s251ex1(sp02)

# s251ex1(sp02) - i y ( t ) = C 1 e 2 t cos t + C 2 e 2 t sin...

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Math 251 Spring 2002 Midterm Exam 1 1. (b) 2. (d) because between equilibrium solutions y = 0 and y = 1 the right-hand side of the equation is negative, therefore the solution starting at 9 10 must decrease to a nearest equilibrium solution, which is 0. 3. (d) since we need to check when ∂y ( e x sin y + bx 2 y 2 ) = ∂x ( e x cos y + x 3 y ) 4. (c) 5. (d) because after dividing the whole equation by t 2 - 1 we will get discontinuities of the coeFcients at - 1 and 1, and the initial point 5 lies to the right of the last point of discontinuity. 6. dy dx = - x y , y (0) = - 1 ydy = - xdx 1 2 y 2 = - 1 2 x 2 + const y 2 = - x 2 + const ( - 1) 2 = - 0 2 + const const = 1 y ( x ) = - - x 2 + 1 7. t 2 dy dt + 3 ty = e t , t > 0 dy dt + 3 t y = 1 t 2 e t μ ( t ) = e R 3 t dt = e 3 ln t = t 3 y ( t ) = R t 3 1 t 2 e t dt t 3 = te t - e t + C t 3 = 1 t 2 e t - 1 t 3 e t + C t 3 8. ±irst not that the equation is exact. Now ψ = Z (2 xy - 3 x 2 ) dx = x 2 y - x 3 + C ( y ) ∂y ( x 2 y - x 3 + C ( y )) = x 2 + C 0 ( y ) = x 2 + 2 y C 0 ( y ) = 2 y C ( y ) = y 2 + const ψ = x 2 y - x 3 + y 2 = const 1

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9. (a) y 00 - 4 y 0 + 4 y = 0 λ 2 - 4 λ + 4 = 0 λ 1 , 2 = 2 y ( t ) = C 1 e 2 t + C 2 te 2 t (b) y 00 - 4 y 0 + 5 y = 0 λ 2 - 4 λ + 5 = 0 λ 1 , 2 = 2
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Unformatted text preview: i y ( t ) = C 1 e 2 t cos t + C 2 e 2 t sin t 10. y 00-4 y = 0 , y (0) = 4 , y (0) = 4 2-4 = 0 1 = 2 2 =-2 y ( t ) = C 1 e 2 t + C 2 e-2 t y (0) = C 1 + C 2 = 4 y (0) = 2 C 1-2 C 2 = 4 C 1 = 3 C 2 = 1 y ( t ) = 3 e 2 t + e-2 t 11. dQ dt = 2 5-2 Q ( t ) 2 , Q (0) = 15 dQ dt + Q ( t ) = 10 Q ( t ) = Ce-t + 10 Q (0) = C + 10 = 15 C = 5 Q ( t ) = 5 e-t + 10 Q (ln 5) = 5 e-ln5 + 10 = 11 2 12. m dv dt = mg-2 v v (0) = 0 dv dt + v = 10 v ( t ) = Ce-t + 10 v (0) = C + 10 = 0 C =-10 v ( t ) =-10 e-t + 10 distance = Z 2 v ( t ) dt = Z 2 (-10 e-t + 10) dt = (10 e-t + 10 t ) 2 = 10 + 10 e-2 3...
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## s251ex1(sp02) - i y ( t ) = C 1 e 2 t cos t + C 2 e 2 t sin...

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