# A text book of engineering mathematics Volume 2 by Rajesh Pandey

By Rajesh Pandey

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**Example text**

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 :. P. 1. = 2 21 sin 2x replacing D2 by - 22 (-2) +D+1 = -1- D-3 . 2x sm 1 (D - 3) (D + 3) 21 D -9 (D + 3) sin 2x (D + 3) sin 2x = ~ 21 -9 (D + 3) sin 2x = -~ (D + 3) sin 2x = -1 [D (sin 2x) + 3 sin 2x] 13 13 = -~ [2 cos 2x + 3 sin 2x] Since D means differentiation with respect to x 13 :.

X = C + fx eX dx, where C is an arbitrary constant fl. 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.

The given equation can be rewritten as 26 Differential Equations o(First Order and First Dei{'ee xy2 (1 + 2xy) dx + x2y (1 - xy) dy = 0 or Y (1+ 2xy) dx + x (1 - xy) dy = 0 or (y dx + x dy) + 2xy2 dx - x2y dy = 0 or d (xy) + 2xy2 dx - x2y dy = 0 Dividing both sides of this equation by x2y2, we get d (xy) 2 1 - + - dx - - dy = 0 2 x y2 X Y or (:2) dz+ (z/x) dx- (1/y) dy=O, wherez=xy Integrating, - .!. + 210g x - log Y = C, z or - (Xy) + log or log ( where C is an arbitrary constant (xX) = C, putting z = xy XX) = C + (Xy) Example 20.