Diff. Eqns.
Problem Set 4
Solutions
1. Consider the linear ODE
y 00 + αy 0 + βy = 0
(1)
where α and β are real constants. Sketch the region of the α-β plane such that
the solutions of the ODE are decaying oscillations.
The solution of (1) is found by substitution ert for y, which gives us the characteristic
equation
r2 + αr + β = 0
p
−α ± α2 − 4β
r=
.
2
So the general solution is
"
! #
"
! #
p
p
−α − α2 − 4β
α2 − 4β
y (t) = c1 exp
t + c2 exp
t
2
2
#
" p
#!
"p
h α i
α2 − 4β
α2 − 4β
.
= exp −
t c1 exp
t + c2 exp −
t
2
2
2
−α +
(2)
(3)
(4)
(5)
For oscillations to occur, the discriminant α2 −4β must be negative, so that we get imaginary
exponentials and can apply Euler’s formula to get sines and cosines. So, if α2 − 4β < 0,
then
#
" p
#!
" p
h α i
i −α2 + 4β
i −α2 + 4β
y (t) = exp −
(6)
t c1 exp
t + c2 exp −
t
2
2
2
!
!!
p
p
h α i
−α2 + 4β
−α2 + 4β
= exp −
t c3 cos
t + c2 sin
t
.
(7)
2
2
2
Our solution is oscillating, but to ensure decaying oscillation, we need the exponential in
front to decay, so −α/2 < 0. Coupled with our requirement that α2 − 4β < 0, we shade the
region below in the α-β plane.
2. Solve the initial-value problem
y 00 − 6y 0 + 9y = 0,
y(0) = 1,
y 0 (0) = −2
Again we solve by finding the roots of the characteristic equation
r2 − 6r + 9 = 0
(8)
2
(r − 3) = 0
(9)
r = 3 (repeated root)
(10)
So our solution is
y (t) = (c1 + c2 t) e3t .
(11)
Applying our initial conditions,
1 = y(0) = c1
(12)
0
−2 = y (0) = 3c1 + c2 ⇒ c2 = −5.
(13)
y (t) = (1 − 5t) e3t .
(14)
Thus our solution is
3. Use the method of undetermined coefficients to find the general solution for
each of the following nonhomogeneous equations
(a) y 00 − 9y = 3t2 − cos 3t,
(b) y 00 + 4y = 2et + sin(2t).
(a) Complementary Solution
First we solve the homogeneous equation.
yc00 − 9yc = 0.
(15)
The characteristic equation is r2 − 9 = 0, and has roots r = ±3. So our solution is
yc (t) = c1 e3t + c2 e−3t .
(16)
Particular Solution
Since the right hand side of our differential equation has two terms, we will have two parts to
our particular solution Y (t), such that Y (t) = Y1 (t) + Y2 (t). First we find Y1 (t) satisfying
Y100 − 9Y1 = 3t2 .
(17)
We assume our solution is of the form
Y1 (t) = At2 + Bt + C.
(18)
2A − 9 At2 + Bt + C = 3t2 .
(19)
Putting this into (17), then
Matching coefficients on the left and right hand sides, we get the following system of equations:
−9A = 3
(20)
−9B = 0
(21)
2A − 9C = 0
(22)
and so A = −1/3, B = 0, and C = −2/27. Plugging these into (18), we have
1
2
Y1 (t) = − t2 − .
3
27
(23)
Y200 − 9Y2 = − cos 3t.
(24)
Next we find Y2 (t) satisfying
We assume our solution is of the form
Y2 (t) = D cos 3t + E sin 3t.
(25)
(−9D cos 3t − 9E sin 3t) − 9 (D cos 3t + E sin 3t) = − cos 3t.
(26)
Putting this into (24), then
Matching coefficients on the left and right hand sides, we get the following system of equations:
−18D = −1
(27)
−18E = 0
(28)
ans so D = 1/18 and E = 0. Plugging this into (25), we have
Y2 (t) =
1
cos 3t.
18
(29)
General Solution
Altogether, we have
1
2
1
y (t) = yc (t) + Y1 (t) + Y2 (t) = c1 e3t + c2 e−3t − t2 −
+
cos 3t.
3
27 18
(30)
(b) Complementary Solution
First we solve the homogeneous equation.
y 00 + 4y = 0.
(31)
The characteristic equation is r2 + 4 = 0, and has roots r = ±2i. So our solution is
yc (t) = c1 e2it + c2 e−2it
= c3 cos(2t) + c4 sin(2t)
(32)
(33)
Particular Solution
Since the right hand side of our differential equation has two terms, we will have two parts to
our particular solution Y (t), such that Y (t) = Y1 (t) + Y2 (t). First we find Y1 (t) satisfying
Y100 + 4Y1 = 2et .
(34)
We assume our solution is of the form
Y1 (t) = Aet .
(35)
Aet + 4Aet = 2et .
(36)
Putting this into (34), then
Matching coefficients on the left and right hand sides, we get the following equation:
5A = 2,
(37)
and so A = 2/5. Plugging this into (35), we have
2
Y1 (t) = et .
5
(38)
Y200 + 4Y2 = sin(2t).
(39)
Next we find Y2 (t) satisfying
In choosing the form of our solution, we must be careful. We might first try
?
Y2 (t) = B cos(2t) + C sin(2t).
(40)
But this is the solution to the homogeneous equation, so this will always yield zero if we plug
it into the left-hand-side of (39), and we will get a contradiction. Thus, we must multiply
our guess by t, and we assume our solution is of the form
Y2 (t) = t (B cos(2t) + C sin(2t)) .
(41)
This solution no longer corresponds to the homogeneous solution. Putting this into (39),
then
(4C − 4Bt) cos(2t) + (−4B − 4Ct) sin(2t) + 4 (t (B cos(2t) + C sin(2t))) = sin(2t)
(42)
4C cos(2t) − 4B sin(2t) = sin(2t)
(43)
Matching coefficients on the left and right hand sides, we get the following system of equations:
4C = 0
(44)
−4B = 1
(45)
ans so C = 0 and B = −1/4. Plugging this into (41), we have
1
Y2 (t) = − t cos(2t).
4
(46)
2
1
y (t) = yc (t) + Y1 (t) + Y2 (t) = c3 cos(2t) + c4 sin(2t) + et − t cos(2t).
5
4
(47)
General Solution
Altogether, we have