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soln_hw6

# soln_hw6 - Solutions to Homework 6 Section 3.6 8 The...

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Solutions to Homework 6 Section 3.6 8. The assumed form is Y = ( At + B ) sin(2 t )+( Ct + D ) cos(2 t ), which is appropriate for both terms appearing on the right hand side of the equation. Since none of the terms appearing in Y are solutions of the homogenous equation, we do not need to modify Y . Calculating Y and Y we have Y prime = A sin(2 t )+ C cos(2 t )+2( At + B ) cos(2 t ) - 2( Ct + D ) sin(2 t ) and Y = - 4 C sin(2 t ) - 4( At + B ) sin(2 t ) - 4( Ct + D ) cos(2 t ). Thus Y + Y = - 3 At sin(2 t ) - (3 B + 4 C ) sin(2 t ) - 3( Ct + D ) cos(2 t ). Equating like coefficients yields A = 0 , - 3 B - 4 C = 3 , - 3 C = 1, and - 3 D = 0. Hence Y ( t ) = - (5 / 9) sin(2 t ) - (1 / 3) t cos(2 t ). 18. The homogeneous solution is y c ( t ) = c 1 e - t cos(2 t )+ c 2 e - t sin(2 t ). Set Y = Ate - t cos(2 t )+ Bte - t sin(2 t ). After comparing coefficients, we get Y = te - t sin(2 t ). Hence the general so- lution is y ( t ) = c 1 e - t cos(2 t ) + c 2 e - t sin(2 t ) + te - t sin(2 t ) . (1) Invoking the initial conditions, c 1 = 1 and c 2 = 1 / 2. Section 3.7 6. The homogeneous solution is y c = c 1 cos(3 t ) + c 2 sin(3 t ) and the two functions y 1 = cos(3 t ) and y 2 = sin(3 t ). So, W ( y 1 , y 2 ) = 3. The particular solution is

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soln_hw6 - Solutions to Homework 6 Section 3.6 8 The...

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