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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
23
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)