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Chapter 4: Problem 3

For \(y^{\prime}=y-y^{2}\), find the general solution corresponding to \(y(0)=y_{0}\). Provide specific solutions for the following initial conditions and sketch them: a. \(y(0)=0.25\), b. \(y(0)=1.5\), and c. \(y(0)=-0.5\).

Short Answer

Expert verified
The specific solutions for the initial conditions are: a. \(\frac{y}{1-y}=\frac{1}{3}e^{x}\), b. \(\frac{y}{1-y}=-3e^{x}\), and c. \(\frac{y}{1-y}=-\frac{1}{3}e^{x}\). The functions are bound between 0 and 1 and approach 1 as \(x\) goes to infinity.

Step by step solution

01

Separating the variables

Write the differential equation in separable form. Rearrange the equation as \(\frac{{dy}}{{(1-y)y}}=dx\). This separates the variables \(y\) and \(x\) on both sides.
02

Obtain the integral

Integrate both sides with respect to \(y\) and \(x\) respectively. The integral gives \(-\log|1-y| + \log|y| = x + C\). Simplifying it gives \(\frac{y}{1-y}=Ce^{x}\) where \(C=\pm e^{C'}\). This is the general form.
03

Solve for initial condition \(y_{0}=0.25\)

Substitute \(y_{0}=0.25\) and \(x=0\) into the general equation \(\frac{y}{1-y}=Ce^{x}\). This gives \(C=\frac{0.25}{1-0.25} = \frac{1}{3}\). So the specific solution for this initial condition is \(\frac{y}{1-y}=\frac{1}{3}e^{x}\).
04

Solve for initial condition \(y_{0}=1.5\)

Substitute \(y_{0}=1.5\) and \(x=0\) into the general equation, giving \(C=\frac{1.5}{1-1.5}=-3\). So the specific solution for this initial condition is \(\frac{y}{1-y}=-3e^{x}\).
05

Solve for initial condition \(y_{0}=-0.5\)

Substitute \(y_{0}=-0.5\) into the general equation and \(x=0\), giving \(C=\frac{-0.5}{1+0.5}=-\frac{1}{3}\). So the specific solution for this initial condition is \(\frac{y}{1-y}=-\frac{1}{3}e^{x}\).
06

Sketching the solutions

The graph will show three curves representing three different initial conditions. All the solutions will gradually approach \(y=1\) from either above or below as \(x\) approaches infinity, depending on whether \(y_{0} > 1\) or \(y_{0} < 1\). The function is not defined at \(y_{0}=1\).

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Most popular questions from this chapter

8\. Find the fixed points of the following systems. Linearize the system about each fixed point and determine the nature and stability in the neighborhood of each fixed point, when possible. Verify your findings by plotting phase portraits using a computer. a. $$ \begin{aligned} &x^{\prime}=x(100-x-2 y) \\ &y^{\prime}=y(150-x-6 y) \end{aligned} $$ b. $$ \begin{aligned} x^{\prime} &=x+x^{3} \\ y^{\prime} &=y+y^{3} \end{aligned} $$ C. $$ \begin{aligned} &x^{\prime}=x-x^{2}+x y \\ &y^{\prime}=2 y-x y-6 y^{2} \end{aligned} $$ d. $$ \begin{aligned} &x^{\prime}=-2 x y \\ &y^{\prime}=-x+y+x y-y^{3} \end{aligned} $$.

.Derive the first integral of the Lotka-Volterra system, \(a \ln y+d \ln x-\) \(c x-b y=C\).

Consider the family of differential equations \(x^{\prime}=x^{3}+\delta x^{2}-\mu x\) a. Sketch a bifurcation diagram in the \(x \mu\)-plane for \(\delta=0\). b. Sketch a bifurcation diagram in the \(x \mu\)-plane for \(\delta>0\). Hint: Pick a few values of \(\delta\) and \(\mu\) in order to get a feel for how this system behaves.

Consider the system $$ \begin{aligned} x^{\prime} &=-y+x\left[\mu-x^{2}-y^{2}\right] \\ y^{\prime} &=x+y\left[\mu-x^{2}-y^{2}\right] \end{aligned} $$ Rewrite this system in polar form. Look at the behavior of the \(r\) equation and construct a bifurcation diagram in \(\mu r\) space. What might this diagram look like in the three-dimensional \(\mu x y\) space? (Think about the symmetry in this problem.) This leads to what is called a Hopf bifurcation.

Plot phase portraits for the Lienard system $$ \begin{aligned} x^{\prime} &=y-\mu\left(x^{3}-x\right) \\ y^{\prime} &=-x \end{aligned} $$ for a small and a not so small value of \(\mu\). Describe what happens as one varies \(\mu\).
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