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Transform the given initial value problem into an

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Transform the given initial value problem into an algebraic equation in { y } and solve for { y }. Do not take inverse transforms and do not attempt to combine terms over a common denominator. Be very careful. 3 y ( t ) 5 y ( t ) 7 y ( t ) sin(2 t ) ; y (0) 4 , y (0) 6 . Applying the Laplace transform operator to BOTH SIDES OF THE ODE, using the two initial conditions, and then solving for the transform of y should reveal that { y ( t )}( s ) 1 3 s 2 5 s 7 12 s 2 2 s 2 4 [A common error: Failure to parenthesize the first and second derivative’s transform correctly.] ______________________________________________________________________ 9. (10 pts.) The solution to a certain linear ordinary differential equation with coefficient functions that are analytic at x 0 = 0 is of the form y ( x ) n 0 c n x n where the coefficients satisfy the following equations: and c 2 0, c 0 3 c 1 6 c 3 0, c n 2 n ( n 2) c n c n 1 ( n 2)( n 1) for all n 2. Determine the exact numerical value of the coefficients c 0 , c 1 , c 2 , c 3 , and c 4 for the particular solution that satisfies the initial conditions y (0) = 0 and y (0) = 1. c 0 y (0) 0 c 1 y (0) 1 c 2 0 c 3 c 0 3 c 1 6 1 2 c 4 8 c 2 c 1 (4)(3) 1 12 _______________________________________________________________________ Silly 10 Point Bonus: Write the function f ( t ) cos 5 ( bt ) as a linear combination of sine and/or cosines actually appearing in the Laplace transform table provided. Say where your work is, for it won’t fit here. From Euler’s equation you should have readily that . cos( bt ) 1 2 e ibt e ibt From the binomial theorem or Pascal’s triangle or whatever, you should get ( a b ) 5 a 5 5 a 4 b 10 a 3 b 2 10 a 2 b 3 5 ab 4 b 5 .
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