Unformatted text preview: cos(8 t ) Now, use initial conditions: u (0) = c 1 = 1 / 4 and u (0) = 8 c 22 = 0 . i.e. c 2 = 1 / 4 . Final answer becomes, u = 1 / 4 cos(8 t ) + 1 / 4 sin(8 t )2 t cos(8 t ) 2. (6 pts.) Find the solution of the given boundary value problem below or show that it has no solution. y 00 + y = x, y (0) = 0 , y ( π ) = 0 SOLUTION: Homogeneous solution: r 2 +1 = 0 and r = ± i . y c = c 1 cos( x )+ c 2 sin( x ). Particular solution: Y ( x ) = Ax + B . Y = A and Y 00 = 0. A = 1 and B = 0. General solution: y = c 1 cos( x ) + c 2 sin( x ) + x . Use initial conditions: y (0) = c 1 = 0. Thus, y ( π ) = c 2 sin( π ) + π = 0. This gives us π = 0. Impossible. Thus there is no solution....
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 Spring '07
 COSTIN
 Math, Differential Equations, Equations, Partial Differential Equations, Sin, Cos, Boundary value problem

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