Download Test 2 Solutions - University of South Alabama

Apr 11, 2011
Differential Equations
Test 2 Solutions
1. Solve the initial value problem: y ′′ + 6y ′ + 9y = 0, y(0) = 3, y ′ (0) = 1.
The characteristic polynomial is r 2 +6r+9, which factors as (r+3)2 ,
and therefore has a repeated root r = −3. Thus the general solution
is
y(x) = c1 e−3x + c2 xe−3x .
Plugging in our initial conditions gives
3 = y(0) = c1 ,
1 = y ′(0) = −3c1 + c2
from which we determine that c1 = 3 and c2 = 10, giving a final solution
of
y(x) = 3e−3x + 10xe−3x .
2. Find the general solution: y ′′ − 6y ′ + 25y = 0.
The characteristic polynomial is r 2 − 6r + 25, which has roots 3 ± 4i
(via the quadratic equation). Thus the general solution is
y(x) = c1 e3x cos(4x) + c2 e3x sin(4x).
3. Find a particular solution: y (3) − 4y ′ = 3x − 1.
The characteristic polynomial is r 3 −4r, which factors as r(r+2)(r−
2), and therefore has roots 0, 2, −2. Thus the complementary solution
is
yc = c1 + c2 e2x + c3 e−2x .
Note that the c1 constant term comes from the zero root, since c1 e0x =
c1 . We now use the method of undetermined coefficients. The ‘f and
its derivatives’ list contains x and 1 (where the 1 represents a constant
term). These lists overlap on the constant term, so we need to multiply
through by x, obtaining x2 and x. Thus a particular solution looks
like yp = Ax2 + Bx.
To solve for A and B we plug in. Note that yp′ = 2Ax + B, yp′′ = 2A,
(3)
and yp = 0. Thus the DE gives −8Ax−4B = 3x−1. Equating coefficients
tells us that A = −3/8 and B = 1/4. Thus the particular solution
is
3
1
yp = − x2 + x.
8
4
4. Use variation of parameters to find the general solution: y ′′ + 3y ′ + 2y = 4ex .
The characteristic polynomial is r 2 + 3r + 2, which has roots −1, −2,
so that the complementary solution is yc = c1 e−x +c2 e−2x . To use variation
of parameters we solve the following system
u′1 e−x + u′2 e−2x = 0,
−u′1 e−x − 2u′2e−2x = 4ex .
Adding the two equations gives −u′2 e−2x = 4ex , from which we find that
u′2 = −4e3x . Plugging this into the first of the two equations gives
u′1 e−x = 4e3x e−2x , so that u′1 = 4e2x . We deduce that u1 = 2e2x and u2 =
− 43 e3x , so that a particular solution is given by
4
2
4
yp = u1 y1 + u2 y2 = 2e2x e−x − e3x e−2x = 2ex − ex = ex .
3
3
3
5. Turn the following differential equation into a system of first order equations:
x(3) + 3x′′ + 2x′ − 5x = sin 2t.
We let x1 = x, so then x2 = x′1 = x′ and x3 = x′2 = x′′1 = x′′ . The
original equation then becomes x′3 + 3x3 + 2x2 − 5x1 = sin 2t. Thus the
final system is

′


x3 = sin 2t − 3x3 − 2x2 + 5x1
x′2 = x3


x′ = x
2
1
6. Find the general solution using the method of elimination x′ = 4x − 3y, y ′ = 6x − 7y.
Solving the first equation for y, we find that 3y = 4x − x′ , or y =
′
′′
4x−x′
. Thus y ′ = 4x −x
. Substituting into the second equation gives
3
3
4x′ − x′′ 28x − 7x′
+
− 6x = 0,
3
3
which can be simplified into
x′′ + 3x′ − 10x = 0.
This has characteristic polynomial r 2 + 3r −10, which factors as (r +
5)(r − 2), and so has roots −5, 2. Thus the general solution is
x(t) = c1 e−5t + c2 e2t .
To find y(t) we plug this into our substitution equation, obtaining
y(t) =
2
4c1 e−5t + 4c2 e2t + 5c1 e−5t − 2c2 e2t
= 3c1 e−5t + c2 e2t .
3
3
7. Use the operational determinant to turn the system x′ = 4x − 3y, y ′ = 6x − 7y into a
system of two second order equations, each involving only a single unknown function.
Rewriting this system gives
(x′ − 4x) + (3y) = 0,
(−6x) + (y ′ + 7y) = 0.
It follows that the operational determinant is
D − 4
3
2
−6 D + 7 = D + 3D − 10.
Since f1 (t) = 0 and f2 (t) = 0, we have equations
x′′ + 3x′ − 10x = 0,
y ′′ + 3y ′ − 10y = 0.
8. A mass of 3 kg is attached to the end of a spring that is stretched 20 cm by a force of 15
N. It is set in motion with initial position x0 = 0 and initial velocity v0 = −10 cm/s. Find a
differential equation describing the resulting motion. Do not solve. (Recall that Hooke’s law
states that FS = −kx, where FS is the restorative force of the spring when it is stretched a
distance x.)
When the spring is stretched 1/5 of a meter, the restorative force is
15 N. Thus by Hooke’s law, the spring constant is k = 15/0.2 = 75.∗
Since there is no friction or forcing, this gives the equation 3x′′ +
75x = 0, or x′′ + 25x = 0.
∗
If you don’t switch to meters, you get a spring constant of 3/4, which leads to the DE 4x′′ + x = 0. The
reason you have to switch to meters is that the unit of force, the Newton, has meters in it, and things need
to cancel out correctly. If you did not do this, and you got this other equation, I’ll still count is as correct,
since we didn’t really say anything about units, and I can’t assume you know anything about what on earth
a Newton is.