3 a Consider y 00 9 y 2 te 3 t sin t 6 t 2 e 3 t The solution of the

# 3 a consider y 00 9 y 2 te 3 t sin t 6 t 2 e 3 t the

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3. a. Consider y 00 - 9 y = 2 te - 3 t sin( t ) + 6 t 2 e 3 t . The solution of the homogeneous equation is y c ( t ) = c 1 e - 3 t + c 2 e 3 t . The guess that one makes for the method of undetermined coefficients is y p ( t ) = e - 3 t (( A 1 t + A 0 ) cos( t ) + ( B 1 t + B 0 ) sin( t )) + t ( C 2 t 2 + C 1 t + C 0 ) e 3 t . b. Consider y 00 + 2 y 0 + 2 y = 4 t 2 e - t + 3 te - t cos( t ). The solution of the homogeneous equation is y c ( t ) = e - t ( c 1 cos( t ) + c 2 sin( t )) . The guess that one makes for the method of undetermined coefficients is y p ( t ) = ( A 2 t 2 + A 1 t + A 0 ) e - t + te - t (( B 1 t + B 0 ) cos( t ) + ( C 1 t + C 0 ) sin( t )) . 4. a. Consider y 00 - 4 y = 16 e - 2 t , y (0) = 1 , y 0 (0) = 2 . Subscribe to view the full document.

Laplace transforms give: s 2 Y ( s ) - s - 2 - 4 Y ( s ) = 16 s + 2 or Y ( s ) = s + 2 s 2 - 4 + 16 ( s + 2)( s 2 - 4) . Partial fractions on the second expression give 16 ( s + 2)( s 2 - 4) = A ( s + 2) 2 + B s + 2 + C s - 2 or 16 = A ( s - 2) + B ( s + 2)( s - 2) + C ( s + 2) 2 . From s = - 2, A = - 4. From s = 2, C = 1. From s 2 , B = - C = - 1. It follows that Y ( s ) = - 4 ( s + 2) 2 - 1 s + 2 + 2 s - 2 . The inverse Laplace transform gives y ( t ) = 2 e 2 t - e - 2 t - 4 te - 2 t . b. Consider y 00 - y 0 - 12 y = t 2 δ ( t - 3) , y (0) = 2 , y 0 (0) = - 6 . Laplace transforms give: s 2 Y ( s ) - 2 s + 6 - sY ( s ) + 2 - 12 Y ( s ) = 9 e - 3 s or Y ( s ) = 2( s - 4) ( s - 4)( s + 3) + 9 e - 3 s ( s - 4)( s + 3) . Partial fractions on the second expression give 9 ( s - 4)( s + 3) = A s - 4 + B s + 3 or 9 = A ( s + 3) + B ( s - 4) . From s = 4, A = 9 7 . From s = - 3, B = - 9 7 . It follows that Y ( s ) = 2 s + 3 + 9 7 e - 3 s s - 4 - e - 3 s s + 3 . The inverse Laplace transform gives y ( t ) = 2 e - 3 t + 9 7 u 3 ( t ) e 4( t - 3) - e - 3( t - 3) . c. Consider y 00 + 2 y 0 + 5 y = 5 , 0 t < 4 0 , t 4 y (0) = 0 , y 0 (0) = 4 . Laplace transforms give: s 2 Y ( s ) - 4 + 2 sY ( s ) + 5 Y ( s ) = 5 s - 5 e - 4 s s or Y ( s ) = 4 ( s + 1) 2 + 4 + 5 - 5 e - 4 s s (( s + 1) 2 + 4) . Partial fractions on the second expression give 5 s (( s + 1) 2 + 4) = A s + B ( s + 1) + 2 C ( s + 1) 2 + 4 or 5 = A (( s + 1) 2 + 4) + Bs ( s + 1) + 2 Cs. From s = 0, A = 1. From s 2 , B = - A = - 1. From s = - 1, 5 = 4 A - 2 C or C = - 1 2 . It follows that Y ( s ) = 4 ( s + 1) 2 + 4 + 1 s - s + 1 ( s + 1) 2 + 4 - 1 2 2 ( s + 1) 2 + 4 (1 - e - 4 s ) . The inverse Laplace transform gives y ( t ) = 1 + e - t 3 2 sin(2 t ) - cos(2 t ) + u 4 ( t ) 1 - e - ( t - 4) cos(2( t - 4)) - 1 2 sin(2( t - 4)) . d. Consider y 00 + 4 y = cos( t ) , y (0) = 1 , y 0 (0) = 0 . Laplace transforms give: s 2 Y ( s ) - s + 4 Y ( s ) = s s 2 + 1 or Y ( s ) = s s 2 + 4 + s ( s 2 + 1)( s 2 + 4) . Partial fractions on the second expression give s ( s 2 + 1)( s 2 + 4) = As + B s 2 + 1 + Cs + 2 D s 2 + 4 or s = ( As + B )( s 2 + 4) + ( Cs + 2 D )( s 2 + 1) . From s = i , i = 3( iA + B ), so A = 1 3 and B = 0. From s = 2 i , 2 i = - 3(2 iC + 2 D ), so C = - 1 3 and D = 0. It follows that Y ( s ) = s s 2 + 4 + 1 3 s s 2 + 1 - 1 3 s s 2 + 4 = 2 3 s s 2 + 4 + 1 3 s s 2 + 1 . The inverse Laplace transform gives y ( t ) = 2 3 cos(2 t ) + 1 3 cos( t ) . e. Consider y 00 + 2 y 0 - 3 y = t 2 , 0 t < 3 0 , t 3 y (0) = 0 , y 0 (0) = 0 . The function on the RHS can be written f ( t ) = t 2 - t 2 u 3 ( t ) = t 2 - ( t - 3) 2 u 3 ( t ) - 6( t - 3) u 3 ( t ) - 9 u 3 ( t ). Subscribe to view the full document. • Fall '08
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