{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Worksheet 9 Solutions

Worksheet 9 Solutions - MA 2930 Worksheet 9 1 Find the...

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

View Full Document Right Arrow Icon
MA 2930, March 30, 2011 Worksheet 9 1. Find the solution (if possible) of y 00 + y = 0 for the following sets of boundary values: (a) y (0) = 0 , y ( π ) = 0 (b) y (0) = 0 , y 0 ( π ) = 0 (c) y 0 (0) = 0 , y ( π ) = 0 (d) y 0 (0) = 0 , y 0 ( π ) = 0 How would you make sense of these results? (Hint: think graphically!) For which of these boundary values would the equation y 00 + y = cos 2 x be solvable? The general solution of the differential equation is y ( x ) = c 1 cos x + c 2 sin x Now let’s apply the boundary conditions: (a) y (0) = c 1 = 0 and y ( π ) = - c 1 = 0. So the solutions are y ( x ) = c 2 sin x . Graphically this makes sense because any sine with period 2 π can satisfy the boundary conditions, but no cosine. Note that there are infinitely many solutions, illustrating that there is no uniqueness result for BVPs. (b) y (0) = c 1 = 0 and y 0 ( π ) = - c 2 = 0. So the only solution is y ( x ) = 0. (c) y 0 (0) = c 2 = 0 and y ( π ) = - c 1 = 0. So the only solution is again y ( x ) = 0. (d) y 0 (0) = c 2 = 0 and y 0 ( π ) = - c 2 = 0. So the solutions are y ( x ) = c 1 cos x . The general solution of the differential equation of y 00 + y = cos 2 x is y ( x ) = c 1 cos x + c 2 sin x - 1 / 3 cos 2 x Now let’s apply the boundary conditions: (a) y (0) = c 1 - 1 / 3 = 0 and y ( π ) = - c 1 - 1 / 3 = 0. So there is no solution. Graphically this makes sense because no cosine can satisfy these
Background image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}