2065s10ex1a

2065s10ex1a - Name: Solutions Exam 1 Instructions. Answer...

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Name: Solutions Exam 1 Instructions. Answer each of the questions on your own paper. Be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without supporting work. Put your name on each page of your paper. A table of Laplace transforms and a short table of integrals are appended to the exam. 1. [17 Points] Find the general solution of: y 0 = 6 t ( y ¡ 1) 2 = 3 . I Solution. This equation is separable. After separating variables, it becomes ( y ¡ 1) ¡ 2 = 3 y 0 = 6 t , which in difierential form is ( y ¡ 1) ¡ 2 = 3 dy = 6 tdt and integration then gives 3( y ¡ 1) 1 = 3 = 3 t 2 + C . Dividing by 3 gives ( y ¡ 1) 1 = 3 = t 2 + K , where K = C= 3 is an arbitrary constant. Cubing both sides and solving for y gives y = ( t 2 + K ) 3 + 1 : J 2. [17 Points] Find the general solution of: y 0 ¡ 4 y = 3 e 4 t + 4 e 3 t . I Solution. This equation is linear with coe–cient function p ( t ) = ¡ 4 so that an integrating factor is given by ( t ) = e R ¡ 4 dt = e ¡ 4 t . Multiplication of the difierential equation by the integrating factor gives e ¡ 4 t y 0 ¡ 4 e ¡ 4 t y = 3 + 4 e ¡ t ; and the left hand side is recognized (by the choice of ( t )) as a perfect derivative: d dt ( e ¡ 4 t y ) = 3 + 4 e ¡ t : Integration then gives e ¡ 4 t y = 3 t ¡ 4 e ¡ t + C; where C is an integration constant. Multiplying through by e 4 t gives y = 3 te 4 t ¡ 4 e 3 t + Ce ¡ 4 t : J 3. [17 Points] Solve the initial value problem: y 0 + 6 t y = 11 t 4 , y (1) = 3.
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2065s10ex1a - Name: Solutions Exam 1 Instructions. Answer...

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