Download A text book of engineering mathematics Volume 2 by Rajesh Pandey PDF

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The given equation can be rewritten as y (y + 2X2) dx + X (2X2 - y) dy = 0 which is of the form as given in method VI . Let xhyk be an integrating factor Multiplying the given equation by xhyk, we get (xh yk+2 + 2Xh + 2 yk + 1) dx + (2Xh +3 yk _Xh+1 yk+1) dy = 0 Here M = Xh yk+2 + 2xh +2 yk+ 1 and N = 2xh+3 yk _xh+ 1 yk+1 :. aM = ay aN and ax (k + 2) xhyk +1 + 2 (k + 1) xh +2yk (1) + 3) Xh +2 yk _ (h + 1) Xh yk +1 (2) = 2 (h . aM If the equation (A) be exact we must have - ay aN = - ax (k + 2) xh yk +1 + 2 (k + 1) xh + 2yk = - (h + 1) Xh yk +1 + 2 (h + 3) xh +2 yk from (1) and (2) Or Equating coefficients of Xh yk +1 and xh + 2yk on both sides we get k + 2 = - (h + 1) and 2 (k + 1) = 2 (h + 3) hence solving we get 5 1 h=-- k=-2' 2 :.

Solve dy + Y cos x = 1:. sin 2x dx 2 Solution. S. 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.

Solution. Here the auxiliary equation is (m -1)2 (m2 + 1)2 = 0 or m = 1, 1, ± i, ±i :. CF. I. = 1 eX (D _1)2 (D2 + 1)2 1 1 eX (D - 1)2 (12 + 1)2 1 1 X -e (D _1)2 (2)2 1 1 X -e (D _1)2 4 = eX 1 1 (D + 1_1)2 4 43 A Textbook ofEn~neerin~ Mathematics Volume - II = eX ~ ! 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.

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