칼큘러스 james stewart 6th solution

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(b) Letting c1 = 1 and c2 = 0 we get x = 5e8t, y = 2e8t. Eliminating the parameter we find y = 2
5x, x > 0.
When c1 = −1 and c2 = 0 we find y = 2
5x, x < 0. Letting c1 = 0 and c2 = 1 we get x = e−10t, y = 4e−10t.
Eliminating the parameter we find y = 4x, x > 0. Letting c1 = 0 and c2 = −1 we find y = 4x, x < 0.
(c) The eigenvectors K1 = (5, 2) and K2 = (1, 4) are shown in the figure in part (a).
18. In Problem 2, letting c1 = 1 and c2 = 0 we get x = −2et, y = et.
Eliminating the parameter we find y = −1
2x, x < 0. When c1 = −1
and c2 = 0 we find y = −1
2x, x > 0. Letting c1 = 0 and c2 = 1 we
get x = e4t, y = e4t. Eliminating the parameter we find y = x, x > 0.
When c1 = 0 and c2 = −1 we find y = x, x < 0.
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10.2 Homogeneous Linear Systems
In Problem 4, letting c1 = 1 and c2 = 0 we get x = −2e−7t/2, y =
e−7t/2. Eliminating the parameter we find y = −1
2x, x < 0. When
c1 = −1 and c2 = 0 we find y = −1
2x, x > 0. Letting c1 = 0 and
c2 = 1 we get x = 4e−t, y = 3e−t. Eliminating the parameter we find
y = 3
4x, x > 0. When c1 = 0 and c2 = −1 we find y = 3
4x, x < 0.

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