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

**Example text**

S. 1998) Ans. eX = y2 (x + C) 5. S. 1997) Ans. (X2 + y2 + 2) = 2e l /2 6. S. 1998) 33 A Textbook of Engineering Mathematics Volume - II 7. The equations of motion of a particle are given by dx dt + wy = 0, dy _ wx = 0, Find the path of the particle and so that it is a circle. 2009) Hint. y(t) = Cl cos wt + C2 sin wt x(t) = C2 cos wt - Cl sin wt so x2 + y2 = C~ + C~ = R 2 Objective Type of Questions Choose a correct answer from the four answers given in each of the following questions: 1. CS. 1999) (a) Circles (b) Ellipses (c) Cycloids (d) Rectangular hyperbolas Ans.

Here a =-1 1 -x =-e 8 :. The required solution is y = CP. I. e. y = e- X(Cl cos 2x + C2 sin 2x) + ? e- x 8 Example 2. Solve (D -1)2 (D2 + 1)2 Y = eX . 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.

2 log (x 2 + y2») = x dxx2 ++ Yy2dy (k) d log [ ~ ) Y (m) d ( ~ ) y = Y dx - x dy xy d (y' ) ~ 2x'y dy - 2xy' dx x2 X4 (h) d ( tan -1 y)_ x dy - Y dx X x 2 + y2 G) d [ _~ ) xy (1) d log ( = x dy + Y dx r) = x x 2y2 x dy - Y dx xy ~ ye' dx - e' dy y2 Example 18. S. S. 1974) Solution. The given eqution can be written as (y dx + x dy) + (xy2 dx - x2y dy) = 0 or d (yx) + xy2 dx - x2y dy = 0 Dividing both sides of this equation by x2y2, we get d (yx) + x 2 y2 or Integrating, ! X dx _ ! z + log x -log Y = C, where C is an arbitrary constant ~ + log x -log Y = C, putting z or - or 1 log (xjy) = C + xy xy = xy Example 19.