s251ex1(sp04)

# s251ex1(sp04) - Math 251 Spring 2004 Midterm Exam I 1 f-1...

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Math 251 Spring 2004 Midterm Exam I 1. Both f ( t, y ) = 1 y - 3 t 2 and ∂f ∂y = - 1 ( y - 3 t 2 ) 2 have to be continuous. The only condition needed is that y 0 - 3 t 2 0 6 = 0 2. a) The equilibrium solutions are y = - 3 , 0 , 1. b) y = - 3 is unstable, y = 0 is (asymptotically) stable, and y = 1 is unstable. c) lim t →∞ y ( t ) = d) lim t →∞ y ( t ) = 0 3. a) M ( x, y ) = y 2 e xy 2 + 4 x 3 , N ( x, y ) = 2 xye xy 2 + 2 ∂M ∂y = 2 xy 3 e xy 2 + 2 ye xy 2 = ∂N ∂x So the equation is an exact equation. b) e xy 2 + x 4 + 2 y = 5 4. a) The roots of the characteristic equation r 2 - 4 r - 5 = 0 are r = - 1 , 5. Therefore, the functions y 1 = e - t and y 2 = e 5 t form a pair of fundamental solutions for this second order linear diFerential equation. b) General solution is y ( t ) = C 1 y 1 + C 2 y 2 = C 1 e - t + C 2 e 5 t . c) y ( t ) = 7 6 e - t + 5 6 e 5 t 5. a) (Solved as a separable equation.) e y = sin t + t + e 3 (implicit solution), or y = ln(sin t + t + e 3 ) (explicit solution).

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## This note was uploaded on 07/23/2008 for the course MATH 251 taught by Professor Chezhongyuan during the Spring '08 term at Penn State.

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s251ex1(sp04) - Math 251 Spring 2004 Midterm Exam I 1 f-1...

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