18.03 Recitation 18, April 13, 2010 �� u(t)e−t (t2 + 1)� �� 1 �� L[e−t ] = s+1 �� L[t2 ] = s23 � �� L[e−t (t2 + 1)] = L[e−t t2 ] + L[e−t ] 2 1 = + 3 (s + 1) s+1 2 s + 2s + 3 = . (s + 1)3 �� f (t) = e−t cos(3t)� �� F (s) = L[f (t)]� �� f � (t) �� t�� L[cos(3t)] = s2s+9 � �� F (s) = L[e−t cos(3t)] = L[cos(3t)](s + 1) s+1 = (s + 1)2 + 9 s+1 = 2 . s + 2s + 10 �� t > 0� f � (t) = −e−t (cos(3t) + 3 sin(3t))� �� f (t) �� f (t)u(t)� �� f (0+) = 1� �� f � (t) = δ(t) − u(t)e−t (cos(3t) + 3 sin(3t)). �� L[sin(3t)] = �� 3 s2 +9 �� L[δ(t)] = 1� �� L[f � (t)] = L[δ(t)] − L[e−t cos(3t)] − 3L[e−t sin(3t)] s+1 3 −3 = 1− 2 (s + 1)2 + 9 s + 2s + 10 s2 + s . = 2 s + 2s + 10 �� L[f � (t)] = sF (s), �� 2s + 1 s2 + 9 s2 + 2 s3 − s , , 2 − 1) s2 (s �� L[a sin(3t)+b cos(3t)] = 3a+bs � �� a = 1/3 �� b = 2� �� L−1 [ 2s+1 ] = 31 sin(3t) + 2 cos(3t)� s2 +9 s2 +9 �� s3 − s = s(s − 1)(s + 1)� �� 1� et � �� e−t � �� a, b �� c� �� a b c + + s+1 s s−1 a(s2 − s) + b(s2 − 1) + c(s2 + s) = s3 − s 2 (a + b + c)s + (c − a)s − b = s3 − s L[ae−t + b + cet ] = �� b = −2� �� a = c = 3/2� L−1 [ ss3+2 ] = 32 (e−t + et ) − 2]� −s �� s2 (s−1)� �� a, b, c �� 2 2 = L[a + bt + cet ] − 1) a b c = + 2+ s s s−1 2 a(s − s) + b(s − 1) + cs2 = s2 (s − 1) (a + c)s2 + (b − a)s − b = . s2 (s − 1) s2 (s �� b = −2� �� a = −c = −2� �� L−1 [ s2 (s2−1) ] = −2(1 + t − et )� �� D +2I � �� ẋ + 2x = u(t)� �� x(t) �� X(s)� �� t�� L[ẋ + 2x] = sX(s) + 2X(s)� �� L[u(t)] = 1s � �� 1 X(s) = s(s+2) = 1/2 + −1/2 � �� s s+2 (1 − e−2t )� �� x = u(t) 2 �� ẋ + 2s = δ(t) �� 1 �� 1� �� X(s) = s+2 � �� x = u(t)e−2t � �� ẋ + 2x = t2 �� x(0+) = 1� �� ẋ + �� 2x = t2 + δ(t)� �� x �� x(0− ) = 0 �� x(0+ ) = 1� �� ẋ �� δ(t) �� t�� (s + 2)X(s) = s23 + 1� �� s3 +2 s3 (s+2) � �� s3 + 2 a b c d = + + + s3 (s + 2) s3 s2 s s + 2 a(s + 2) + b(s2 + 2s) + c(s3 + 2s2 ) + ds3 = s3 (s + 2) (c + d)s3 + (b + 2c)s2 + (a + 2b)s + 2a = s3 (s + 2) �� a = 1� b = −1/2� c = 1/4� �� d = 3/4� �� x = 14 u(t)(2t2 − 2t + 1 + 3e−2t )�

18.03 Differential Equations, Recitation 18 (Solutions to Problems)

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