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Thomas Calculus 13th [Solutions]

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1024 Chapter 17 Second-Order Differential Equations<br />

∞ ∞ ∞ ∞<br />

ww <br />

3. y 4y œ 0 Ê n 2<br />

n n 1 c x 4 n<br />

c x œ 0 Ê n 2<br />

nn 1c x n<br />

4c x œ 0<br />

n n n n<br />

nœ2 nœ0 nœ2 nœ0<br />

power of x coefficient equation<br />

0<br />

x 21c2 4c0 œ 0 Ê c2 œ 2c0<br />

1<br />

2<br />

x 32c 3 4c1 œ 0 Ê c3 œ 3c1<br />

2<br />

1 2<br />

x 43c 4 4c2 œ 0 Ê c4 œ 3c2 œ<br />

3c0<br />

3<br />

1 2<br />

x 54c 5 4c3 œ 0 Ê c5 œ 5c3<br />

œ<br />

15c<br />

1<br />

4<br />

2 4<br />

x 65c 4c œ 0 Ê c œ 15c œ 45c<br />

ã ã ã<br />

n<br />

4<br />

x n 2n 1c 4c œ 0 Ê c œ n2n1c<br />

6 4 6 4 0<br />

n 2 n n 2 n<br />

0 1 0<br />

2 2 3 2 4 2 5 4<br />

3 1 3 0 15 1 45 0<br />

6<br />

0 0<br />

2 2 4 4 6 2 3 2<br />

3 0 45 0 1 3 1 15 1<br />

5<br />

0<br />

2 4 6 3 5 ∞ n ∞ n<br />

2x 2x 2x 2x 2x 1 2n 1 2n1<br />

2x 4x 6x 2 3x 5x 0 2nx 2 2n1x<br />

nœ0 nœ0<br />

0<br />

c1 c1<br />

2 0<br />

2<br />

y œ c c x2c x c x c x c x c x á<br />

œ c 2cx cx cx á cx cx cx á<br />

c1 c<br />

œ c Š 1 á‹ Š 2x á‹ œ c <br />

1<br />

2x <br />

<br />

2x<br />

œ c cos 2x sin 2x œ a cos 2x b sin 2x, where a œ c and b œ<br />

∞ ∞ ∞<br />

ww w <br />

4. y 3 y 2y œ 0 Ê n 2<br />

n n 1 c x 3 n 1<br />

<br />

n c x 2<br />

n<br />

c x œ 0<br />

n n n<br />

nœ2 nœ1 nœ0<br />

∞ ∞ ∞<br />

n2 nn 1c x n1 n 3ncnx n<br />

2cnx 0<br />

nœ2 nœ1 nœ0<br />

Ê œ<br />

power of x<br />

coefficient equation<br />

0<br />

3<br />

x 21c 2 31c 1 2c0 œ 0 Ê c2 œ<br />

2c1 c0<br />

1<br />

1 7<br />

x 32c 3 32c 2 2c1 œ 0 Ê c3 œ c2 <br />

3c1 œ<br />

6c1 c0<br />

2<br />

3 1 5 7<br />

x 43c 4 33c 3 2c2 œ 0 Ê c4<br />

œ<br />

4c3 <br />

6c2 œ<br />

8c1 <br />

12c0<br />

3<br />

3 1 31 1<br />

x 54c 5 34c 4 2c3 œ 0 Ê c5 œ<br />

5c4 <br />

10c3 œ<br />

120c1 <br />

4c0<br />

4<br />

1 1 7 31<br />

x 65c 35c 2c œ 0 Ê c œ<br />

2c <br />

15c œ<br />

80c <br />

360c<br />

ã ã ã<br />

n<br />

x n<br />

2n1c 3n1c 2c œ 0 Ê c œ 3 c <br />

2 c<br />

6 5 4 6 5 4 1 0<br />

n2 n1 n n2 n2 n1 n2n1<br />

n<br />

y œ c c xˆ 3c c ‰ 2<br />

x ˆ 7c c ‰ 3<br />

x ˆ 5 7<br />

c c ‰ 4<br />

x ˆ 31 1 5 7 31 6<br />

c c ‰ x ˆ c c ‰ x á<br />

0 1 2 1 0 6 1 0 8 1 12 0 120 1 4 0 80 1 360 0<br />

2 3 7 4 1 5 31 6 3 2 7 3 5 4 31 5 7 6<br />

0 0 0 12 0 4 0 360 0 1 2 1 6 1 8 1 120 1 80 1<br />

0ˆ 2 1 3 7 4 1 5 31 6 ‰ 2 3 4 5 6<br />

1ˆ 3 7 5 31 7 ‰<br />

3 12 4 360 2 6 8 120 80<br />

x 2x<br />

œ œ ∞<br />

x<br />

∞<br />

n<br />

2x n<br />

<br />

nx<br />

nx<br />

nœ0 nœ0<br />

3 2 7 3 5 7 4 31 1 5<br />

0 1 2 1 0 6 1 0 8 1 12 0 120 1 4 0<br />

œ c c x c x c x c x c x á c x c x c x c x c x c x á<br />

œ c 1x x x x x á c x x x x x x á<br />

Note that if we use the techniques of Section 17.2, our solution is y a e b e a b . Thus<br />

aba2bx á œ c c xˆ c c ‰ x ˆ c c ‰ x ˆ c c ‰ x ˆ c c ‰ x á<br />

If the series are equivalent, then take c0 œ a b and c1 œ a 2b and substitute in our first series for c 0 and c 1 to obtain<br />

y œ a ba 2bx ˆ 3a 2ba b‰ 2<br />

x ˆ 7a 2ba b‰ 3<br />

x ˆ 5 7<br />

a 2b a b‰<br />

4<br />

2 6 8 12<br />

x<br />

ˆ 31 1<br />

a2b ab ‰ 5<br />

x ˆ 7 31<br />

a2b ab‰<br />

6<br />

120 4 80 360<br />

x á<br />

œ aba2bxˆ 1a2b‰ 2<br />

x ˆ 1 4<br />

a b‰ 3<br />

x ˆ 1 2<br />

a b‰ 4<br />

x ˆ 1 4<br />

a b‰ 5<br />

x ˆ 1 4<br />

a b‰<br />

6<br />

x á<br />

2 6 3 24 3 120 15 720 45<br />

1 2 1 3 1 4 1 5 1 6 2 4 3 2 4 4 5 4 6<br />

2 6 24 120 720 3 3 15 45<br />

1 2 1 3 1 4 1 5 1 6<br />

2<br />

2x 3<br />

2x 4<br />

2x 5<br />

2x 6<br />

2x<br />

2x 3x 4x 5x 6x 2x 3x 4x 5x 6x<br />

œ a ax ax ax ax ax ax á b 2bx 2bx bx bx bx bx á<br />

œ aˆ 1x x x x x x በbŠ 12x á‹<br />

∞ ∞ n<br />

2x x 2x<br />

nx<br />

nœ0 nœ0<br />

nx<br />

n<br />

œ a x<br />

b<br />

<br />

œ ae be<br />

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

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