Your work so far is fine (watch the capital P on the LHS, though). On the RHS, you can get a common denominator and add. The LHS factors, and then you can solve for L(y). What do you get?

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- January 3rd 2011, 02:10 AM #1

- January 3rd 2011, 05:05 AM #2

- January 3rd 2011, 12:28 PM #3

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Oh, dear, oh dear, oh, dear! I really dislike the Laplace Transform! It can be used in two diametrically opposite ways- to "look up" solutions to non-homogeneous equations or to assert a form of a solution in very abstract theorems. The first is more easily done in other ways and the second, undergraduates never see.

I would look at that equation and immediately write down it characteristic equation: . It's easy to see, by inspection, that r= 1 is a root of that and that so that r= 1 is a double root and r= -1 is a root. That tells us that the general solution to the associated homogeneous equation is .

The "non-homogeneos" part is . Since is already a solution to the associated homogeneous equation, we try a solution of the form . Then , , and .

Putting those into the differential equation will give you equations for A and B.

- January 3rd 2011, 04:50 PM #4

- January 3rd 2011, 05:07 PM #5

- January 3rd 2011, 05:38 PM #6

- January 3rd 2011, 10:29 PM #7

- January 4th 2011, 02:26 AM #8

- January 4th 2011, 02:13 PM #9

- January 6th 2011, 02:42 AM #10