Chapter 8: Problem 12
Make a sketch of the population function (as a function of time) that results from the following growth rate functions. Assume the population at time \(t=0\) begins at some positive value.
Chapter 8: Problem 12
Make a sketch of the population function (as a function of time) that results from the following growth rate functions. Assume the population at time \(t=0\) begins at some positive value.
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Get started for freeAnalytical solution of the predator-prey equations The solution of the predator-prey equations $$x^{\prime}(t)=-a x+b x y, y^{\prime}(t)=c y-d x y$$ can be viewed as parametric equations that describe the solution curves. Assume that \(a, b, c,\) and \(d\) are positive constants and consider solutions in the first quadrant. a. Recalling that \(\frac{d y}{d x}=\frac{y^{\prime}(t)}{x^{\prime}(t)},\) divide the first equation by the second equation to obtain a separable differential equation in terms of \(x\) and \(y\) b. Show that the general solution can be written in the implicit form \(e^{d x+b y}=C x^{c} y^{a},\) where \(C\) is an arbitrary constant. c. Let \(a=0.8, b=0.4, c=0.9,\) and \(d=0.3 .\) Plot the solution curves for \(C=1.5,2,\) and \(2.5,\) and confirm that they are, in fact, closed curves. Use the graphing window \([0,9] \times[0,9]\)
Determine whether the following equations are separable. If so, solve the initial value problem. $$\frac{d y}{d x}=e^{x-y}, y(0)=\ln 3$$
What is a carrying capacity? Mathematically, how does it appear on the graph of a population function?
The equation \(y^{\prime}(t)+a y=b y^{p},\) where \(a, b,\) and \(p\) are real numbers, is called a Bernoulli equation. Unless \(p=1,\) the equation is nonlinear and would appear to be difficult to solve-except for a small miracle. By making the change of variables \(v(t)=(y(t))^{1-p},\) the equation can be made linear. Carry out the following steps. a. Letting \(v=y^{1-p},\) show that \(y^{\prime}(t)=\frac{y(t)^{p}}{1-p} v^{\prime}(t)\). b. Substitute this expression for \(y^{\prime}(t)\) into the differential equation and simplify to obtain the new (linear) equation \(v^{\prime}(t)+a(1-p) v=b(1-p),\) which can be solved using the methods of this section. The solution \(y\) of the original equation can then be found from \(v\).
Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection. A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. Use the following steps to find the orthogonal trajectories of the family of ellipses \(2 x^{2}+y^{2}=a^{2}\) a. Apply implicit differentiation to \(2 x^{2}+y^{2}=a^{2}\) to show that $$ \frac{d y}{d x}=\frac{-2 x}{y} $$ b. The family of trajectories orthogonal to \(2 x^{2}+y^{2}=a^{2}\) satisfies the differential equation \(\frac{d y}{d x}=\frac{y}{2 x} .\) Why? c. Solve the differential equation in part (b) to verify that \(y^{2}=e^{C}|x|\) and then explain why it follows that \(y^{2}=k x\) Therefore, the family of parabolas \(y^{2}=k x\) forms the orthogonal trajectories of the family of ellipses \(2 x^{2}+y^{2}=a^{2}\)
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