Applied Differential Equations I
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16. The curves x^2 + c y^2 = 1 are ellipses (c>0), vertical lines (c=0), and hyperbolas (c<0), all going through (-1,0) or (1,0). (The ellipses go through both.)
22. (a) c_1=5/3, c_2=1/3. (b) c_1=(2/3)e^(-1), c_2=(1/3)e^2.
28. No, the initial value problem is not guaranteed to have a unique solution. f(x,y) is continuous at the point in question, but the partial derivative of f with respect to y is not.
14. The solutions are y=x (isocline for slope 1), y=-x (isocline for slope -1), and a lot of hyperbolas.
18. y approaches 0 as x approaches infinity.
(d) Equilibria at y = 1, 2, 3. Arrows point right in (1, 2) and in (3, infinity). Arrows point left in (-infinity, 1) and in (2, 3).
(e) Equilibria at y = 1, 2, 3. Arrows point right in (1, 2) and in (2, 3). Arrows point left in (-infinity, 1) and in (3, infinity). Since arrows point right in (2, 3), the solution that starts at y=2.1 approaches 3 as t increases.
(f) Equilibria at y = any multiple of pi. Arrows point right in (0, pi), (2pi, 3pi), (4pi, 5pi), ..., and in ..., (-5pi, -4pi), (-3pi, -2pi), (-pi,0). They point left in (pi, 2pi), (3pi, 4pi), (5pi, 6pi), ..., and in ..., (-6pi, -5pi), (-4pi, -3pi), (-2pi,-pi). The equilibrium at 0 is of mixed type. Attractors: pi, 3pi, ..., and ..., -4pi, -2pi. Repellers: 2pi, 4pi, ..., and ..., -5pi, -3pi, -pi.
25. (a) Why rewrite the law of gravitational attraction this way? Because when you write it this way, dv/dt when r=R (i.e., at the Earth's surface) is just g, which is familiar quantity to us (see part (e)).
(b) Hint: dv/dt = dv/dr times dr/dt. What is the point of this substitution? In the differential equation of part (a) there were three changing quantities, t, v, and r. Now there are only two, v and r.
(c) Solve the differential equation of part (b) with the initial condition v(R)=v_0. Your answer will give you (implicitly) v as a function of r, i.e., it will tell you the object's velocity should it reach the distance r from the center of the earth.
2. c1 cos (2t) + c2 sin (2t) - (1/4) cos (2t) ln | sec (2t) + tan (2t) |
4. c1 cos (4 theta) + c2 sin (4 theta) + (theta/4) sin (4 theta) + (1/16) cos (4 theta) ln | cos (4 theta) |
1. The eigenvector u_2 should be col(2s,-s). The book has an extra s.
6. Eigenvectors are r_1 = r_2 = -1 and r_3 = 2 with associated eigenvectors u_1 = s col(-1,1,0), u_2 = u col(-1,0,1), and u_3 = s col(1,1,1).
16. c_1 e^(-10t) col(2,0,-1) + c_2 e^(5t) col(0,1,0) + c_3 e^(5t) col(1,0,2)
12. 9x^2 + 4y^2 = c. The phase portrait consists of concentric ellipses. The flow direction is counterclockwise.
29. (a) y=Kx^3, or x=ky^(1/3). The phase portrait looks like Figure 5.10, but the arrows are reversed.
The problem dx/dt = x/2, dy/dt = -2y: y=Kx^(-4), or x=ky^(-1/4). The phase portrait looks like Figure 12.6(a).
3. Spiral attractor.
5. Attracting node.
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