A text book of engineering mathematics Volume 2, Edition: by Rajesh Pandey

By Rajesh Pandey

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Example 24. 2001) Solution. Here M = 3x2 y4 + 2xy , N = 2x3 y3 - x2 aN, so the equation is not exact in this form. Thus, we have to find As oM ~ Oy Ox the integrating factor by trial. In the present case, we see that - 1 [ON -- - M Ox oM Oy I -) = 6X 2y3 - 2x _12x 2y3 - 2x 3X 2y 4 + 2xy . The integrating factor is e -J~ dy y 1 = 2 Y Thus, the differential equation becomes (3x2y2 + 2xjy) dx + (2X3 Y - x2jy2) dy = 0 Which is an exact, as oM = 6x 2y _ 2xjy2 = oN Oy Ox 30 2. the function of y alone y = - - , IS Differential Equations of First Order and First Dewee Its solution is J(3x 2 y2 + 2x/y) dx = 0 or x3 y2 + X2/y = c Where C is an arbitrary constant.

CS. 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.

1994) (a) h =-h . (c) h = X2 h (b) (d) hh=x hh=l Ans. (d) 22. 1994) (a) Circles (b) Parabolas (c) hyperbolas (d) ellipse Ans. (c) 38 Chapter 3 Linear Differential Equations with Constant Coefficients and Applications INTRODUCTION A differential equation is of the form dny dn- l dn- 2 + ........... X2 dxn dxndx n - (I) where aI, a2, ....... an are constants and Q is a function of x only, is called a linear differential equation of nth order. Such equations are most important in the study of electro-mechanical vibrations and other engineering problems.

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