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**Unformatted text preview: **Exercises 5.4
6. We see by Deﬁnition 1 on page 266 of the text that we must solve the system of equations given by y 2 − 3 y + 2 = 0, (x − 1)(y − 2) = 0. By factoring the ﬁrst equation above, we ﬁnd that this system becomes (y − 1)(y − 2) = 0, (x − 1)(y − 2) = 0. Thus, we observe that if y = 2 and x is any constant, then the system of diﬀerential equations given in this problem will be satisﬁed. Therefore, one family of critical points is given by the line y = 2. If y = 2, then the system of equations above simpliﬁes to y − 1 = 0, and x − 1 = 0. Hence, another critical point is the point (1, 1). 7. Here f (x, y ) = y − 1, g (x, y ) = ex+y . Thus the phase plane equation becomes dy ex+y ex ey = = . dx y−1 y−1 Separating variables yields (y − 1)e−y dy = ex dx ⇒ −ye−y + C = ex ⇒ or (y − 1)e−y dy = ex + ye−y = C. ex dx 9. The phase plane equation for this system is g (x, y ) ex + y dy = = . dx f (x, y ) 2y − x We rewrite this equation in symmetric form, −(ex + y ) dx + (2y − x) dy = 0, and check it for exactness. ∂ ∂M = [−(ex + y )] = −1, ∂y ∂y 295 ...

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