MA 2327
Assignment 3
Due 29 October 2015
Id:
2327-f2015-3.m4,v 1.2 2015/10/27 16:50:38 john Exp john
1. Solve the following initial value problems
(a)
x′ (t) = − tan(t) x(t) + sec t,
x(0) = 1.
(b)
(1 − t2 )x′ (t) + 2tx(t) = 1,
x(0) = 1.
2. Assume, as we will see later in lecture, that for all x0 , x1 , . . . , xn−1 the
homogeneous initial value problem
cn (x)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = 0
x(t0 ) = x0 ,
x′ (t0 ) = x1 ,
...
x(n−1) (t0 ) = xn−1
has a unique solution in the interval I when t0 ∈ I, ck is continuous
in I and cn has no zeros in I. Here x(k) denotes the k’th derivative of x.
Show the following:
(a) If y and z are solutions of
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = 0
then so is αy + βz for any constants α and β. In other words, the
set of solutions is a vector space.
1
Id:
2327-f2015-3.m4,v 1.2 2015/10/27 16:50:38 john Exp john 2
(b) If y and z are solutions of
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = f (t)
then y − z is a solution of
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = 0
(c) If y is a solution of
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = 0
and z is a solution of
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = f (t)
then y + z is a solution of
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = f (t)
(d) The initial value problem
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = f (t)
x(t0 ) = x0 ,
x′ (t0 ) = x1 ,
...
x(n−1) (t0 ) = xn−1
has at most one solution.
(e) Suppose that the differential equation
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = f (t)
has at least one solution. Then the initial value problem
cn (t)x(n) (t) + · · · + c1 (t)x′ (t) + c0 (t)x(t) = f (t)
x(t0 ) = x0 ,
x′ (t0 ) = x1 ,
...
x(n−1) (t0 ) = xn−1
has a solution for any values of x0 , x1 , . . . , xn−1 .