ q- / (+) ] d = 0 [ ( / - ) q- / (/(-)) ]. / (-). d = 0 [ q- / (-) -q ] d = B(q, -q) = Γ(q) Γ(-q) Proving that Γ(/ ) = π Γ(/) = 0 /- e - d = 0 -/ e - d Let = ½ = d = d Γ(/) = 0 - e ^ d = 0 e ^ d 4

140 = (π / ) = π 5

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157 0 DIFFERENTIAL EQUATIONS UNIT STRUCTURE 6. Objective 6. Introduction 6. Differential Equation 6.4 Formation of differential equation 6.5 Let Us Sum Up 6.6 Unit End Eercise 6. OBJECTIVE After going through this chapter ou will able to i. Define differential equation ii. Order & degree of differential equation iii. Formulate the differential equation 6. INTRODUCTION We have alread learned differential equation in XII th. Hence we are going to discuss differential equation in brief. In this chapter we discuss onl formulation of differential equation. 6. DIFFERENTIAL EQUATION Definition:- An equation involving independent and dependent variables and the differential coefficients or differentials is called a differential equation. e.g. d =9 d =independent variable = depedent variable d +=0 d n d d n d d =0

158 04 These are all eamples of differential equations. The differential equation is said to be ordinar if it contains onl one independent variable. All the eamples of above are of ordinar differential equations. Order and Degree of a Differential Equations:- (i) Order:- The order of the differential equations is the order of the highest orderal derivatives present in the function or equation. If = f () is a function, then d is the first order derivative, d d d d is the second order derivative. d d d e.g ) d Order = d d 0 d ) E = Ri + L di dt Order = Degree:- The degree of differential equation is the degree of the highest ordered derivative in the equation when it is made free from radicals and fractions. e.g. 4 d k 0 d order =, degree = d d Y 0 d d Order =, degree = d d d d Order=, degree= d d d d d d Cubing both sides d d

159 05 d d d d Squaring both sides d d d d Order=, degree= Solved eamples: Eample : Find the order and degree of the following i) e Solution: d d d d d e d d d Squaring both sides ii) d d d d d e d d order =, degree = sin( ) e d Solution: iii) 4 d d d sin( ) 4 d d d Order = 4, degree= d 5 d d d e Solution: d d 5 d d Order =, degree=

160 06 iv) Solution: d d 5 d d Squaring on both sides Order =, degree= d 5 d d d d - 5 d d d d d 5 d d d d Check our progress: ) u u Ans : order =, degree= ) ) u 4 7 d d d d 5 7 d d d d Ans : order=, degree=7 4 e Ans : order=, degree= 4) Ans : order=, degree= 5) Ans : order=, degree= Ans : order =, degree= e 6.4 FORMATION OF DIFFERENTIAL EQUATION Formation of differential equation involves elimination of arbitrar consonants, in the relation of the variables. Consider

161 07 a Where = independent variable = dependent variable Differentiating equation () with respect to we have d d a () From equation () we have a Put value of a in equation (), we have d d d d d d d d 0 This is the required differential equation Note:- To eliminate two arbitrar constants, three equations are required. To eliminate three arbitrar constants, four equations are required. In general to eliminate n arbitrar constants. (n+) equations are required. In other words elimination of n arbitrar consonants will bring us to differential equation of nth order. Solved Eamples:- Eample : Form the differential equations if c cos c sin Solution: We have Y= c cos + c sin () This equation contains two arbitrar constants, therefore we shall require three equations to eliminate c and c. Differentiating equation () with respect to d = -c cos c cos. d Again differentiate with respect to d = -c cos c sin d

162 08 d = (c cos +c sin ) d d = d from eq ----() d + = 0 d This is the required differential equation. Eample : Form the differential equation from = a sin (wt+c) where a and c are arbitrar constants. Solution: We have, = a sin (wt+c) () Differentiate equation () with respect d + acos (cot+c) w dt d + aw cos(cot+c) dt Again differentiating w.r.t. 't' d dt d dt d dt d -a wsin cot c w -w a sin cot c - w... u sin g equation w 0 dt This is the required differential equation Eample 4: From the differential equation if = log (a) Solution: log( a ) () Differentiate equation () with respect to. d d A A d d d d This is the required differential equation.

163 09 Eample 5: Obtain the differential equation for the equation Y=c+ c Solution: we have, c c Differentiate equation () with repect to d c d Put value of c in equation () d d = d d This is the required differential equation. Eample 6: Obtain the differential equation for the relation = a e b e Where a,b are constants. Solution: we have, () = a e b e (). Here the number of arbitrar constants is two Hence we shall require three equations to Eliminate and b. So we differentiate the given equations twice. d = a e b e (). d d = 4a e 9 b e () d From equation () () & () elimination of a & b gives directl d d d d In the determinant st column is LHS Column nd column nd column contains coefficients Epanding the determinant nd column contains coefficients of b e d d d d d d d d a e

164 0 d d d d This is the required differential equation. Eample 7: Find the differential equation of all circles touching ais at the origin and centers on -ais Solution: The equation of such a circle is i.e. a a a a a a 0 () Where a is the onl arbitrar contents Differentiate equation () with respect to We have d a 0 d d d 0 d d 0 d d - 0

165 d 0 d Which is the require differential equation. Check Your Progress: ) Form the differential equation of all circles of radius a. d Ans. d d a d ) Obtain the differential equation whose general solution is given b e Acos Bsin Ans d d 0 d d ) Find the differential equation whose general solution is given b c e c e c e Ans d d d d d d ) Obtain the differential equations for the following: i) A e B e Ans d d 5 d d 6 0 t t ii) s ce c e Ans d s ds dt dt s 0 iii) A cos t+ B sin t Ans d dt 4 0 iv) a b Ans d d 4 d d 6 o v) Acos( nt Ams vi Soln Vii d n 0 dt Y A B C d d 0 Y sin c

166 d Soln cos d Viii ( c c ) e Ans d d 0 d d 6.5 LET US SUM UP In this chapter we have learn Equation in term d of is called differential equation. d Degree & order of differential equation. Formation of differential equation while removing arbitrar constant likes A&B,&C. i. ii. iii. iv. 6.6 UNIT END EXERCISE ) Find the order 7 degree of Differential equation given below d d d d d d i d d 5 d d d k d d d d d 0 d d d v. d d d ) Formulate the differential equation i. Y A Blog ii. X = asin(w++c) iii. iv. v. vi. vii. viii. i. Y = c (Acos+Bsin) mcos Y = e Y = a b Y = c+c c X +Y = a Y = 4a e +Ce =

167 . Y = ACos+ BSin ***** SOLUTION OF DIFFERENTIAL EQUATION UNIT STRUCTURE 7. Objectives 7. Introduction 7. Solution of Differential equation 7.4 Solution of Differential Equation of first order and first degree 7.5 Let Us Sum Up 7.6 Unit End Eercise 7. OBJECTIVES After going through this chapter ou will able to Find general & particular solution of differential equations. Classification of differential equation. Appl particular method first find the solution of differential equation. 7. INTRODUCTION We have alread formed differential equation in previous chapter. Here we are going to find solution of differential equation with different method. It is ver useful in different field. 7. SOLUTION OF DIFFERENTIAL EQUATION General Solutions:- The general Solution of a differential equation is the most general relation between the dependent and the independent variable occurring in the equation which satisfies the given differential equation. Particular Solutions:-

168 4 An particular solution that satisfies the given equation is called a particular solution e.g. d 5 d d=5d Integrating both sides we get d=5 d constant Y= 5+C This is called as general solution Suppose C=7 is given Then particular solution is given b putting of c in the general solution = 5+7 Check Point:- ) Find the general solution and particular solution of the differential equation d When 4 at 0 d Solution: = c 4 Differential equations of first order and of First Degree :- An equation of the form, d M+N 0 d differential equation of first order and first degree. This equation can also be written as Md Nd 0 called 7.4 SOLUTION OF DIFFERENTIAL EQUATION OF FIRST ORDER AND FIRST DEGREE There are man methods that can be used to solve the differential equations. Important one among those are listed below. ) Variable seperable form. ) Equations reducible to variable seperable form. ) Homogeneous equations. 4) Eact differential equations. 5) linear differential equations. 6) differential equation)

169 7) Methods of substitution. 5 We will eplain all these methods one b one in detail Variable Separable form:- Working Rule ) Consider the differential equation Md+ Nd=0 ) If possible rearrange the terms and get f() d +g().d=0 ) Integrate and write constant of integration in suitable form, usuall C. 4) Simplif if possible. Solved Eamples:- Eample : Solve tan d e Sec. d 0 Solution: tan d e Sec. d 0 throughout b -e tan we get e sec d -e tan This is in variable separable form Integrate equation (), we get d 0 ) e Sec d d constant e tan e Sec - e tan d d c -log e log tan log log e log tan log log e - tan e - Removing log both side - = c tan = c e - tan = logc This is the general solution of a given differential equation. d Eample : Solve d c c

170 6 Solution: d d d = + + d d = + + d d = + d This is in variable separable form Integrate equation () d d c f \ d = f +c f n f f d f = n+ n+ + = + +c + = + +c This is in required general solution. d Eample : Solve e d Solution: The given equation is d e d d = d - e - + e e - + d = d e - e d d e e 0= d + e d

171 7 d e e This is in variable separable form, Integrate equation (), we get d 0 () e d e d =log c log log e log c log + e log c e c This is the required general solution. Eample 4: Solve e tan d e sec d 0 given when =0 4 Solution: The given equation is e tan d e sec d 0 through out b +e tan e sec d +e tan This is in variable separable form, Integrate equation () we get e +e tan d sec d d log c e sec d d log c +e tan log e log tan=log c 0 () log +e log tan=log c log +e tan log +e tan =c () This is the required general solution To final particular solution:- put = at =0 in equation () 4 c