Practice exam 3 2019 PDF

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Practice exam 3 2019...

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Math 2552 – Practice problems for the final exam (Fall 2019) Note: The following practice problems only contain materials after the second midterm (Sec 5.5– Sec 5.8 and Sec 7.1–7.4). The final exam will be cumulative, with about a half of the points coming from these sections. To practice on earlier materials, please refer to the two sets of practice problems for the midterms. 1.

True or false: (1) For any continuous function f , we have

R5 3

f (t)δ(t − 1)dt = f (1).

TRUE

FALSE

(2) If f, g are both non-negative functions, then f ∗ g must be a non-negative function too. TRUE FALSE (3) For a nonlinear autonomous system, the type and stability of a critical point is always the same as those of its approximating linear system. TRUE FALSE (4) If the point (1, 2) is a center for the system x′ = F (x, y), y ′ = G(x, y), then the system x′ = −F (x, y ), y ′ = −G(x, y ) must have (1, 2) as a center. TRUE FALSE (5) If (x0 , y0 ) is an asymptotically stable critical point for an autonomous system, then every solution (x(t), y(t)) must converge to (x0 , y0 ) as t → +∞. TRUE FALSE 2.

Compute the Laplace transform of the following functions: ( 2, 0 ≤ t < 4; (1) f (t) = t 2 , t ≥ 4. (2) f (t) =

Z

t

e2(t−τ ) cos(3τ )dτ. 0

( cos t, 0 ≤ t < π; (3) f (t) = 0, t ≥ π. (4) f (t) = δ(t) − δ(t − 1) + δ(t − 2) − δ(t − 3) + δ(t − 4) − δ(t − 5) · · · (5) Let f be a periodic function given by the following graph. Find its Laplace transform. f(t) 1 0

1

2

3

−1

1

4

5

6

t

3.

Find the inverse Laplace transform for the following functions. 2s2 (1) F (s) = 2 s −4 (2) F (s) =

e−3s s(s2 + 1)

(3) F (s) =

2 2 − + s s3



5 2 3 + + 3 s s2 s



(4) F (s) = G(s)H (s), where G(s) =

4.

1 s2

and H(s) =

1 . s+3

Use convolution to find L−1 {F }.

Use the Laplace transform to solve the initial value problem y′′ − 2y′ − 3y = 2δ(t − 1),

5.

e−2s . Then sketch the graph of f (t) = L−1 {F (s)}(t).

y(0) = 2, y ′ (0) = 6.

( 2 1 π/4, the displacement function is given by A π A y(t) = 3 cos(2t) + sin(2t − ) = 3 cos(2t) − cos(2t) for t > π/4. 2 4 4 Since we want stop the movement for t > π/4, we can set A = 12. Y (s) =

(7) By setting y(t) = x′ (t), the 2nd order equation can be transformed into a first order system ( x′ (t) = y, y′ (t) = −c sin x − y.

If c 6= 0, its critical points are (kπ, 0) for any integer k. (If c = 0, all points of the form (a, 0) with any a would be a critical point).     0 1 0 1 Its Jacobian is J (x, y) = , hence its Jacobian at (0, 0) is J (0, 0) = , −c cos x −1 −c −1 √ whose eigenvalues are given by λ2 + λ + c = 0. So its eigenvalues are λ = − 1 2 ± 12−4c . In order for (0, 0) to be a nodal sink, we need two different negative real eigenvalues, hence we need to have 1 − 4c > 0 and c > 0, which gives 0 < c < 14 . (8) (1) There are four critical points in total, and they are (2, 2), (−2, 2), (2, −2) and (−2, −2). (2) The only critical point in  the {x < 0, y < 0} quadrant is (−2, −2). The Jacobian at this point is  −4 4 . Its eigenvalues are λ = −4 ± 4i, which corresponds to a spiral sink. To J (−2, −2) = −4 −4 sketch the we plug in a point (0, 1) into the approximating linear system   phase portrait, −4 4 u, so the trajectory at (0, 1) is going in the direction (4, −4). It is going towards u′ (t) = −4 −4 7

the right, so it means the trajectories of the spiral is spinning clockwise. So for the original nonlinear system, the trajectories near (−2, −2) is a spiral sink spinning clockwise. (The sketch is omitted here). (9) (1) The relationship is predator-prey, and polar bear (with corresponding population x) is the predator, while penguin is the prey. To see this, note that y has a positive effect on the growth of x, while x has a negative effect on the growth of y. (2) By setting the right hand sides equal to zero, we have ( x(−4 + 0.2y) = 0 y(4 − 0.1y − 0.1x) = 0, which has three solutions (0, 0), (0, 40), (20, 20). (3) The only  . The Jacobian is  have positive population  is (20, 20)  critical point where both species 0 4 −4 + 0.2y 0.2x , hence J (20, 20) = . Its eigenvalues are J (x, y) = −0.1y 4 − 0.2y − 0.1x −2 −2 √ λ = −1 ± 7i, hence it is a (stable) spiral sink. By plugging in a point (0, 1) into the linearized system, we know the linear system is a spiral sink spinning clockwise. Therefore for the nonlinear system, the phase portrait near (20, 20) is also spiral sink spinning clockwise. (The sketch is omitted here).

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