285hw10 - HW 10 1 Sec 9.3 5 We have f(t = 0 if t 2(0 1 f(t...

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a 0 = 2 3 Z 3 0 f ( t ) dt = 2 3 Z 2 1 dt = 2 3 t j 2 1 = 2 3 : a n = 2 3 Z 2 1 cos n&t 3 dt = 2 3 & 3 n& & sin n&t 3 j 2 1 = 2 n& (sin 2 &n 3 ± sin &n 3 ); n = 6 k = ) a n = 2 n& (sin 0 ± sin 0) = 0 ; n = 1 + 6 k = ) a n = 2 n& & sin 2 & 3 ± sin & 3 ± = 0 ; n = 2 + 6 k = ) a n = 2 n& & ± sin 2 & 3 ± sin 2 & 3 ± = 2 n& & ± 2 & p 3 2 ! = ± 2 p 3 n& ; n = 3 + 6 k = ) a n = 2 n& (sin 0 ± sin 0) = 0 ; n = 4 + 6 k = ) a n = 2 n& & sin 2 & 3 ± ( ± sin 2 & 3 ) ± = 2 n& & 2 & p 3 2 = 2 p 3 n& ; n = 5 + 6 k = ) a n = 2 n& ² ± sin & 3 ± ( ± sin & 3 ) ³ = 0 : Thus f ( t ) = 1 3 ± 2 p 3 & ´ 1 2 cos 2 &t 3 ± 1 4 cos 4 &t 3 + 1 8 cos 8 &t 3 ± ::: µ : 2 case: sine series. See the sketch. 2
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Thus b n = 2 3 Z 3 0 f ( t ) sin n&t 3 dt = 2 3 Z 2 1 sin n&t 3 dt = & 2 3 ± 3 n&t ± cos n&t 3 j 2 1 = & 2 n& & cos 2 &n 3 & cos n& 3 ± ; n = 6 k = ) b n = & 2 n& (cos 0 & cos 0) = 0 ; n = 1 + 6 k = ) b n = & 2 n& ² & cos & 3 & cos & 3 ³ = 2 n& ; n = 2 + 6 k = ) b n = & 2 n& ² & cos & 3 & ( & cos & 3 ) ³ = 0 ; n = 3 + 6 k = ) b n = & 2 n& (cos 0 & cos & ) = & 4 n& ; n = 4 + 6 k = ) b n = & 2 n& ² & cos & 3 & ( & cos & 3 ) ³ = 0 ; n = 5 + 6 k = ) b n = & 2 n& ² & cos & 3 & cos & 3 ³ = 2 n& : f ( t ) = 2 & ´ sin &t 3 & 2 3 sin &t + 1 5 sin 5 &t 3 + ::: µ : 2. Sec. 9.3: 15. We have x 00 + 2 x = t;x 0 (0) = x 0 ( & ) = 0 : We need to &nd 3
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the cosine series of f ( t ) = t: a 0 = 2 & Z & 0 tdt = 2 & & 1 2 t 2 j & 0 = 1 & & & 2 ± 0 ± = &: a n = 2 & Z & 0 t cos ntdt = 2 n& Z & 0 td sin nt = 2 n& ² t sin nt j & 0 ± Z & 0 sin ntdt ³ = 2 n& & 1 n & cos nt j & 0 = 2 n 2 & (cos n& ± cos 0) = 2 n 2 & (( ± 1) n ± 1) : Thus f ( t ) = & 2 ± 2 & 1 X n =1 1 ± ( ± 1) n n 2 cos nt: Now we are looking for the solution in the form x ( t ) = A 0 2 + 1 X n =1 A n cos nt; then x 0 ( t ) = ± 1 X n =1 nA n sin nt; x 00 ( t ) = ± 1 X n =1 n 2 A n cos nt: So we get ± 1 X n =1 n 2 A n cos nt + A 0 + 2 1 X n =1 A n cos nt = & 2 ± 2 & 1 X n =1 1 ± ( ± 1) n n 2 cos nt; whence A 0 = & 2 ; 2 A n ± n 2 A n = ± 2 &n 2 (1 ± ( ± 1) n ) = ) A n = 2 (1 ± ( ± 1) n ) &n 2 ( n 2 ± 2) : Thus x ( t ) = & 4 + 2 & 1 X n =1 1 ± ( ± 1) n n 2 ( n 2 ± 2) cos nt: 4
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3. Sec. 9.3: 19. We have t = 2 1 X n =1 ( & 1) n +1 n sin nt;t 2 ( & &;& ) : 1 step. We integrate termwise the series for the &rst time. We get t 2 2 = & 2 1 X n =1 ( & 1) n +1 n 2 cos nt + C 1 : Now let t = 0 = ) 0 = & 2 1 X n =1 ( & 1) n +1 n 2 + C 1 = ) C 1 = 2 ± & 2 12 = & 2 6 : So we have t 2 2 = & 2 1 X n =1 ( & 1) n +1 n 2 cos nt + & 2 6 : 2 step. We integrate termwise the series for the second time. We get t 3 6 = & 2 1 X n =1 ( & 1) n +1 n 3 sin nt + & 2 6 t + C 2 : Now let t = 0 = ) 0 = & 2 P 1 n =1 ( & 1) n +1 n 3 0 + 0 + C 2 = ) C 2 = 0 : We have t 3 6 = & 2 1 X n =1 ( & 1) n +1 n 3 sin nt + & 2 6 t: 3 step. We integrate termwise the series for the third time. We get t 4 24 = & 2 1 X n =1 ( & 1) n n 4 cos nt + & 2 12 t 2 + C 3 : Now let t = 0 = ) 0 = & 2 1 X n =1 ( & 1) n n 4 + 0 + C 3 = ) C 3 = 2 1 X n =1 ( & 1) n n 4 : So we have t 4 24 = & 2 t 2 12 & 2 1 X n =1 ( & 1) n n 4 cos nt + 2 1 X n =1 ( & 1) n n 4 : 5
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4. Sec. 9.4: 3. We have x 00 + 3 x = F ( t ) ; where F ( t ) is the odd function of period 2 & with F ( t ) = 2 t;t 2 (0 ;& ) . We are looking for the sine series of F ( t ) = 2 t: b n = 2 & Z & 0 2 t sin ntdt = & 4 &n Z & 0 td cos nt = & 4 &n & t cos nt j & 0 & Z & 0 cos ntdt ± = & 4 &n & & cos n& & 1 n sin nt j & 0 ± = & 4 &n & & cos n& & 1 n (0 & 0) ± = & 4 n ( & 1) n : Thus F ( t ) = 4 1 X n =1 ( & 1) n +1 n sin nt: Now we are looking for the solution in the form x ( t ) = 1 X n =1 B n sin nt: Then x 0 ( t ) = 1 X n =1 nB n cos nt;x 00 ( t ) = & 1 X n =1 n 2 B n sin nt: So we get & 1 X n =1 n 2 B n sin nt + 3 1 X n =1 B n sin nt = 4 1 X n =1 ( & 1) n +1 n sin nt; that is, 3 B n & n 2 B n = 4 n ( & 1) n +1 = ) B n = 4 ( & 1) n n ( n 2 & 3) : Thus x sp ( t ) = 4 1 X n =1 ( & 1) n n ( n 2 & 3) sin nt: 6
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5. Sec. 9.4: 7. We have x 00 + 9 x = F ( t ) ;
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285hw10 - HW 10 1 Sec 9.3 5 We have f(t = 0 if t 2(0 1 f(t...

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