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By Pandey, Rajesh.

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P. I. A particular integral of the differential equation f(D) y = Q is given by _1_ Q f(D) Methods of finding Particular integral (A) Case I. , when Q is of the form of eax, where a is any constant and f(a) :1; 0 we know that 41 A Textbook of Engineering Mathematics Volume - II D2 (eax) = a2 eax (e ax) = a 3 e ax Dn (e ax) = an e ax :. I. : f(a) is constant e ax f(D) e ax = _1_ f(a) eax if f(a) "# 0 , Case II. I. I. = - - e ax f (D) = eax x2 - - if f" (a) "# 0 f" (a)' Example 1. Solve (D2 - 2D + 5) Y = e- X 42 Linear Differential Equations with Constant Coefficients and Applications Solution.

Here the auxiliary equation is (m + 2) (m - 1)3 = 0 m = - 2 and m = 1 (thrice) or Therefore C. 1. = = 1 eX (D + 2) (D _1)3 1 eX (1 + 2) (D - 1)3 1 1eX1 __ = _ eX 1 1 3 (D _1)3 3 {(D + 1) -I} ~ :. I. I. 1. P. P. P. P. P. P. of %(-;) (i cos ax - sin ax) 1 x =- - - cos ax 2 a 1. --:---:- sm ax = .. I. P. of = - x 2a 2 1 D +a . 2 e,ax -~ (~) (i cos ax - sin ax) . sIn ax 1 x . ---::---:-- cos ax = - sm ax .. D2 + a 2 2a Example 4. Solve (D2 + D + 1) Y = sin 2x Solution. Here the auxiliary equation is m 2 + m + 1 = 0 which gives m :.

EX dx or v. x = C + x eX - or (log y) x = C + x eX - ex . Exact Differential Equations A differential equation which can be obtained by direct differentiation of some function of x and y is called exact differential equation, consider the equation Mdx + Ndy = 0 is exact 23 A Textbook of Engineering Mathematics Volume - II where M and N are the functions of x and y. Solution of a exact differential equation is fMdx fNdy =c + Regarding yasa constant only those termsofN not containing x Example 16.

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