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AMATH 351 Homework 2 Keys July 8, 2009 Section2.1 14,16,19,31,40 Section2.4 3,12,15,29 Section2.6 8,21,23,24 Section 2.1 In each of Problems 13 through 20 nd the solution of the given initial value problem. 14. y + 2 y = te- 2 t , y (1) = 0 Multiply ( t ) = e 2 t , then e 2 t y + 2 e 2 t y = t ( e 2 t y ) = t e 2 t y = t 2 2 + C y = e- 2 t t 2 2 + C Apply initial condition y (1) = ( 1 2 + C ) e- 2 = 0 = C =- 1 2 . So the solution is y = e- 2 t t 2 2- 1 2 16. y + (2 /t ) y = (cos t ) /t 2 , y ( ) = 0 , t > Integrating factor ( t ) = e 2 t dt = t 2 . Multiply it to the DE, we get y t 2 + 2 ty = cos t ( t 2 y ) = cos t t 2 y = sin t + C y = t- 2 (sin t + C ) 1 Apply initial condition, we have y ( ) = - 2 (sin + C ) = C 2 = 0 = C = 0 Plug C back into the general solution, y ( t ) = t- 2 sin t 19. t 3 y + 4 t 2 y = e- t , y (- 1) = 0 , t < First write it in the standard form y + 4 t y = t- 3 e- t ( t ) = e 4 t dt = t 4 then multiply it, t 4 y + 4 t 3 y = te- t ( t 4 y ) = t e- t t 4 y = te- t dt =- te- t- e- t + C y = e- 4 (- te- t- e- t + C ) Apply initial condition y (- 1) = e- 4 ( e- e + C ) = e- 4 C = 0 = C = 0 Plug it into the general solution, we get y ( t ) =- e- 4- t ( t + 1) 31. Consider the initial value problem y- 3 2 y = 3 t + 2 e t , y (0) = y . Find the value of y that separates solutions that grow positively as t from those that grow negatively. How does the solution that corresponds to this critical value of y behave as t ?... View Full Document

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