A text book of engineering mathematics Volume 2 by Rajesh Pandey

By Rajesh Pandey

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Dx 1 2 1 + y2 sm 2x - - - cos x = - dy Y Y putting -cos2x = v or -2 cos x (-sinx) dx = dv or sin 2x dx = dv 21 (1) A Textbook of Enzineerinz Mathematics Volume - II the equation (1) reduce to dv 1 1 + y2 -+-v=-dy Y Y (2) Which is linear in v with y as independent variable . Its integrating factor = e J~ dy Y = e 10g Y = Y Multiplying both sides of (2) by Y and integrating, we get vy = C + f(1 + y2) dy, where C is constant of integration or 1 Y cos2 X + C + Y + - y3 = 0 3 or Example 14. S. 1995) Solution.

Solve (2x + y - 3) dy = (x + 2y - 3) dx dy Solution. The given differential equation is dx 14 = x + 2y-3 ---=-2x + Y - 3 (1) Di(ferential Equations o(First Order and First Degree pu tting x = X + hand y = Y + k (2) dY (X + h) + 2 (Y + k) -3 the equation (1) reduce to - = -"----'----------'-dX 2 (X + h) + (Y + k) -3 dY = X + 2Y + (h + 2K - 3) dX 2X + Y + (2h + K - 3) or (3) Choose hand k such that h + 2k - 3 = 0 and 2h + k - 3 = 0 (4) Solving the equations (4) we have h = 1 = k :. From (2) x = X + 1 and y = Y + 1 or X = x - 1 and Y = Y - 1 (5) dY X + 2Y Also (3) reduce to - = - - dX 2X+ Y .

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 = ..

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