Download MATH 107.01 HOMEWORK #11 SOLUTIONS Problem 4.1.2

MATH 107.01
HOMEWORK #11 SOLUTIONS
Problem 4.1.2. Determine if the differential equation
ey y 00 − 4y 0 + 5xy = 0
(1)
is linear or not. If it is linear, then give its order.
Solution. This is not linear.
Problem 4.1.3. Determine if the differential equation
5 00
y − 2y 0 + x3 y
x
is linear or not. If it is linear, then give its order.
(2)
5x sin x = 3y (4) − 4xy 000 +
Solution. This is a linear fourth-order differential equation.
Problem 4.1.10. Show that {1/x, x} is a fundamental set of solutions to the homogeneous differential equation
(3)
x2 y 00 + xy 0 − y = 0.
Find the general solution and find the solution with initial values y(1) = 0 and
y 0 (1) = −1.
Solution. It is easily verified that both y = 1/x and y
1/x and x are linearly independent, note that
1
1/x
x
= w(1/x, x)x=1 = 2
−1
−1/x 1 x=1
= x solve (3). To see that
1
= 2 6= 0.
1
Since the solution space of (3) is two-dimensional, we see that {1/x, x} is a fundamental set of solutions to (3). The general solution is then given by
y = c1
1
+ c2 x.
x
Imposing the initial conditions y(1) = 0 and y 0 (1) = −1 gives the solution
11 1
− x.
2x 2
Problem 4.1.11. Show that ex , e2x , e−3x is a fundamental set of solutions to
the homogeneous differential equation
y=
(4)
y 000 − 7y 0 + 6y = 0.
Find the general solution and find the solution with initial values y(0) = 1, y 0 (0) =
0, and y 00 (0) = 0.
1
2
MATH 107.01 HOMEWORK #11 SOLUTIONS
Solution. It is easily verified that each of y = ex , y = e2x , and y = e−3x solves (4).
To see that ex , e2x , e−3x are linearly independent, note that
x
e
e2x
e−3x 1 1 1 w ex , e2x , e−3x x=0 = ex 2e2x −3e−3x = 1 2 −3 = 20 6= 0.
ex 4e2x 9e−3x 1 4 9 Since the solution space of (4) is three-dimensional, wee see that ex , e2x , e−3x is
a fundamental set of solutions to (4). The general solution is then
y = c1 ex + c2 e2x + c3 e−3x .
Imposing y(0) = 1, y 0 (0) = 0, and y 00 (0) = 0 gives
y=
3 x 3 2x
1
e − e + e−3x .
2
5
10
Problem 4.1.15. Show that yp = e−x is a particular solution to the differential
equation
(5)
y 000 − 7y 0 + 6y = 12e−x .
Use Problem 4.1.11 to find the solution to (5) with initial conditions y(0) = 1 and
y 0 (0) = y 00 (0) = 0
Solution. It is easily verified that yp = e−x solves (5). By Problem 4.1.11, the
general solution to (5) is
y = yp + yH = e−x + c1 ex + c2 e2x + c3 e−3x .
Imposing y(0) = 1 and y 0 (0) = y 00 (0) = 0 gives
1
1
y = e−x + e2x − e−3x .
5
5
2
Problem 4.1.17. Show that x, x is a fundamental set of solutions to
1 2 00
x y − xy 0 + y = g(x).
2
Then, determine g(x) so that yp = ex is a particular solution to (6). Finally,
determine the solution to (6) with initial values y(1) = 1 and y 0 (1) = 0.
(6)
Solution. It is easily verified that y = x and y = x2
(6). Furthermore, x and x2 are independent since
x x2 1
2
w x, x x=1 = =
1 2xx=1 1
solve the homogeneous part of
1
= 1 6= 0.
2
Since
the
homogeneous part of (6) has two-dimensional solution space, we see that
x, x2 is a fundamental set of solutions to (6).
Next, to determine g(x) so that yp = ex is a particular solution to (6), plug
yp = ex into (6) to obtain
1 2
x
x −x+1 .
g(x) = e
2
Finally, we see that the general solution to (6) is
y = yp + yH = ex + c1 x + c2 x2
MATH 107.01
HOMEWORK #11 SOLUTIONS
3
and imposing y(1) = 1 and y 0 (1) = 0 gives
y = (2 − e) x − x2 + ex .