Chapter 4, Section 5

Answers and solutions to selected problems

2. Here we consider the DE

x' + 3x = 5 cos t

a) The associated operator is L(x) = x' + 3x and this is a linear, homogeneous operator

b) The homogeneous solution, , is found by solving L(x) = ) or
' + 3x = 0. It should be relatively clear to us at this stage that
x = is a solution, i.e. (If this is not clear, separate varialbes, integrate, etc). We should also note that is also a solution

c) Suppose , then




d)


e)


f) For x' + 3x = 10 cos t or L(x) = 10 cos t, we have the same since the homogeneous DE si the same, namely L(x) = 0 or x' + 3x = 0. The function for will change, since we now need ; so the new should be 2* the original one or

4. The DE is x" + 4x = 2 which is linear by Theorem 4.7

a) The operator is L(x) = x" + 4x

b) sin 2t and cos 2t both satisfy the DE x" + 4x = 0 (Problem #8 from section 4.4 is very important in this respect) Since this is a linear homogeneous DE, the superposition principle applies and
is also a solution

c) If then is a solution to x" + 4x = 2, since x" = 0 and

d) Again we have

e)
,

so is not s solution.

6. The DE is v'= - g - k*v or v' + k*v = -g

a) The homogeneous solution solves v' + k*v = 0 and is the function The general solution is (we can get this ourselves by using integrating factors)

We know this is the sum of the particular and homogeneous solutions, so we can see that a particular solution is where the homogeneous solution(from above) is

b) The position function satisfies the DE x' = v, so we simply integrate v(t) to get to x(t)

Remembering to add a constant of integration we get

d) The constant is the initial velocity, i.e. v(0) and the second constant is the initial position, x(0).

8. For the DE x' + x/t = 1 we have an integrting factor of for t>0.

Multiplying by this factor gives us t*x' + x = t or

and integrating with respect to t produces

or

Using x(1) = 1, solving for c yields that and the solution is

The interval of existence is (0, ∞), since the initial value is x(1) = 1 or more specifically t = 1 and has a discontinuity at t = 0 which divided the reals into two intervals ( -∞, 0 ) and ( 0, ∞); the initial value of t is in the second of these intervals, so that's our interval of existence.