MAT 274
HW 3 Solutions
c
Bin
Cheng
MAT 274 HW 3 Solutions
( 50 points total )
1. (20’) Find the general solution in trigonometric form
y 00 (x) + 4y 0 (x) + 8y(x) = 0
and then
y 00 (x) + 4y 0 (x) + 8y(x) + 10e−x = 0.
(1)
Note: to accommodate the 10e−x term,
use ae−x as your “guess” for a particular solution.
Solution. Homogeneous part
y 00 (x) + 4y 0 (x) + 8y(x) = 0
(2)
Char. equation: r2 + 4r + 8 = 0
=⇒ r1 = −2 + 2i,
r2 = −2 − 2i
=⇒ yh (x) = e−2x (C1 cos(2x) + C2 sin(2x))
“Guess” a nonhomogeneous solution
ynh = ae−x
so that
0
ynh
= −ae−x ,
00
= ae−x .
ynh
Plug them into the nonhomogeneous DE (1)
ae−x + 4(−ae−x ) + 8ae−x + 10e−x = 0
Factor out e−x
e−x (a − 4a + 8a + 10) = 0
The coef of e−x has to equal zero
(a − 4a + 8a + 10) = 0 =⇒ a = −2
Thus, we found a particular solution ynh = −2e−x to the nonhomogeneous DE (1).
Finally, superimpose yh and ynh to obtain the general solution to the original DE (1)
y = yh + ynh = e−2x (C1 cos(2x) + C2 sin(2x)) − 2e−x .
P age
1
MAT 274
HW 3 Solutions
c
Bin
Cheng
2. (10’) Find the general solution, using your answer from the previous problem,
y 00 (x) + 4y 0 (x) + 8y(x) + 10e−x = 16x2 .
(3)
Solution. The homogeneous part is the same as (2) in the previous problem. So we
know the homogeneous solution is also the same
=⇒ yh (x) = e−2x (C1 cos(2x) + C2 sin(2x))
For the nonhomogeneous solution, “guess”
ynh = ae−x + bx2 + cx + d
so that
0
ynh
= −ae−x + 2bx + c
00
= ae−x + 2b
ynh
Plug them into the nonhomogeneous DE (3)
(ae−x + 2b) + 4(−ae−x + 2bx + c) + 8(ae−x + bx2 + cx + d) + 10−x = 16x2
Gather like terms on both sides (i.e. the e−x terms, the x2 terms, the x terms and the
constant terms)
(a − 4a + 8a + 10)e−x + 8bx2 + (8b + 8c)x + (2b + 4c + 8d) = 16x2
Equate the coef of each group of like terms
e−x terms, a − 4a + 8a + 10 = 0
(4)
x2 terms, 8b = 16
(5)
x terms, 8b + 8c = 0
(6)
constant terms, 2b + 4c + 8d = 0
(7)
Solve this algebraic system for a, b, c, d
(4) =⇒ a = −2
(5) =⇒ b = 2
(6) =⇒ c = −b =⇒ c = −2
1
1
(7) =⇒ d = − (2b + 4c) =
8
2
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2
MAT 274
HW 3 Solutions
c
Bin
Cheng
Thus, we found a particular solution
ynh = −2e−x + 2x2 − 2x +
1
2
to the nonhomogeneous DE (3).
Finally, superimpose yh and ynh to obtain the general solution to the original DE (3)
1
y = yh + ynh = e−2x (C1 cos(2x) + C2 sin(2x)) − 2e−x + 2x2 − 2x + .
2
3. (20’) Find the general solution to
d2 y
dy
−8
− 9y = 0
2
dx
dx
and then
dy
d2 y
−8
− 9y = 100 sin(2x).
2
dx
dx
Note: to accommodate the 100 sin(2x) term,
(8)
use a cos(2x) + b sin(2x) as your “guess” for a particular solution.
Solution. Homogeneous part
dy
d2 y
−8
− 9y = 0
dx2
dx
Char. equation
r2 − 8r − 9 = 0 =⇒ r = −1, 9
=⇒ yh = C1 e−x + C2 e9x
“Guess” a nonhomogeneous solution
ynh = a cos 2x + b sin 2x
so that
0
ynh
= −2a sin 2x + 2b cos 2x
00
ynh
= −4a cos 2x − 4b sin 2x
Plug them into the nonhomogeneous DE (8)
(−4a cos 2x − 4b sin 2x) − 8(−2a sin 2x + 2b cos 2x) − 9(a cos 2x + b sin 2x) = 100 sin 2x
Gather like terms on both sides (i.e. the cos 2x terms, the sin 2x terms)
(−4a − 16b − 9a) cos 2x + (−4b + 16a − 9b) sin 2x = 100 sin 2x
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3
MAT 274
HW 3 Solutions
c
Bin
Cheng
Equate coef of like terms
cos 2x terms: − 4a − 16b − 9a = 0
sin 2x terms: − 4b + 16a − 9b = 100
(9)
(10)
To solve for a, b, first
13
a
16
Then, plug it into (10) i.e. plug into 16a − 13b = 100
(9) =⇒ b = −
13 16a − 13 − a = 100
16
=⇒ a =
1600
= 3.7647...
425
Plug this back into b = − 13
16 a
b=−
1300
= −3.0588...
425
Thus, a particular, nonhomogeneous solution is
ynh =
1300
1600
cos 2x −
sin 2x
425
425
and the general, nonhomogemeous solution is
y = yh + ynh = C1 e−x + C2 e9x +
1600
1300
cos 2x −
sin 2x
425
425
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4