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Unformatted text preview: Math 20D Solutions: HW 3 Section 2.5 2. For this problem we do not need to solve the DE, only determine qualitative information about the solutions. We are given dy dt = ay + by 2 ,a > ,b > . To find equilibrium solutions, we set y = 0 = ay + by 2 and solve for y , getting y = 0 and y = a/b as the solutions. Since a and b are both positive, a/b is negative. So the plot of y with respect to y looks like an upwardopening parabola with zeros at a/b and 0. In general for an autonomous equation, an equilibrium y = y is stable if y = f ( y ) is decreasing as it crosses the y axis at y , and unstable if it is increasing at y . Looking at the graph reveals that ay + by 2 decreasing at y = a/b and increasing at y = 0, so a/b is stable and 0 is unstable. With this information it is not hard to draw the phase line and qualitative plots of solutions. 7. a) To find equilibrium points we set y = 0 = k (1 y ) 2 . As long as k is nonzero, this equation has only one solution, y = 1. b) The plot of y = f ( y ) = k (1 y ) 2 is clearly going to be an upward opening parabola with a vertex at y = 1. (Since k > 0 it is upward opening.) If y 6 = 1, y is always positive. This means on either side of the equilibrium y = 1, solutions increase. In other words, a solution with initial value less than 1 will converge to the equilibrium, while a solution with initial value greater than 1 will diverge to ∞ . This is a socalled “semistable” equilibrium. c) This is a separable equation, so we rearrange and integrate: Z dy k (1 y ) 2 = Z dt. Using usubstitution with u = 1 y and integrating gives 1 k (1 y ) = t + C. The initial condition is y (0) = y , so solving for C gives C = 1 / ( k (1 y )). Now we must solve 1 k (1 y ) = t + 1 k (1 y ) for y . This is just algebra and we eventually get y = y + (1 y ) kt 1 + (1 y ) kt ....
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 Spring '10
 Edgars
 Derivative, Constant of integration, ﬁnal implicit solution

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