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Winter 2006 - Reynold's Class - Final Exam (Version B)

# Winter 2006 - Reynold's Class - Final Exam (Version B) -...

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1. [ 8 points ] Classify each of the following equations as (i) either ordinary or partial differential equations, (ii) either first, second or third order, (iii) either linear or nonlinear: a. [ 4 pts ] d 2 y dt 2 + y cos( t ) = ln( t 2 ) Solution: ordinary (1 pt), 2nd-order (1 pt), linear (2 pts). b. [ 4 pts ] dy dt + e 2 t cosh( y ) = 4 Solution: ordinary (1 pt), 1st-order (1 pt), nonlinear (2 pts).
2. [ 8 points ] Determine whether each of the two functions is a solution to the given differential equation: a. [ 4 pts ] y + 4 y + 3 y = t ; y 1 ( t ) = t 3 , y 2 ( t ) = e - t + t 3 Solution: y 1 is a solution to the differential equation (2 pts). y 2 is a solution to the differential equation (2 pts). b. [ 4 pts ] 4 2 y ∂x 2 = ∂y ∂t ; y 1 ( x, t ) = e - 4 t sin( x ) , y 2 ( x, t ) = e - t sin x 2 Solution: y 1 is a solution to the differential equation (2 pts). y 2 is a solution to the differential equation (2 pts).

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3. [ 9 points ] Solve the following linear first-order initial value problem y + 2 y = e - 2 t , y (1) = 0 Solution: We first compute the integrating factor (2 pts setup, 1 pt correct μ ), μ ( t ) = exp 2 dt = e 2 t . We then compute the general solution y as (2 pts setup, 1 pt correct general solution) y ( t ) = 1 μ μe - 2 t dt + c = e - 2 t e 2 t e - 2 t dt + c = te - 2 t + ce - 2 t . We insert the initial condition to solve for the constant c , 0 = y (1) = e - 2 + ce - 2 obtaining c = - 1 (2 pts for correct c), so the particular solution to the differential equation is (1 pt for final solution) y ( t ) = te - 2 t - e - 2 t .
4.

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