Math 2280
Quiz 6
1. Write the solution to the initial value problem
x′′ + 25x = 90 cos 4t x(0) = 0, x′ (0) = 90
as a linear combination of two oscillations, one of the form cos(ω0 t − α) and one of the form cos 4t.
The characteristic polynomial for this equation is
r2 + 25 = 0 ⇒ r2 = −25 ⇒ r = ±5i,
so the complimentary solution is
xc (t) = c1 cos 5t + c2 sin 5t.
We need to find any solution to the non-homogeneous problem. One way to do this is to use the method of
undetermined coefficients. We assume that xp (t) has the form
xp (t) = A cos 4t + B sin 4t,
which should work since none of the terms in xp are repeated in xc . We have that
x′p (t) = −4A sin 4t + 4B cos 4t
x′′p (t) = −16A cos 4t − 16B sin 4t,
so
x′′p + 25xp = −16 cos 4t − 16B sin 4t + 25A cos 4t + 25B sin 4t
= (−16A + 25A) cos 4t + (−16B + 25B) sin 4t = 9A cos 4t + 9B sin 4t,
which implies
Thus our solution is
9A = 90, 9B = 0 ⇒ A = 10, B = 0.
x(t) = c1 cos 5t + c2 sin 5t + 10 cos 4t ⇒ x′ (t) = −5c1 sin 5t + 5c2 cos 5t − 40 sin 4t.
Applying the initial conditions x(0) = 0 and x′ (0) = 90, we find that
c1 + 10 = 0 ⇒ c1 = −10, 5c2 = 90 ⇒ c2 = 18.
All that is left to determine the values of C and α. We have that
q
√
√
C = c21 + c22 = 100 + 324 = 424 = 20.591
and
α = tan−1 −
18
= 5.219,
10
so that
x(t) = 18 cos(5t − 5.219) + 10 cos 4t.
2. Write the system of second order equations
x′′ − 5x + 4y = 0, y ′′ + 4x − 5y = 0
as an equivalent system of first order equations.
This system of two second order equations should be equivalent to a system of four first order equations. To that
end, we write
x1 = x, x2 = x′ , y1 = y, y2 = y ′ .
Then
This is our system of first order equations.
 ′
x = x2


 1′
x2 = 5x1 − 4y1
.
y ′ = y2


 1′
y2 = −4x1 + 5y1
1