Question

Consider the equation

5/4 5/4

1. Find the trace ? and the determinant ?. Sketch the graphs of ? vs. b and ? vs. b. Then sketch the graph of the curve ? = ? (b), ? = ?(b) in the ??-plane along with ?^2 = 4?, ? = 0 and ? = 0(? ? 0).

2. Find the intersection points of the the curve ? = ?(b),? = ?(b) with ?^2 = 4?, ? = 0 and ? = 0(? ? 0) and the corresponding values of b. The corresponding values of b are called the critical values.

3. Find the general solution for each critical value.

4. Draw a phase portrait for a value slightly below the smallest critical value and a value slightly above the largest critical value. (You can pick any value slightly below the smallest critical value and any value slightly above the largest critical value)

Answer 1

1) the trace is (5/4 + b) + 5/4 = 5/2 + b. Determinant is (5/4+b)*5/4 - 5/4*b = 25/16.

The sketches:

(i)

Sketches of T vs b -10 -5 0 10

(ii) Determinant value

Sketches of Delta vs b CN -4 -2 0 4

(2) The intersection of T² = 4$\Delta$ => (5/4+b)² = 4*25/16 = 25/4

=> 25/16 + 5/2b + b² = 25/4

=> 16b² + 40b - 75 = 0

the roots are b = $\frac{-40+\sqrt{40^2-4*16*(-75)}}{2*16}$ ; b = 5/4 , 15/4

The graph is,

Sketches of T vs b CN 寸 -10 -5 0 10

(3) For any general T² = k*$\Delta$, we have,

(5/4 + b)² = k*25/16

=> 16b² + 40b + 25(1-k) = 0

the general solution will be, b = $\frac{-40+\sqrt{40^2-4*16*\{25(1-k)\}}}{2*16}$ .