By Pandey, Rajesh.

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**Additional info for A Text Book of Engineering Mathematics. Volume II**

**Example text**

2004) 1:. sin 2x = sin x cos x 2 :. Integrating factor = e fPdx = efcos x dx = e sin x Multiplying the given equation by the integrating factor esin with respect to x, we get 18 x and integrating Differential Equations of First Order and First Degree y. e sin x = C + JeSin x sin x cos x dx, where C is an arbitrary constant. y. e sin x = C + Jet t dt, or = C + t. e t = C + esinx (sin x -1) where t = sin x et - y. e sin x = C + esin x (sin x - 1) or Equations Reducible to the Linear Form. The equation (1) Where P and Q are constants or functions of x alone and n is a constant other than zero or unity is called the extended form of linear equation or Bernoulli's Equation.

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 :. 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 :.

D2 4 = ! eX ~ (1) = 4 D2 2 eX x 4 2 ! = .!. x2 eX 8 :', The required solution is y = C. 1. y = (Cl or X + C2) ex + (C3 x + ~) cos x + (Cs x + C6) sinx + .!. 8 x2 eX Example 3. Solve (D + 2) (D -1)3 Y = eX 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 = ..