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**Unformatted text preview: **Math 415 Name: Quiz 3A For the diﬀerential equation y = y 2 (1 − y 2 ), sketch the graph of f (y ) versus y , determine the equilibrium
points, and classify them. Draw the phase line and several solutions for −∞ < y0 < ∞ (where y (0) = y0 ). 1 .5 -1 -0.5 0 0.5 1 1. -1 Here we see that y (t) = −1 is an unstable solution, y (t) = 0 is semistable, and y (t) = 1 is stable. Solutions
below the line y = −1 will drop to −∞, those between 0 and 1 will asymptotically approach y = 1 as t → ∞
(and approach y = 0 as t → −∞), and those between −1 and 0 will approach 0. Solve the initial value problem
y = (3x2 + 2y − 1)dx + (4y + 2x)dy = 0, y (1) = 0, and determine (approximately) where the solution is valid.
The equation is exact as (3x2 +2y − 1)y = 2 = (4y +2x)x . So there is a function Ψ(x, y ) with Ψx = 3x2 +2y − 1
and Ψy = 4y + 2x. Integrating one of these we get
Ψ(x, y ) = 4y + 2x dy = 2y 2 + 2xy + h(x). So, on the one hand Ψx = 3x2 + 2y − 1, and on the other Ψx = 2y + h (x). Comparing the two tells us that
h (x) = 3x2 − 1, or h(x) = x3 − x (ignoring the constant for now). Since our solution should be Ψ(x, y ) = C ,
we get
2y 2 + 2xy + x3 − x = C. 1 ...

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