Let y = y(x) be the solution of the differential equation x ( y x )dy = (y ( y x ) - x )dx, y(1) = 2 and let = ( y(e^ 12 ) e^ 12 ). Then the number of integral values of p, for which the equation x^2 + y^2 - 2px + 2py + + 2 = 0 represents a circle of radius r 6, is __________.
A. A
Given the differential equation: x ( y x )dy = (y ( y x ) - x )dx Rearranging the terms, we get: x ( y x )dy - y ( y x )dx = -xdx Dividing both sides by x^2: ( y x ) ( xdy - ydx x^2 ) = - dx x ( y x )d ( y x ) = - dx x Integrating both sides: - ( y x ) = - |x| - C ( y x ) = |x| + C Using the initial condition y(1) = 2 : ( 2 ) = (1) + C C = 0 Thus, the solution to the differential equation is: ( y x ) = |x| We need to find = ( y(e^ 12 ) e^ 12 ). Substituting x = e^ 12 : = (e^ 12 ) = 12 The given equation of the circle is: x^2 + y^2 - 2px + 2py + + 2 = 0 Substituting = 12: x^2 + y^2 - 2px + 2py + 14 = 0 The radius r of the circle is given by: r = (-p)^2 + p^2 - 14 = 2p^2 - 14 For the equation to represent a real circle, r > 0: 2p^2 - 14 > 0 p^2 > 7 We are given that r 6: 2p^2 - 14 6 2p^2 - 14 36 2p^2 50 p^2 25 Combining the inequalities: 7 Since p is an integer, the possible values for p^2 are 9, 16, 25. This gives p \ -5, -4, -3, 3, 4, 5\ . Therefore, there are 6 integral values of p. Answer: 6
Related: Mathematics — Differential Equations · All PYQ Banks