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UNIT-I
SOLUTION FOR LINEAR SYSTEMS
Elementary row(and column)transformations
Rank of a matrix-Echelon form-Normal form
Solution of linear systems-Direct methods
L-U-Decomposition
L-U-Decomposition from Gauss elimination
Solution of Tridiagonal system
Summary
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Summary1. Rank of a matrix: The rank of a matrix is the order r of the largest no
vanishing minor of the matrix.
2.Elementary transformation of a matrix:a)Row Transformations :
i) Interchange of ith and jth row .R ijii)Multiplication of ith row by non-zero scalar l.R i(l)iii)Addition of l times the elements of jth row to corresponding elements of
ith row---R ij(l)b)Column Transformations are similar to above to (a)---replace R by
3.Computation of Rank of matrix:Method I:Echelon Form: Transform the given matrix to an echelon form usingelementary transformations. The rank of the matrix is equal to the nuof non-zero rows of echelon form.
Method II:Canonical Form OR Normal Form:Reduce the given matrix A to one of the normal formsIr 0 , Ir , Ir 0 or Ir , using elementary transformation,0 0 0 Then Rank of A = r
4.Simulataneous Linear Equations-Methods of Solution.
1.System for m linear equations in n unknowns can be written in a mform as AX =B,where A = a11 a12 a1n , X = x1 , B = b1
a21 a22 a2n x2 b2.. .. . .
am1 am2 amn xn bnIf bi = 0 ,the system is homogeneous i.e. B=0 ,otherwise ,it isnon- homogeneous.
2.Condition for the consistency: A system of linear equations AX= Bconsistent iff the rank of A is equal to the rank of the augmented ma[A/B]
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3.Solution of AX= B Working rule:
i)Find r(A) and r(A/B)by applying elementary transformations.ii)Ifr(A)=r(A/B)=n ,(n,being number of unknowns).The system isconsistent and has unique solution.[In case A 0]
iii)If r(A)=r(A/B)
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7.Tri-Diagonal matrix:
Matrices of the type a11 a12 0 0 are Tri Diagonal matricesa21 a22 a23 00 a32 a33 a340 0 a43 a44
8.Solution of Tri diagonal system procedure is like above method.
9.Homogeneous linear equations:AX=0 , where A =(aij)mxn:X=[x1,x2,,.xn]1: solution of AX=0 can be done by elementarytransformations.Conclusions:
i)The system AX=0 is always consistent. Since the trival solution x1= x2=x3==xn=0 always exist.
ii)If rank of (A/B) = rank of A = n , [A 0] , then the trival solutthe only solution.
iii)If the rank of (A/B)= rank of A = r < n, [A 0] the solution has
infinite number of non-trival solutions involving (n-r) arbitraryconstants.
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UNIT-II
EIGEN VALUES & EIGEN VECTORS& THEIR APPLICATIONS
Eigen values,eigen vecors - Properties
Cayley-Hamilton theorem-Inverse & powers of matrix by CayleyHamilton theorem
Diagonalization of a matrix
Calculations of powers of a matrix-modal & spectral matrices
Summary
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Summary
1. Eigen values & Eigen vectors : Let A = (aij)mxn
(a)Characteristic equation of A is given by A- I =0(b)Roots of this equation are 1, 2 , 3,.., n. They arecalled Eigen values of A.
(c)A non-zero vector X = [x1,x2,x3,..xn]1 which satisfiesthe relation [A- I]X=0 (or) AX= X, is called the Eigen veof A corresponding to . This each Eigen value has an Eigevector.
2. Properties of Eigen values & Eigen vectors:1.The sum of the Eigen values of a square matrix A is its trac
their product is A.2.The Eigen values of A and AT are equal.3.If A is non-singular matrix and is an eigen values of A,
then 1/ is an Eigen values of A-1.4.If is an eigen values of A , then is an eigen values o
A where is a non-zero scalar.5.If is an eigen values of A, then m is an eigen values of Am,
m being any positive integer.6.The eigen values of a diagonal matrix are its diagonal elem
7.If B be a non-singular matrix, and A,B are matrices of samorder,then A and B-1AB have same eigen values.8. is a characteristic root of a square matrx A iff their exists
non- zero vector X such that AX= X.9.If X be an eigen vector of A corresponding to the eigen va
,then c X is also an eigen vector of A corresponding to , being a non-zero scalar.
10.If X is an eigen vector of a square matrix A, then X cannocorresponding to more than one eigen values of A.
11.Zero is an eigen value of a matrix iff it is singular.12.If is an eigen value of a non-singular matrix A, thenA / is an eigen values of Adj A.
3. Cayley Hamilton Theorem-State thatEvery square matrix satisfies its own characteristic equatio
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4. To find the inverse of a square matrix A; by using C-H theorLet A be a square matrix and and n+a1 n-1+a2 n-2++an= 0.(1)be the characteristic
equation of A. (ai,i=1 to n are constants).Then C-H theorem gives An +a1An-1+anI=0(2)(2) x A-1 = An-1+a1An-2+an-1I+anA-1 =0
A-1 = (-1)/an[An-1+An-2+an-1I]
5. To find positive integral powers of A using C-H theorem:Let m n, be a positive integer.Then Am-nx (2) ==Am+a1Am-1+..anAm-n= 0,from which we canfind Am interms of powers of lower order of A.
6. Diagonalization of a square matrix:Let A be a square matrix of order n having n linearlyindependent Eigen vectors. Then there exists a non singular matrix P such that P-1AP =D is a diagonal matrix, andD=Diag[ 1, 2,.., n ]
7. Working rule to diagonalise A=(aij)nxnStep1:Find Eigen values i (i=1,2,.,n)of A.Step2:Find Eigen vectors Xi corresponding to i ( i,i=1,2,.,n are
distinct).Step3:Form the matrix P=[X1 X2 X3 Xn]where columnvectors Xi are the Eigen vectors of i.(The matrix P is known as the Modal matrix of A)
Step4:Find D= P-1AP=Diag[ 1 2 3 n].This is thediagonalisation of A.The matrix D is known as the Spectral matrix of A.Computation of positive powers of A:If m is a positive integer of A: Then,
Am=(PDP-1)m= [P DmP-1]
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UNIT-III
LINEAR TRANSFORMATIONS
Real Matrices Symmetric, Skew-symmetric, Orthogonal
Linear transformations- Orthogonal Transformation Complex Matrices- Hermitian, Skew-Hermitian and Unitary
Eigen Values and Eigen Vectors of Complex matrices and Their Properties
Quadratic forms- Reduction to Canonical Form
Rank- Positive, Negative Definite; Semi definite Index, SignatuSylvester Law
A Summary
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Summary1. Definitions: and properties of some real and complex matrices are
following
2. Properties of eigen values of real and complex matrices are given
1.If is a characterstic root of an orthogonal matrix, then 1/ is alcharacterstic root.
2.The eigen values of an orthogonal matrix are of unit modulus.3. The eigen values of a hermitian matrix are all real.4. The eigen values of a real symmetric matrix are all real.5. The eigen values of a skew hermitian matrix are either purely
imaginary or zero.6.The eigen values of a real skew symmetric matrix are purelyimaginary or zero.
7. The eigen values of a unitary matrix are of unit modulus.8. If A is nilpotent matrix, then 0 is the only eigen value of A9. If A is involuntary matrix its possible eigen values are 1 and -110.If A is an idempotent matrix its possible eigen values are 0 and
3. Transformations:
(a) The transformations X = AY where A = (aij)nXn; X = [x x2. xn];Y = columns of [y y2. yn]; transforms vector Y to vector X over thematrix A.
The transformations is linear.
(b) Non-singular transformation:(i) If A is non-singular, (A 0 ) then Y = AX is non-singular
transformation.(ii) Then, X = A-1Y is inverse transformation of Y = AX.
(c) Orthogonal transformation: If A is an orthogonal matrix, thenY = AX is an orthogonal transformation;A is orthogonal , A1= A- 1 => Y1Y = X1Xi.e., Y = AX transforms ( x12 + x22 +.+ xn2) to (y12 + y22 +..
+yn2)
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4. Quadratic forms: A homogeneous polynomial of 2nd degree in n
variables x1, x2,xn is called of quadratic form.
Thus , q = aijxix j from i , j = 1 to n(or) q = [a11x12 + a22x22 ++ annxn2 + (a12+a21)x1x2 +(a13+a31)x1x3 ++]
is a quadratic form in n variables x1,x2xn.
5. Matrix of a quadratic form q: If A is a symmetric matrix,
q = X1AX is the matrix representation of q and A is the matrixof q where ,(aij+a ji)=2aij is coefficient of xix j.
[i.e. aij=a ji=1/2 coefficient of xix j]Then q = X1AX = [x1x2.xn] A columns of[x1 x2.. xn]
6. Rank of quadratic form: If q = X1AX, then rank of A is the rank of quadratic form q
(a) If rank of A = r = n , q is non-singular form(b) If r < n , q is singular
7. Canonical form or Normal form of q: A real quadratic form q in
which product terms are missing (i.e. all terms are square terms ois called the canonical form of q.i.e. q = a1x12 + a2x22 + + anxn2 is canonical form.
8. Reduction to canonical form: If D = Diag [d1, d2,.dr ] is thediagonalization of A, then q1= d1x12 + d2x22 + . + dr xr 2 ,(where r = rank of A) is canonical form of q = X1AX.
9. Nature of a quadratic form:1. If q= X1AX is the given quadratic form (in n variables) of rank r
then, q1=d1x12 + d2x22 +.+ dr xr 2 is the canonical for of q.[di is +ve, -ve, or zero]
(a) Index: The number of +ve terms in q1 is called the index s of quadratic form q
(b) The number of non +ve terms = r-s(c) Signature = S- (r-s)= 2s-r.
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2. The quadratic form q is said to be
(a) +ve definite if r=n, and s=n(b) ve definite if r=n, and s=0(c) +ve semi-definite if r
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10.Methods of Reduction of quadratic form to the canonical form.
(a) Lagranges method: A quadratic form can be reduced by th
method to a canonical form by completion of squares.(b) Diagonalization method: Write A=I3AI3 [if A=(aij)3x3] apply
elementary row transformation on L.H.S and on prefactor oR.H.S. Apply corresponding column transformations on L.Has well as the post-factor of R.H.S continue this process till equation is reduced to the form,
D = P1A P , where D is a diagonal matrix D = [d1 0 0][0 d2 0][0 0 d3]
Then the canonical form is q1=y1(P1AP)Y=Diag(d1 d2 d3) whereY = [y1 y2 y3], i.e., if q = X1A X, X = [x1 x2 x3] ,q1= d1y12 + d2y22 + d3y32.
Here X=PY is corresponding transformation.
(C) Orthogonal Reduction of q = X1AX:
(i) Find eigen values i and corresponding eigen vectors Xi,(i=1,2,n) of A.(ii) Find modal matrix B = [X1 X2 Xn](iii) Normalize each column vector Xi of B by dividing it with it
magnitude and write the normalized modal matrix P which iorthogonal (i.e. P1= P-1)
(iv) Then X = PY reduces q to q1 where q1 = 1 y12 + 2 y22 + + n yn2
= Y1(P1AP)Y( X=PY is know as orthogonal transformation)
11.Sylvesters law of inertia: The signature of real quadratic form isinvariant for all normal reductions.
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Symmetric matrix
Inlinear algebra, asymmetric matrix is asquare matrix, A, that is equal to its transpose
The entries of a symmetric matrix are symmetric with respect to themain diagonal(topleft to bottom right). So if the entries are written as A = (a ij), then
for all indicesi and j. The following 33 matrix is symmetric:
A matrix is called skew-symmetricor antisymmetric if its transpose is the same as itsnegative. The following 33 matrix is skew-symmetric:
Skew-symmetric matrix
Inlinear algebra, askew-symmetric (or antisymmetric or antimetric [1]) matrix is asquare matrix A whose transposeis also its negative; that is, it satisfies the equation:
or in component form, if : for all and
For example, the following matrix is skew-symmetric:
Compare this with asymmetric matrixwhose transpose is the same as the matrix
or an orthogonal matrix, the transpose of which is equal to its inverse:
http://en.wikipedia.org/wiki/Linear_algebrahttp://en.wikipedia.org/wiki/Square_matrixhttp://en.wikipedia.org/wiki/Transposehttp://en.wikipedia.org/wiki/Transposehttp://en.wikipedia.org/wiki/Main_diagonalhttp://en.wikipedia.org/wiki/Skew-symmetric_matrixhttp://en.wikipedia.org/wiki/Skew-symmetric_matrixhttp://en.wikipedia.org/wiki/Linear_algebrahttp://en.wikipedia.org/wiki/Skew-symmetric_matrix#cite_note-0http://en.wikipedia.org/wiki/Square_matrixhttp://en.wikipedia.org/wiki/Transposehttp://en.wikipedia.org/wiki/Transposehttp://en.wikipedia.org/wiki/Symmetric_matrixhttp://en.wikipedia.org/wiki/Orthogonal_matrixhttp://en.wikipedia.org/wiki/Orthogonal_matrixhttp://en.wikipedia.org/wiki/Square_matrixhttp://en.wikipedia.org/wiki/Transposehttp://en.wikipedia.org/wiki/Main_diagonalhttp://en.wikipedia.org/wiki/Skew-symmetric_matrixhttp://en.wikipedia.org/wiki/Linear_algebrahttp://en.wikipedia.org/wiki/Skew-symmetric_matrix#cite_note-0http://en.wikipedia.org/wiki/Square_matrixhttp://en.wikipedia.org/wiki/Transposehttp://en.wikipedia.org/wiki/Symmetric_matrixhttp://en.wikipedia.org/wiki/Orthogonal_matrixhttp://en.wikipedia.org/wiki/Linear_algebra8/14/2019 Solution for Linear Systems
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The following matrix is neither symmetric nor skew-symmetric:
Every diagonal matrixis symmetric, since all off-diagonal entries are zero. Similarly,each diagonal element of a skew-symmetric matrix must be zero, since each is its onegative.
Orthogonal matrixInlinear algebra, anorthogonal matrix is asquare matrixwithrealentries whosecolumns (or rows) are orthogonal unit vectors(i.e., orthonormal). Because the columnsare unit vectors in addition to being orthogonal, some people use the termorthonormalto describe such matrices.
Equivalently, a matrixQ is orthogonal if itstransposeis equal to itsinverse:
alternatively,
(OR)
Definition: Ann n matrix A is called an orthogonal matrix whenever T A A I = .
EXAMPLE:1 0 1 0 1 0 cos sin
, , ,0 1 0 1 0 1 sin cos
http://en.wikipedia.org/wiki/Diagonal_matrixhttp://en.wikipedia.org/wiki/Diagonal_matrixhttp://en.wikipedia.org/wiki/Linear_algebrahttp://en.wikipedia.org/wiki/Matrix_(mathematics)#Square_matriceshttp://en.wikipedia.org/wiki/Real_numberhttp://en.wikipedia.org/wiki/Orthogonalhttp://en.wikipedia.org/wiki/Orthogonalhttp://en.wikipedia.org/wiki/Unit_vectorhttp://en.wikipedia.org/wiki/Unit_vectorhttp://en.wikipedia.org/wiki/Orthonormalityhttp://en.wikipedia.org/wiki/Orthonormalityhttp://en.wikipedia.org/wiki/Transposehttp://en.wikipedia.org/wiki/Inverse_matrixhttp://en.wikipedia.org/wiki/Diagonal_matrixhttp://en.wikipedia.org/wiki/Linear_algebrahttp://en.wikipedia.org/wiki/Matrix_(mathematics)#Square_matriceshttp://en.wikipedia.org/wiki/Real_numberhttp://en.wikipedia.org/wiki/Orthogonalhttp://en.wikipedia.org/wiki/Unit_vectorhttp://en.wikipedia.org/wiki/Orthonormalityhttp://en.wikipedia.org/wiki/Transposehttp://en.wikipedia.org/wiki/Inverse_matrix8/14/2019 Solution for Linear Systems
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Conjugate transpose
"Adjoint matrix" redirects here. An adjugate matrix is sometimes called a "classical adjoint matrix".
In mathematics , the conjugate transpose , Hermitian transpose , or adjoint matrix of an m -by-
n matrix A with complex entries is the n-by- m matrix A * obtained from A by taking
the transpose and then taking the complex conjugate of each entry (i.e. negating their imaginary
parts but not their real parts). The conjugate transpose is formally defined by
where the subscripts denote the i , j -th entry, for 1 i n and 1 j m , and the overbar denotes a
scalar complex conjugate . (The complex conjugate of a +bi , where a and b are reals, is a bi .)
This definition can also be written as
where denotes the transpose and denotes the matrix with complex conjugated entries.
Other names for the conjugate transpose of a matrix are Hermitian conjugate , or transjugate .
The conjugate transpose of a matrix A can be denoted by any of these symbols:
or , commonly used in linear algebra
(sometimes pronounced " A dagger "), universally used in quantum mechanics
, although this symbol is more commonly used for the Moore-Penrose
pseudoinverse
In some contexts, denotes the matrix with complex conjugated entries, and thus the
conjugate transpose is denoted by or .
EXAMPLE:
then
http://en.wikipedia.org/wiki/Adjugate_matrixhttp://en.wikipedia.org/wiki/Adjugate_matrixhttp://en.wikipedia.org/wiki/Mathematicshttp://en.wikipedia.org/wiki/Matrix_(mathematics)http://en.wikipedia.org/wiki/Matrix_(mathematics)http://en.wikipedia.org/wiki/Complex_numberhttp://en.wikipedia.org/wiki/Transposehttp://en.wikipedia.org/wiki/Complex_conjugatehttp://en.wikipedia.org/wiki/Complex_conjugatehttp://en.wikipedia.org/wiki/Complex_conjugatehttp://en.wikipedia.org/wiki/Linear_algebrahttp://en.wikipedia.org/wiki/Dagger_(typography)http://en.wikipedia.org/wiki/Dagger_(typography)http://en.wikipedia.org/wiki/Quantum_mechanicshttp://en.wikipedia.org/wiki/Moore-Penrose_pseudoinversehttp://en.wikipedia.org/wiki/Moore-Penrose_pseudoinversehttp://en.wikipedia.org/wiki/Moore-Penrose_pseudoinversehttp://en.wikipedia.org/wiki/Adjugate_matrixhttp://en.wikipedia.org/wiki/Mathematicshttp://en.wikipedia.org/wiki/Matrix_(mathematics)http://en.wikipedia.org/wiki/Complex_numberhttp://en.wikipedia.org/wiki/Transposehttp://en.wikipedia.org/wiki/Complex_conjugatehttp://en.wikipedia.org/wiki/Complex_conjugatehttp://en.wikipedia.org/wiki/Linear_algebrahttp://en.wikipedia.org/wiki/Dagger_(typography)http://en.wikipedia.org/wiki/Quantum_mechanicshttp://en.wikipedia.org/wiki/Moore-Penrose_pseudoinversehttp://en.wikipedia.org/wiki/Moore-Penrose_pseudoinverse8/14/2019 Solution for Linear Systems
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Hermitian matrix
A Hermitian matrix (or self-adjoint matrix ) is a square matrix with complex entries which isequal to its own conjugate transpose that is, the element in the i th row and j th column is equal
to the complex conjugate of the element in the j th row and i th column, for all indices i and j :
If the conjugate transpose of a matrix is denoted by , then the Hermitian property can be
written concisely as
Hermitian matrices can be understood as the complex extension of a real symmetric matrix .
For example,
is a Hermitian matrix
Skew-Hermitian matrix
Inlinear algebra, asquare matrixwithcomplex entries is said to beskew-Hermitian or antihermitian if itsconjugate transposeis equal to its negative.[1]That is, the matrix A isskew-Hermitian if it satisfies the relation
where denotes the conjugate transpose of a matrix. In component form, this mean
for alli and j, wherea i, j is thei, j-th entry of A, and the overline denotes complexconjugation.
Skew-Hermitian matrices can be understood as the complex versions of realskew-symmetric matrices, or as the matrix analogue of the purely imaginary numbers.
[2]
Unitary matrix
Inmathematics, aunitary matrix is ann byn complex matrix U satisfying the condition
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where is theidentity matrixin n dimensions and is the conjugate transpose(alsocalled theHermitian adjoint) of U . Note this condition says that a matrixU is unitary if and only if it has aninversewhich is equal to its conjugate transpose
A unitary matrix in which all entries are real is an orthogonal matrix. Just as anorthogonal matrixG preserves the (real) inner productof two realvectors,
so also a unitary matrixU satisfies
for allcomplex vectors x and y, where stands now for the standardinner producton
.
http://en.wikipedia.org/wiki/Identity_matrixhttp://en.wikipedia.org/wiki/Conjugate_transposehttp://en.wikipedia.org/wiki/Conjugate_transposehttp://en.wikipedia.org/wiki/Hermitian_adjointhttp://en.wikipedia.org/wiki/Inverse_matrixhttp://en.wikipedia.org/wiki/Orthogonal_matrixhttp://en.wikipedia.org/wiki/Orthogonal_matrixhttp://en.wikipedia.org/wiki/Realhttp://en.wikipedia.org/wiki/Inner_producthttp://en.wikipedia.org/wiki/Vectorshttp://en.wikipedia.org/wiki/Inner_producthttp://en.wikipedia.org/wiki/Identity_matrixhttp://en.wikipedia.org/wiki/Conjugate_transposehttp://en.wikipedia.org/wiki/Hermitian_adjointhttp://en.wikipedia.org/wiki/Inverse_matrixhttp://en.wikipedia.org/wiki/Orthogonal_matrixhttp://en.wikipedia.org/wiki/Realhttp://en.wikipedia.org/wiki/Inner_producthttp://en.wikipedia.org/wiki/Vectorshttp://en.wikipedia.org/wiki/Inner_product8/14/2019 Solution for Linear Systems
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UNIT-IV
SOLUTIONS OF NON-LINEAR SYSTEMS
Solution of Algorithm and Transcendental Equations
1. Bisection Method2. Method of False Position3. The Iteration Method4. Newton Raphson Method
Interpolation
- Finite Differences- Forward Differences- Backward Differences- Central Differences
Newtons Forward Interpolation Formula
Newtons Backward Interpolation Formula Gauss Forward Interpolation Formula
Gauss Backward Interpolation Formula
Lagranges Interpolation Formula
Spline Interpolation and Cubic Splines
Summary
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Summary
Solution of algebraic and transcendental equations1. The numerical methods to find the roots of f(x)=0
(i) Bisection method :If a function F(x) is continuous between a and b,f(a) & f(b) are of opposite sign then there exists at least one root between a and b. The approximate value of the root between themXo=(a+b)/2
If F(Xo)=0 then the Xo is the correct root of F(x)=0.If F(Xo)0, then the root either lies in between [a, (a+b)/2] or
[(a+b)/2,b] depending on whether F(xo) is negative or positive Again bisection the interval and repeat same method until theaccurate root is obtained.
(ii) Method of false position; (Regular falsemethod) :
This is another method to find the root of F(x)=0. in this method,we choose two points and taking the point of intersection of thechord with the x-axis as an approximate root (using y=0 on x-axi
X1=[aF(b)- b(F(a)]/[F(b)-F(a)]
Repeat the same process till the root is obtained to the desiredaccuracy.
(iii) Iteration method :If a function F(x) is continuous between a and b,f(a) & f(b) are of opposite sign then there exists at least one root
between a and b. The approximate value of the root between themXo=(a+b)/2We can use this method,if we can express f(x)=0 , as
X =(X0) such that 1(X0)< 1 thenThe successive approximate roots are given by
Xn =(Xn-1), n=1,2----
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(iv) Newton Raphson method : The successive
approximate roots aregiven by Xn+1=Xn- F(Xn) /F1(Xn) , n=0,1,2----
Provided that the initial approximate root Xo is chosen sufficientlyclose to root of F(x)=0
2. Interpolation
(i) Newtons forward interpolation formula :Let y=F(x) be the function which take the values Yo,Y1,Y2,.Yn corresponding to the equally spaced values Xo, X1, X2..Xn of Xwith h as the interval length between two consecutive points.
The Newtons forward interpolation formula isF(Xo+ph)=Y p=Yo+pYo=[P(P-1)]/ 2! 2Yo+ [P(P-1)(P-2)]/3! 3Yo++[P(P-1)(P-2).(P-n+1)]/ n! nYo
Where X= Xo+ph i.e., P=[X-Xo]/hThis is also called Newton- Gregory forward interpolation formul
(ii) Newtons backward interpolation formula;
Y p=Yn+Pyn + [P(P-1)]/ 2! 2Yn + [P(P-1)(P-2)]/3! 3Yn +.
Where P=[X-Xn]/h(iii) Gauss forward interpolation formula:
using central differences, delta as an operator the Gauss forward interpolation formula isY p=Yo + P Y1/2+ [(P)(P-1)] /2! 2yo+ [(P+1)(P-1)]/3! 3Y1/2
+ [(P+1)P(P-1)(P-2)] /4! 4Yo +
Where P=[X-Xo]/w(iv) Gauss backward interpolation formula:
Y p=Yo + PY-1 + [(P+1)P] /2! 2Y-1 + [(P+1)P(P-1) ]/3! 3Y-2 +[(P+2)(P+1)(P-1)]/4! 4Y-2 +..
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(v) Lagranges Interpolation formula:
Let Yo, Y1, Y2,..Yn be the values of Y=(x) corresponding toXo, X1, X2,Xn (not necessarily equispaced)
Lagranges Interpolation formula is
Y= (x)= [(X-X1)(X-X2)..(X-Xn)] Y0+(X-Xo)(X-X2)(X-X3)(X-Xn) Y1[(Xo-X1)(Xo-X2)..(X-Xn)] (X1-Xo)(X1-X2)(X1-X3)(X1-Xn)
+..(X-Xo)(X-X1)(X-X3)(X-Xn) Yn (X2-Xo)(X2-X1)(X2-X3)(X2-Xn)3. Spline interpolation and cubic Splines
Let the given interval [a,b] is sub divided into n subintervals[Xo,X1], [X1,X2],..[Xn-1,Xn] where a = Xo
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OR
1)Bisectionor Bolzanos method:-i) f(a) = +ve , f(a) = -ve then c(a,b)ii) f(x0) = 0 x0
0 f(x0) = +ve the root lies in a & x0 , x0 = (a+b)/20 f(x0) = -ve the root lies in x0& b
iii) f(x0) = +ve then second approximation x1 = (a+x0)/2
= -ve then x1 = (x0+b)/2till we get an repeated end values.
2)Regula Falsi Methodor False position method:-i) f(a) = +ve, f(b) = -ve c (a, b)ii) let x1 = a f(b) b f(a)
f(b) f(a)a) f(x1) & f(a) are opposite then x2 = af(x1) x1f(a)
f(x1) f(a) b) f(x1) & f(a) are same then x2 = x1f(b) b f(x1)
f(b) f(x1)up to accuracy root i.e. repeated end values.
3)Iterationor Successive Approximation method:-i) f(a) = +ve, f(b) = -ve c (a,b)ii) f(x) = 0 x= (x) since | 1(x) | < 1
let x0 = a+b x1= (x0)2 x2 = (x1)
x3 = (x2)and so on up to accuracy root i.e., repeated end values
4)Newton Raphson methodOr Tangents method:-i) f(a) = +ve, f(b) = -ve c (a,b)ii) let x0 = a+b x1 = x0 f(x0)
2 f 1(x0)
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x2= x1 f(x1)f 1(x1)
and so on up to accuracy root i.e. repeated end values.
FINITE DIFFERENCE :-
1. Forward Difference operator: f (x) = f(x+h) f(x)
y0 = y1 y0
2. Backward Difference operator:f (x) = f (x) f (x h)
y1 = y1 y0
3. Central Difference operator: f (x) = f (x+h/2) f (x h/2)
y1/2= y1 y0
4. Shift operator : E f(x) = f(x+h)Eyx = yx + h
5. Inverse operator : E-1 = E-1f(x) f (x h)
6. Averaging/Mean operator: f(x) = f(x + h/2) + f( x h/2)2
yx = 1/2[yx + h/2+ yx - h/2]7. E = 1 +
8. E = 1/2 [E1/2+ E-1/2]
9. = E1/2 E-1/2
10. = E = E = E1/2
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11. 2 = =
12. 1 = (1+ ) (1- )
(INTERPOLATION WITH EQUAL & UNEQUAL INTERVALS)
I. 1. Newton Gregory Forward Interpolation Formula:-y = f(x) = y0 + p y0 + p (p-1) 2y0 + p (p-1)(p-2) 3y0+..
2! 3! Where p = x - x0h
2. Newton Gregory Backward Interpolation Formula:-y =f(x) = yn+ p yn + p(p+1) 2yn + p (p+1)(p+2)3yn +...Where p = x-xn2! 3! h
II Central Difference Interpolation Formula: -1. Gauss Forward: -
y p = y0 + p y0 + p(p-1) 2y-1 + (p+1)p(p-1) 3y-1+.. Where p = x-x0
2! 3! hy0 --- 2y-1 --- 4y-2 --- 6y-3 --- 8y-4
y0 3
y-1 5
y-2 7
y-3 2. Gauss Backward:- y p = y0 + p y-1 + (p+1)p 2y-1 + (p+1)p(p-1) 3y-2+(p+2)(p+1)p(p-1) 4y-1+..
2! 3! 4!Where p = x-x0
h y-1 3y-2 5y-3 7y-4
y0 ---- 2y-1 ---- 4y-2 ---- 6y-3 ---- 8y-4
3.Stirlings: -y p = y0+p y0+ y-1+ p2 2y-1 + p(p2-1) 3 y -1 + 3 y -2 +p2(p2-1) 4y-2+
.2 2! 3! 2 4!
y-1 3y-2 5y-3 7y-4
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y0 ---- 2y-1 ---- 4y-2 ---- 6y-3 ---- 8y-4 y0 3y-1 5y-2 7y-3
4.Lagranges Interpolation:(Unequal Intervals ):-
y = f(x) = (x-x1)(x-x2)(x-xn) f(x0)+ (x-x0)(x-x2)..(x-xn) f(x1)+ (x0-x1)(x0-x2)..(x0-xn) (x1-x0)(x1-x2)(x1-xn)
(x-x0)(x-x1)..(x-xn) f(x2) +.+ (x-x0)(x-x1)(x-xn-1) f(xn)(x2-x0)(x2-x1)(x2-xn) (xn-x0)(xn-x1)(xn-xn-1)
UNIT-V
Curve Fitting & Numerical Integration
Curve Fitting
1. Fitting a straight line
2. Fitting Quadratic Polynomial or parabola
Numerical differentiation Numerical Integration
Trapezoidal Rule
Simpsons 1/3 Rule and Simpsons 3/8 Rule
Gaussian Integration
Summary
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Summary
1. Curve Fitting
Fitting a straight line
Let Y(x)=ax+b be the straight line approximation for the data.
The normal equations are
axi2 + bxi = yixi axi + b1 = yi i from 1 to n
Solving above equations we get a and b
(ii) Fitting quadratic polynomial or parabola
Let Y(x)= aX2+ bX + c be the quadratic polynomial
The normal equations are
axi4 + bxi3 +cxi2 = xi2yi
axi3 + bxi2 +cxi = xiyi
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axi2 + bxi +c1 = yi i from 1 to n
Solving the above equations we get the values of a, b, c
2. Interpolation
Derivates using Newtons forward difference interpolation
Formula
(i) [dy/dx]x=xo = [Yo- (1/2)2Yo + (1/3)3Yo-(1/4)4Yo +]
(ii) [d2y/dx2]x=xo = (1/h2)[2Yo-3Yo + (11/12)4Yo+..]
3. Derivates using Newtons backward Interpolation Formula
(i) [dy/dx]x=xo = (1/h)[yn-(1/2)2
yn+(1/3)3
yn+](ii) [d2y/dx2]x=xo = (1/h2)[2yn-3yn+(11/12)4yn+(5/6)5yn+]
4. Trapezoidal Rule
The integral I=(x)dx in-between a to b
I=(h/2)[(Yo+Yn)+2(Y1+Y2++Yn-1)]Where yo, y1, yn i.e.,yi= (xi) are the values corresponding to theargument xo=a,X1,=Xo+hXn=Xo+nh=b
5. Simpsons 1/3 Rule
The integral I=(x)dx = y dx in-between a to b
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I = (x)dx=( h/3)[(yo + yn) + 4(y1+y3+..+yn-1)+ 2(y2+ y4++ yn-2)]in-between a to b
This rule can be applied when the given internal (a,b) is divided i
even number of sub intervals of length h
6. Gaussian Integration
The definite integral I= (x)dx is expressed as I=(x)dx=w1 (x1) + w2 (x2) ++wn (xn)
= wi (xi) i from 1 to n
Limit integral in-between a to b
Which is called Gaussian integral formula where Wi are called
weights and xi are called abscissa. The weights and abscissa aresymmetrical with respect to the middle points of the interval.
OR
(PART-A) (CURVE FITTING)
1. Fitting of a Straight Line (y=a+bx): -
y = na + b x
xy = a x + b x2
2. i) Parabola (y= a+bx+cx2)
y = na + b x + c x2
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xy = a x + b x2 + c x3
x2y = a x2 + b x3 + c x4
ii) Parabola (y = a+bx2)
y = na + b x2
x2y = a x2 + b x4
4. y= ae bx(y=a+bx) : logy = loga+xbloge10 Y=A+Bx
logy = Y, loga=A, B= b/loge10
5.y=abx(y=a+bx) : logy = loga+xlogbY=A+Bx
logy = Y, loga = A, logb = B
6. y = ax b(y=a+bx): logy = loga+blogxY=A+bX
since logx=X, logy=Y, loga=A
Weighted least square Approximation:-
1.Straight line(y=a0+a1x): Wy = a0W + a1Wx
Wxy = a0 Wx + a1 Wx2
2.Parabola(y=a0+a1x+a2x2): Wy = a0 W + a1 Wx + a2 Wx2
Wxy = a0 Wx + a1Wx2 + a2Wx3
Wx2y = a0 Wx2 + a1 Wx3 + a2 Wx4
(NUMERICAL DIFFERENTIATION)
1. Newton Forward:
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+(n5-2n4+35n3-50n2+12n) 5 y 0+(n6-15n5+17n4-225n3+274n2-60n) 6 y 0+]6 4 3 5! 6 6 4 3 6!
1. Trapezoidal Rule (n=1): -
y dx = (h/2) [(y0+yn) + 2(y1+y2+..+yn-1] Note: Number of subintervals odd or even.
2. Simpson 1/3 Rule (n=2): -
y dx = (h/3) [( y0+yn) + 4(y1+y3+y5+.) + 2(y2+y4+y6+)] Note: Number of subintervals should be even.
3. Simpson 3/8 Rule (n=3)
y dx = (3h/8) [ (y0+yn) + 3(y1+y2+y4+y5+y7+y8+)+2(y3+y6+y9+.)] Note: Sub intervals should be multiples of 3.
4. Booles Rule (n=4): -
y dx = (2h/45) [7y0+32y1+12y2+32y3+14y4+32y5+12y6+.] Note: Subintervals should be multiples of 4.
5. Weddles Rule (n=6): -
y dx = (3h/10) [(y0+yn) +(y2+y4+y8+y10+y14+..+yn-4+yn-)+5(y1+y5+y7+y11+..+yn-5+yn-1)+ 6(y3+y9+y15+..+yn-3)+2(y6+y12++yn-6)]
Note: - Subintervals should be multiples of 6.
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UNIT-VI
Numerical solutions of Initial Value Problems inOrdinary Differential Equations
Numerical Solution of Ordinary Differential equations
Taylors series method
Picards method
Eulers method
Modified Eulers method
Runge kutta method
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Predictor corrector method
Adams Bashforth method
Summary
Summary
The most important methods of solving ordinary differential equationsnumerically are
1. Taylors series method
2. Picards method
3. Eulers modified method
4. Runge-kutta method
5. Predictor corrector method
1. Taylors series method: The numerically solution of the differentialequation
dy/dx = (x,y) with the given initial condition y(xo) = yo is
yn-1= yn +(h/1!)y1n + (h2/2!)y11n + (h3/3!)y111n +
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2. Picard s method
To solve the differential equation dy/dx = (x,y) , y(xo) = Yousing Picards method of successive approximations with the help of y
yo + xo to x (x,y) dx which is called an integral equation.It can be solved by a process of successive approximations y(1)(x), y(2 The first approximation y(1)(x)= yo + xo to x (x,xo0dx
The second approximation y(2)(x)= yo + xo to x (x,y(1))dx
OR
Consider dy/dx = f(x,y) and the initial condition is y(x0)= y01. Taylors:- y(x0) = y0+(x-x0) y01+(x-x0)2 y011 + (x-x0)3 y0111+
2! 3!
2. Picards: - y1= y0+ f (x,y0) dxy2= y0 + f(x, y1) dxy3= y0+ f(x, y2) dx
Similarly yn= y0 + f(x,yn-1) dx3. Eulers :- y1 = y0 +h f(x0, y0)
y2 = y1+ h f(x1,y1)
y3 = y2+h f(x2,y2)
Similarly yn+ 1 = yn + h f(xn,yn)
4.Runge-Kutta Order 4: -
y1 = y0 + (1/6)[k1+2k2+2k3+k4]
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where k1 = h f(x0,y0) , k2= h f(x0+h/2, y0+k1/2)
k3= h f(x0+h/2, y0+k2/2) , k4= h f(x0+h, y0+k3)
yn+1 = yn+ (1/6) [k1+2k2+2k3+k4]
where k1= h f(xn,yn) , k2 = h f(x0+h/2, y0+k1/2)
k3= h f(x0+h/2, y0+k2/2) , k4= h f( x0+h, y0+k3) k=0,1,2.
5.Milnes Predictor-corrector :-
Predictor:- y4= y0 + 4h [2y11-y21+2y31] where yk1 = f(xk,yk), k=0,1,23
Corrector:- y4 =y2+h [y21+4y31+y41] where yk1= f(xk, yk), k=0,1,2
36.Adams Moulton Predictor-Corrector:-
Predictor: y4 = y3+ (h/24)[ 55y31-59y21+37y11-9y01]
where yk1 = f(xkyk) , k = 0,1,2,3...
Corrector: y4 = y3+ (h/24) [9y41+19y31-5y21+y11]
where yk1 = f(xk,yk) , k = 0,1,2,3..
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UNIT-VII
FOURIER SERIES
Periodic Functions
Even and odd Function
Fourier Series
Eulers Formulae Fourier Series in an arbitrary Interval (change of interval)
Fourier Series of Even and odd Functions
Half-Range Fourier Sine and Cosine Series
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Summary
Summary1. Periodic functions
Definition: A function f: R R is said to be periodic if there exists a positinumber T such that (x+T)= (x)for all x belongs to R, T is called per (x).
2. Even and odd functions
(i) A function (x) is said to be even if (-x)= (x)(ii) A function (x) is said to be odd if (-x)= - (x)
3. Definition : The Fourier series for (x)in the interval (C,C+2) is
(x)= ao/2 + [ancos nx + bn sin nx ] where n from 1 to
where ao= 1/ (x)dx limits in between c to C+2
an= 1/ (x)cos nx dx in between c to C+2
bn= 1/ (x)sin nx dx in between c to C+2 where C isconstant
ao, an, bn, are called Fourier coefficients (Fourier constant ),these formulae are called Eulers formula
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Note:
(i) If C =0 , than the interval becomes (0,2)
The Fourier coefficients are
ao= 1/ (x)dx limits in between 0 to 2
an= 1/ (x)cos nx dx limits in between 0 to 2
bn= 1/ (x)sin nx dx limits in between 0 to 2
(ii) If C=- then the interval becomes (-, )The Fourier coefficients are
ao= 1/ (x)dx limits in between - to
an= 1/ (x)cos nx dx limits in between - to
bn= 1/ (x)sin nx dx limits in between - to
4. Dirichlets conditions:
A function (x) defined in the interval a1 x a2 can berepresented as a Fourier series if (x) satisfies the followingconditions in the interval
(i) (x) and its integrals are finite and single valued(ii) (x) has a finite number of discontinuities(iii) (x) has finite number of maxima and minima . Then the
Fourier series converges to (x) at all points where (x) iscontinuous. Also the series converges to the average of theleft limit and the right limit of (x) at each point of discontinuity of (x).
5. Change of interval (Arbitrary interval): If a function (x) is defined in (C, C+21). The Fourier expansion
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(x) is
(x)= ao/2 + [ancos(nx)/l + bn sin(nx)/l ] where n from 1to
ao=1/l (x) dx limits in between c to c+2l
an= 1/l (x)cos(nx)/l dx limits in between c to c+2l
bn= 1/l (x)sin(nx)/l dx limits in between c to c+2l
Note: If c = 0 then the interval becomes (0,2l)If c = -l then the interval becomes (-l,l)
6. Fourier Series for even and odd functions.
(i) If (x) is an even function in (0,2) or (-;) the Fourier Series(x) is
(x) = ao/2 + [ancos nx ] where n from 1 to
where ao= 2/ (x) dx limits in between 0 to 2
an= 2/ (x)cos nx dx in between 0 to 2
here ,if f(x) is an even function, the Fourier coefficient bn= 0
(ii) If (x) is an odd function in (0,2) or (-;) the Fourier Series (x) is
f(x) = [bnsin nx ]
where bn= 2/ (x)sinnx dx limits in between 0 to 2here, the coefficients a0 = 0 , an = 0
similarly in the case of the intervals (0,2l) or (-l,l)
7. Half range Fourier series:
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(i) Half range fourier sine series for f(x) in (0, )
f(x) = [bnsin nx ]
when bn= 2/ (x)sinnx dx limits in between 0 to
(ii) Half range fourier sine series for f(x) in (0, l )
f(x) = [bn sin(nx)/l]
when bn= 2/l (x)sinnx dx limits in between 0 to
(iii) Half range fourier cosine series
(x)= ao/2 + [ancos nx ] where n from 1 to
where ao= 2/ (x) dx limits in between 0 to
an= 2/ (x)cos nx dx in between 0 to
(iv) Half range fourier cosine series
(x)= ao/2 + [ancos(nx)/l + bn sin(nx)/l ]
where n from 1 to
ao= 2/l (x) dx limits in between 0 to l
an= 2/l (x)cos(nx)/l dx limits in between 0 to l
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UNIT-VIII
PARTIAL DIFFERENTIAL EQUATIONS
Formation of Partial Differential Equation by eliminating arbitraryconstants and arbitrary functions
First Order Linear (Lagranges) Equations
Non-Linear (Standard Types) Equations
Method of Separation of variables for second order
One Dimensional Wave equation
One Dimensional Heat equation
Laplaces equation
Two Dimensional Wave equation
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Summary
Summary
1. Formation of Partial equations by the elimination of arbitraryconstants and arbitrary function.
(a) Elimination of arbitrary constants
Let (x,y,z,a,b)=0 (1) be the equation where a,b are arbitrary constants.Differentiating partially w.r.to x and yf + f . z = 0 f + p f = 0 .(2)y z x x z
f + f . z = 0 f + q f = 0 ..(3)y z y y z
Elimination of two constants a,b from (1) (2) (3) gives anequation of the form (x, y, z, p, q)=0 which is the first ordP.D.E.
If the numbers of constants are more than the number of independent variables, then the result of eliminating theconstants will give rise to a P.D.E of higher order than thefirst.
(b) Elimination of arbitrary functions:
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Let (u,v)=0 ..(1)Be the equation where u,v are functions of x, y, z and be arbitrary function.Differentiating (1) partially with respect to x and y
u + u . z = 0 v + vz = 0u x z x v x zx
And
u + u . z = 0 v + vz = 0u y z y v y zy
Eliminating , from 2 nd 3 we get u vu v + u v = 0 x y y x
P = z , q = zx y
Pp+Qq=R is the required P.D.E
Where, P =u v - u vx z z y
Q = u v - u v
z x x z
R = u v - u vx y y x
2. Lagranges Linear P.D.E
The P.D.E Pp+Qq=R ..(1)Is called Lagranges first order partial differential equation
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P = u v - u vx z z y
Q = u v - u v
z
x x z
R = u v - u vx y y x
To solve (1),first write Lagranges auxiliary equation (subsidiary equation)
x = y = z .(2)P Q R
Auxiliary equation gives two independent solutions u=c1and v=c2 where u, v are functions of x, y, zFrom these two solutions, the general solutions is (u,v)=0
3. Non linear partial differential equations of order one
(i) Complete integral:If F(x, y, z, p, q)=0 (1)Is the non linear partial differential equation of first order then equation
(x, y, z, a, b)=0 ..(2)Which contains as many constants as the number of independenvariables is called the complete integral.
(ii) Partial integral : A partial integral of (1) is obtained by giving particular values is called the complete integral.(iii) Singular integral: Differentiating the complete integral (x,y, z, a, b)=0 .(2)Partially w. r. to a and b and then equate to zero
/a=0 ..(3)
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/b=0 ............(4)
Elimination of a and b from (2) (3) (4) gives an equation of form f(x, y, z)=0 is called singular integral.
There are four standard forms of non-linear first order partialdifferential equations.
(i) Standard Form I: The equation of the form f(p,q) = 0 (i.e., the equation in terms of p and q only)is called standard tyThe solution of equation is Z=ax+by+c
a = z , b = zx y
Now replacing p= z =a ,and q = z = b in the givenP.D.E x y
F(a,b) = 0 b = (a)
Substituting b = (a) in (1)Z = ax + (a)y + c is called complete integral
(ii)Standard Form II:The equation of the form (x, y, p, q) = 0(1) is called standtype II
Arrange (1) in the form f 1(x, p) = f 2(y, q) = a(constant)From these two equations we get p=1(x, a) and q=2(y, a)
Substituting in
Integrating dz= pdx + qdy + cdz =1(x, a)dx + 2(y, a)dy
z=1(x, a)dx + 2(y, a)dy + c is the complete integral
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(iii) Standard form III:The equation of the form (z, p, q)=0..(1)
Substituting q=ap ..(2)In (1) we get
P = (z) (3)from (2) , (3) q = a (z) .(4)
From (3) , (4) substituting p, q values in
dz = p. dx + q dydz = (z) dx + a (z) dy
z / (z) = dx + a dy
Integrating F(z) = x + ay +c is complete integral
v) Standard form IV : (Clairauts equation)
The P.D.E of form z = px + qy + (p, q) (1)is called clairauts equation
The complete integral of (1) is
z = ax + by + (a ,b) .(2)
To find the singular integral difference (2) partially with respeca and b
0 = x +f ..(3) a 0 = y +f ..(4)
bThe eliminate of a, b from (3) (4) gives the singular integral
4. Application of P.D.E s (Method of separation of variables)
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(1) One- dimensional wave equation :u/t = c 2 2u/x 2(2) One- dimensional wave equation :
2u/x 2 + 2u/y 2 = 1/c 2. 2u/t 2
(3) One- dimensional wave equation :u/t = c 2 2u/x 2
(4) Laplaces equation :2u/x 2 + 2u/y 2 = 0Problems which satisfy certain initial and boundary conditions called boundary value problems. The suitable method to solve s problems is the method of separation of variables also known a product method.