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FP2 – Second Order Differential Equations – Exam Questions June 2009 dx d2x +5 + 6x = 2e–t. 2 d t dt 8. dx = 2 at t = 0, dt (a) find x in terms of t. (8) Given that x = 0 and The solution to part (a) is used to represent the motion of a particle P on the x-axis. At time t seconds, where t > 0, P is x metres from the origin O. (b) Show that the maximum distance between O and P is 23 m and justify that this distance 9 is a maximum. (7) June 2010 8. (a) Find the value of λ for which y = λx sin 5x is a particular integral of the differential equation d2 y + 25y = 3 cos 5x. dx 2 (4) (b) Using your answer to part (a), find the general solution of the differential equation d2 y + 25y = 3 cos 5x. dx 2 (3) Given that at x = 0, y = 0 and dy = 5, dx (c) find the particular solution of this differential equation, giving your solution in the form y = f(x). (5) (d) Sketch the curve with equation y = f(x) for 0 x π. (2) June 2011 8. The differential equation dx d2x + 6 + 9x = cos 3t, dt dt 2 t 0, describes the motion of a particle along the x-axis. (a) Find the general solution of this differential equation. (8) (b) Find the particular solution of this differential equation for which, at t = 0, x = 1 dx and 2 dt = 0. (5) On the graph of the particular solution defined in part (b), the first turning point for T > 30 is the point A. (c) Find approximate values for the coordinates of A. (2) June 2012 4. Find the general solution of the differential equation dx d2x + 5 + 6x = 2 cos t – sin t. dt dt 2 (9) June 2013 7. (a) Show that the transformation y = xv transforms the equation 4 x2 d2 y dy 8 x (8 4 x 2 ) y x 4 2 dx dx (I) into the equation 4 d 2v 4v x dx 2 (II) (6) (b) Solve the differential equation (II) to find v as a function of x. (c) Hence state the general solution of the differential equation (I). (6) (1) June 2013 (R) 7. (a) Find the value of λ for which λt2e3t is a particular integral of the differential equation d2 y dy 6 9 y 6e3t , 2 dt dt t≥0 (5) (b) Hence find the general solution of this differential equation. (3) Given that when t = 0, y = 5 and dy =4 dt (c) find the particular solution of this differential equation, giving your solution in the form y = f(t). (5)