{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

quiz6-sol

# quiz6-sol - Name MA 366 Spring 2010 Quiz 6(1(5 points...

This preview shows pages 1–2. Sign up to view the full content.

Name: MA 366, Spring 2010, Quiz 6 (1) (5 points) Determine the Laplace transform of f ( t ) = t × u π ( t ). Solution: To make it match line 13 in the table we write f ( t ) = t × u π ( t ) = ( t - π + π ) u π ( t ) = ( t - π ) u π ( t ) + πu π ( t ) . So we can use lines 13 and 3 (for the Laplace transforms of 1 and t ) to get F ( s ) = e πs 1 s 2 + πe πs 1 s . (5 points) Determine the inverse Laplace transform of F ( s ) = s s 2 + 4 s + 13 . Solution: We can’t factor the denominator, so we write F ( s ) = s ( s + 2) 2 + 9 . This looks a bit like lines 9 and 10, but F doesn’t have the right numerator for either of them. So we write F ( s ) = s + 2 - 2 ( s + 2) 2 + 9 = s + 2 ( s + 2) 2 + 9 - 2 ( s + 2) 2 + 9 = s + 2 ( s + 2) 2 + 9 - 2 3 3 ( s + 2) 2 + 9 , and now the first term matches line 10 and the second matches line 9, so f ( t ) = e 2 t cos 3 t - 2 3 e 2 t sin 3 t. (2) (10 points) Determine the solution y ( t ) of the differential equation y ( t ) - 2 y ( t ) = u 2 ( t ) , y (0) = 1 . Solution: We let Y ( s ) = L [ y ( t )], so L [ y ( t )] = sY ( s ) - y (0). Taking the Laplace transform of the equation we get from line 12 e 2 s s = sY ( s ) - y (0) - 2 Y ( s ) = Y ( s )( s - 2) - 1 , so Y ( s ) = 1 s - 2 + e 2 s s ( s -

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

quiz6-sol - Name MA 366 Spring 2010 Quiz 6(1(5 points...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online