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2oo1) (a) x2 + y2 - 2xy = C (b) . (c) X2 + 4xy + y2 = C (d) xy+y2+ x2 = C xy + y2 - x2 = C Ans. (d) 10. In the differential equation x dy + my = e- x , if the integrating factor is -\-' dx x then the value of m is (a) 2 (b) -2 (c) 1 (d) -1 35 A Textbook of Engineering Mathematics Volume - II Ans. (b) 11. S. 1988) (b) log (y2 -1) (d) none of these = Clog (X2 + 1) _x2 e (c) y2= 1 + C 2 x Ans. (c) 12. S. S. S. 1995) 1 (a) eY = ex + - x3 + C 3 (b) (c) eY = ex + x3 + C (d) 1 eY = e-X+ - x3 + C 3 Ans.

F(a) is constant e ax f(D) e ax = _1_ f(a) eax if f(a) "# 0 , Case II. I. I. = - - e ax f (D) = eax x2 - - if f" (a) "# 0 f" (a)' Example 1. Solve (D2 - 2D + 5) Y = e- X 42 Linear Differential Equations with Constant Coefficients and Applications Solution. Here auxiliary equation is m 2 - 2m + 5 = 0, whose roots are m=-1±2i :. CP. = e- X [Cl cos 2x + C2 sin 2x], where Cl and C2 are arbitrary constants and P.. : here a =-1 1 -x =-e 8 :. The required solution is y = CP. I. e. y = e- X(Cl cos 2x + C2 sin 2x) + ?

Solve dy + 2y tan x = sin x, given that y = 0 when x = rt/3. S. 2003) Solution. Here P = 2 tan x and Q = sin x ... Integrating factor = e lpdx = e l2tan xdx = e210g sec x Multiplying the given equation by sec2x, we get 2 sec x (~~ 2 + 2y tan x) = sin x sec x d or ' - (y sec 2x) = sec x tan x dx Integrating both sides with respect to x, we get J sec x tan x dx, y sec2 x = C + ysec 2 x=C+secx or it is given that when x = rtj3, y ... from (1) 0 where C is an arbitrary constant x = (1) 0 2 rt rt sec - = C + sec 3 3 17 A Textbook of Engineering Mathematics Volume - II or 7t 0=C+2 :.

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