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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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