### MA 341H-040 Applied Differential Equations I

Mathematics

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#### Section 1.2

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.
26. 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.

#### Section 1.3

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.

#### Section 2.4

16. e^(xy) - x/y = C.

#### Section 9.5

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)

#### Section 5.4

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).