quiz2-sol

# quiz2-sol - Name MA 366 Spring 2010 Quiz 2(1(6 points...

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Unformatted text preview: Name: MA 366, Spring 2010, Quiz 2 (1) (6 points) Determine the values of r for which y ( t ) = t r , t > 0, is a solution of the equation t 2 y ′′- ty ′- 3 y = 0 . Solution: If y ( t ) = t r , then y ′ ( t ) = rt r − 1 and y ′′ ( t ) = r ( r- 1) t r − 2 . So we have 0 = t 2 × r ( r- 1) t r − 2- t × rt r − 1- 3 t r = t r [ r ( r- 1)- r- 3] = t r ( r 2- 2 r- 3) . t r is never zero for t > 0, so we require 0 = r 2- 2 r- 3 = ( r- 3)( r +1), so r =- 1 or r = 3. (2) (6 points) For each number a , determine the solution of y ′ + 2 y = t, y (0) = a. Solution: The integrating factor is μ ( t ) = exp( integraltext 2 dt ) = exp(2 t ) = e 2 t , so e 2 t y ′ + 2 e 2 t y = ( e 2 t y ) ′ = te 2 t . Integrating, we see that e 2 t y = integraldisplay te 2 t dt = t 1 2 e 2 t- integraldisplay 1 2 e 2 t dt = t 2 e 2 t- 1 4 e 2 t + C, so y = t 2- 1 4 + Ce − 2 t . The initial condition y (0) = a implies a = 2- 1 4 + Ce 2 × , so C = a + 1 4 . So y = t 2- 1 4 + parenleftBig a +...
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quiz2-sol - Name MA 366 Spring 2010 Quiz 2(1(6 points...

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