By Rajesh Pandey

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1995) Solution. The given equation can be written as 1 dy 1 1 1 - - + - - = - log X y2 dx x Y x . 1 putting - - = v y or (1) ~ dy = dv in (1), we get y2 dx dx dv 1 1 - - - v= -logx dx x x (2) which is in the standand form of the linear equation and integrating factor dx =e- 1og x 1 e x =- -p x Multiplying both sides of (2) by the integrating factor and integrating, we get v. or ! Y 1 = (1 + log x) Y - C xy Example 15. S. 1996) Solution. Dividing both sides of the given equation by y, we get x dy + (log y) = x eX y dx - - or 1 dy 1 X - - + - (log y) = e y dx x putting v = log Y or dv 1 dy .

Mn are distinct roots of the auxiliary equation, then the general solution of (1) is Y = C I e ffijX + C 2 e ffi2X + ................ + C n em"x where CI, C2, ............ Cn are arbitrary constants Illustration. Solve the differential equation d 2y dy - 2 +3 - -54y=O dx dx Solution. The given equation is (D2 + 3D - 54) Y = 0 Here auxiliary equation is m 2 +3m-54= 0 or (m + 9) (m - 6) = 0 => m = 6,-9 Hence the general solution of the given differential equation is y = Cle6x + C2 e-9x Case II. When Auxiliary Equation has real and some equal roots.

1999) (a) Circles (b) Ellipses (c) Cycloids (d) Rectangular hyperbolas Ans. (d) 2. 1999) (a) x2y2 + 2X2 + 2y2 = C (b) x2y2 + x2 + y2 = C (c) x2y2 + X + Y = C (d) x2y2 + 2x + 2y = C Ans. (d) 3. CS. 1999) (a) 2x (y')2 + 1 = 2yy' (b) 2xy + 1 = 2yy' (c) 2x2y' + 1 = 2yy' (d) 2 (y')2 + x = 2yy' Ans. (a) 4. 2oo0) (a) xy = C eX-Y (b) x+y=Ce xy (c) xy = C eY-X (d) x-y=Ce xy Ans. (c) 34 Diiferential Equations of First Order and First 5. 2000) (a) 1 + x2 (b) (c) log (1 + X2) (d) -log (1 + x2) Ans. (a) 6. 2oo0) (a) (e6 + 9)/2 (b) (c) log e6 (d) Ans.