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CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
4.2 Reduction of Orderq
4.2 Reduction of Orderq
In Problems 1-8 we use reduction of order to find a second solution. In Problems 9-16 we use
formula (5) from the text.
1. Define y = u(x)e2x so
y 00 = e2x u00 + 4e2x u0 + 4e2x u,
and y 00 − 4y 0 + 4y = e2x u00 = 0.
Therefore u00 = 0 and u = c1 x + c2 . Taking c1 = 1 and c2 = 0 we see that a second solution is
y2 = xe2x .
2. Define y = u(x)xe−x so
y 0 = (1 − x)e−x u + xe−x u0 ,
y 00 = xe−x u00 + 2(1 − x)e−x u0 − (2 − x)e−x u,
and
y 00 + 2y 0 + y = e−x (xu00 + 2u0 ) = 0
If w = u0 we obtain the linear first-order equation w0 +
e2
R
dx/x
= x2 . Now
d 2
[x w] = 0
dx
or u00 +
2 0
u = 0.
x
2
w = 0 which has the integrating factor
x
gives x2 w = c.
Therefore w = u0 = c/x2 and u = c1 /x. A second solution is y2 =
1 −x
xe = e−x .
x
3. Define y = u(x) cos 4x so
y 0 = −4u sin 4x + u0 cos 4x,
y 00 = u00 cos 4x − 8u0 sin 4x − 16u cos 4x
and
y 00 + 16y = (cos 4x)u00 − 8(sin 4x)u0 = 0
or u00 − 8(tan 4x)u0 = 0.
If w = u0 we obtain the linear first-order equation w0 −8(tan 4x)w = 0 which has the integrating
R
factor e−8 tan 4x dx = cos2 4x. Now
d
[(cos2 4x)w] = 0
dx
gives
(cos2 4x)w = c.
Therefore w = u0 = c sec2 4x and u = c1 tan 4x. A second solution is y2 = tan 4x cos 4x = sin 4x.
4. Define y = u(x) sin 3x so
y 0 = 3u cos 3x + u0 sin 3x,
y 00 = u00 sin 3x + 6u0 cos 3x − 9u sin 3x,
and
y 00 + 9y = (sin 3x)u00 + 6(cos 3x)u0 = 0
or u00 + 6(cot 3x)u0 = 0.
© 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed
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© Cengage Learning. All rights reserved. No distribution allowed without express authorization.
y 0 = 2ue2x + u0 e2x ,
4.2
Reduction of Orderq
If w = u0 Rwe obtain the linear first-order equation w0 +6(cot 3x)w = 0 which has the integrating
factor e6 cot 3x dx = sin2 3x. Now
d
[(sin2 3x)w] = 0 gives (sin2 3x)w = c.
dx
Therefore w = u0 = c csc2 3x and u = c1 cot 3x. A second solution is y2 = cot 3x sin 3x = cos 3x.
5. Define y = u(x) cosh x so
y 0 = u sinh x + u0 cosh x,
y 00 = u00 cosh x + 2u0 sinh x + u cosh x
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and
y 00 − y = (cosh x)u00 + 2(sinh x)u0 = 0
or u00 + 2(tanh x)u0 = 0.
If w = u0 Rwe obtain the linear first-order equation w0 +2(tanh x)w = 0 which has the integrating
factor e2 tanh x dx = cosh2 x. Now
d
[(cosh2 x)w] = 0 gives (cosh2 x)w = c.
dx
Therefore w = u0 = c sech2 x and u = c tanh x. A second solution is y2 = tanh x cosh x = sinh x.
6. Define y = u(x)e5x so
y 0 = 5e5x u + e5x u0 ,
y 00 = e5x u00 + 10e5x u0 + 25e5x u
and
y 00 − 25y = e5x (u00 + 10u0 ) = 0
or u00 + 10u0 = 0.
If wR = u0 we obtain the linear first-order equation w0 +10w = 0 which has the integrating factor
e10 dx = e10x . Now
d 10x
[e w] = 0 gives e10x w = c.
dx
Therefore w = u0 = ce−10x and u = c1 e−10x . A second solution is y2 = e−10x e5x = e−5x .
7. Define y = u(x)e2x/3 so
2
y 0 = e2x/3 u + e2x/3 u0 ,
3
and
4
4
y 00 = e2x/3 u00 + e2x/3 u0 + e2x/3 u
3
9
9y 00 − 12y 0 + 4y = 9e2x/3 u00 = 0.
Therefore u00 = 0 and u = c1 x + c2 . Taking c1 = 1 and c2 = 0 we see that a second solution is
y2 = xe2x/3 .
8. Define y = u(x)ex/3 so
1
2
1
y 0 = ex/3 u + ex/3 u0 , y 00 = ex/3 u00 + ex/3 u0 + ex/3 u
3
3
59
and
6y 00 + y 0 − y = ex/3 (6u00 + 5u0 ) = 0 or u00 + u0 = 0.
6
If w =R u0 we obtain the linear first-order equation w0 + 56 w = 0 which has the integrating factor
e(5/6) dx = e5x/6 . Now
d 5x/6
[e
w] = 0 gives e5x/6 w = c.
dx
Therefore w = u0 = ce−5x/6 and u = c1 e−5x/6 . A second solution is y2 = e−5x/6 ex/3 = e−x/2 .
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© 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed
with a certain product or service or otherwise on a password-protected website for classroom use.
145