Question

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

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

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.

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.

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 \(\frac{y}{1-y}=\frac{1}{3}e^{x}\).

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 \(\frac{y}{1-y}=-3e^{x}\).

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 \(\frac{y}{1-y}=-\frac{1}{3}e^{x}\).

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 \(y_{0}=1\).

Key concepts

Separable Differential Equations

Separable differential equations are a special type of differential equation in which the variables can be separated on either side of the equation. This is incredibly useful because it allows us to integrate each side independently, simplifying the process of finding a solution.
In the exercise, the equation \( y' = y - y^2 \) was transformed into \( \frac{dy}{(1-y)y} = dx \).
This rearrangement splits the equation into components of \( y \) and \( x \), making it easier to integrate.

Initial Conditions

Initial conditions are an important aspect of solving differential equations since they allow us to find specific solutions from a general one. After finding the general solution, you can substitute the initial conditions to get a particular solution for a specific scenario.
In the example, over three separate cases with initial conditions \( y(0) = 0.25 \), \( y(0) = 1.5 \), and \( y(0) = -0.5 \), the solution changes based on these initial values, represented by different constants \( C \). This illustrates how different starting points influence the behavior of the solution.

General Solution

The general solution to a differential equation is a family of solutions that contains arbitrary constants. This general form can be adapted to any specific solution by applying initial conditions.
In this case, the general solution is expressed as \( \frac{y}{1-y}=Ce^{x} \), which comes from integrating both sides of the separable equation.
These solutions are not complete until the constants are determined using given initial conditions.

Integral Calculus

Integral calculus is an essential mathematical tool used in solving differential equations. It involves finding the antiderivative or integral of a function. This process is crucial for solving separable differential equations, as seen in the exercise where both sides were integrated to derive the general solution.
The right-hand side was integrated with respect to \( x \), while the left was integrated with respect to \( y \), resulting in the simplified equation \( -\log|1-y| + \log|y| = x + C \).
Understanding integral calculus is key to understanding how solutions to differential equations are formed.