칼큘러스 james stewart 6th solution

소개글

챕터11

목차

없음

본문내용

1. The corresponding plane autonomous system is
x
= y, y
= −9 sin x.
If (x, y) is a critical point, y = 0 and −9 sin x = 0. Therefore x = ±nπ and so the critical points are (±nπ, 0)
for n =0, 1, 2, . . . .
2. The corresponding plane autonomous system is
x
= y, y
= −2x − y2.
If (x, y) is a critical point, then y = 0 and so −2x − y2 = −2x = 0. Therefore (0, 0) is the sole critical point.
3. The corresponding plane autonomous system is
x
= y, y
= x2 − y(1 − x3).
If (x, y) is a critical point, y = 0 and so x2 − y(1 − x3) = x2 = 0. Therefore (0, 0) is the sole critical point.
4. The corresponding plane autonomous system is
x
= y, y
= −4 x
1 + x2 − 2y.
If (x, y) is a critical point, y = 0 and so −4x/(1 + x2) − 2(0) = 0. Therefore x = 0 and so (0, 0) is the sole
critical point.

참고 자료

없음