{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Worksheet 9 Solutions

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

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

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

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 ]}