Question

Solve the initial-value problem $$ \begin{aligned} y^{\prime} &=\alpha y-\beta y^{2}, \\ y(0) &=y_{0} \end{aligned} $$ where \(\alpha\) and \(\beta\) are small positive numbers. Show that $$ \lim _{x \rightarrow \infty} \phi(x)=\left\\{\begin{array}{ll} \alpha / \beta & \text { for } y_{0}>0, \\ 0 & \text { for } y_{0}=0 . \end{array}\right. $$ For \(y_{0}<0, \phi\) is unbounded as \(x\) approaches a certain value depending on \(y_{0}\).

Short answer

In summary, when solving the given initial-value problem, we found that the limit of the solution as \(x\) approaches infinity is different depending on the value of the initial condition \(y_0\). Specifically, if \(y_0 > 0\), the limit is equal to \(\frac{\alpha}{\beta}\); if \(y_0 = 0\), the limit is 0; and for \(y_0 < 0\), the solution is unbounded as \(x\) approaches a certain value depending on \(y_0\).

Step-by-step solution

Solve the Differential Equation

First, we rewrite the given differential equation as a separable equation: $$\frac{dy}{dx}=\alpha y-\beta y^2$$ Now, let's separate the variables: $$\frac{dy}{\alpha y-\beta y^2} = dx$$ Next, we integrate both sides: $$\int \frac{dy}{\alpha y-\beta y^2} = \int dx$$

Integration and Implicit Solution

For the left side of the equation, we can perform a change of variables to simplify the integral. Let \(u=\alpha y - \beta y^2\). Then, \(du/dy = \alpha - 2\beta y\) or \(du=(-2\beta)dy(\frac{1}{2}\alpha - \beta y)\), so: $$\int \frac{dy}{\alpha y-\beta y^2} = -\frac{1}{2\beta} \int \frac{du}{u}$$ Now, we can continue integrating both sides: $$-\frac{1}{2\beta} \ln |u| =-\frac{1}{2\beta} \ln |\alpha y-\beta y^2 | = x+C$$ Exponentiating both sides, we get: $$|\alpha y-\beta y^2| = e^{2\beta x+C} = Ke^{2\beta x}$$ Where \(K\) is a constant, and we don't need the absolute value sign anymore, since \(K\) can take any value. Now, we solve for \(y\): $$\alpha y-\beta y^2 = Ke^{2\beta x}$$ $$y\left(\alpha - \beta y\right) = Ke^{2\beta x}$$ $$y = \frac{Ke^{2\beta x}}{\alpha - \beta y}$$ Now we can simplify the equation by multiplying both sides by \((\alpha - \beta y)\): $$y(\alpha - \beta y) = Ke^{2\beta x}$$

Apply the Initial Condition

We now apply the initial condition \(y(0) = y_0\) to find the value of the constant \(K\): $$y_0\left(\alpha - \beta y_0\right) = Ke^{2\beta (0)}$$ $$y_0\left(\alpha - \beta y_0\right) = K$$ Now we can substitute the value of \(K\) back into our solution for \(y(x)\): $$y = \frac{y_0\left(\alpha - \beta y_0\right)e^{2\beta x}}{\alpha - \beta y}$$

Find the Limit of the Solution

We now find the limit of the solution \(\phi(x)\) as \(x \rightarrow \infty\) for different values of \(y_0\): $$\lim _{x \rightarrow \infty} \phi(x)=\lim_{x \rightarrow \infty} \frac{y_0\left(\alpha - \beta y_0\right)e^{2\beta x}}{\alpha - \beta y}$$ Case 1: \(y_0 > 0\) As \(x \rightarrow \infty\), \(e^{2\beta x} \rightarrow \infty\). Thus, the numerator becomes infinitely large, and the only way for the limit to converge is for the denominator to become infinitely large as well. To make the limit convergent, we set the denominator equal to zero: $$\alpha - \beta y = 0$$ Solving for \(y(x)\), we get: $$y(x) = \frac{\alpha}{\beta}$$ So $$\lim _{x \rightarrow \infty} \phi(x) = \frac{\alpha}{\beta}, \ \text{for} \ y_{0} > 0$$ Case 2: \(y_0 = 0\) If \(y_0 = 0\), the whole expression for \(y(x)\) becomes zero: $$y(x) = \frac{0}{\alpha - \beta y} = 0$$ So, $$\lim _{x \rightarrow \infty} \phi(x) = 0, \ \text{for} \ y_{0} = 0$$ For \(y_{0}<0, \phi\) is unbounded as \(x\) approaches a certain value depending on \(y_{0}\).

Key concepts

Separable Differential Equations

A differential equation is termed separable if it can be rewritten so that all terms involving the dependent variable are on one side of the equation while the independent variable terms are on the other. In our given problem, the equation\[ \frac{dy}{dx} = \alpha y - \beta y^2 \]is separable. By rearranging, we can write this as:\[ \int \frac{dy}{\alpha y - \beta y^2} = \int dx \]This separate arrangement allows us to integrate each side independently.
Separable equations are often among the simplest differential equations to solve, making them an introductory topic in differential equations courses. They allow for direct integration once the variables are divided, making them a valuable tool in handling initial value problems.

Integration Techniques

Integration is a fundamental operation in solving differential equations. In the given problem, we integrated the term \( \frac{dy}{\alpha y - \beta y^2} \). This requires a technique often named as partial fraction decomposition or substitution, due to its form.
We introduce a substitution such as \( u = \alpha y - \beta y^2 \) to simplify our integral. This allows handling complex algebraic denominators by breaking them down into simpler constituent parts, which can be more easily integrated. Integration techniques like these are crucial: they help solve complex integrals by converting them into forms that are known and can be easily resolved.

Asymptotic Behavior

Asymptotic behavior refers to the behavior of a function as the input values become very large or very small. In the context of the initial value problem, we analyze the solution as \( x \rightarrow \infty \).
The solution shows that as \( x \) approaches infinity, the function \( \phi(x) \) approaches specific limits that depend on the initial value \( y_0 \). For instance, when \( y_0 > 0 \), the solution converges to \( \alpha/\beta \). This implies that the population or quantity being modeled matures to a stable state.
Asymptotic analysis helps predict long-term behavior of systems modeled by differential equations, a key area in applied mathematics.

Constants of Integration

When integrating to solve differential equations, adding a constant of integration \( C \) is vital because it accounts for all possible solutions of the indefinite integral. This constant helps form the general solution and is adjusted to meet initial conditions.
Once we integrate both sides, the constant appears in the expression. In our equation, after integration, the expression \( Ke^{2\beta x} \) involves such a constant. By applying the initial conditions, we can find the exact value of this constant and thus a particular solution to the problem.

• This process ensures the solution fits specific problem constraints, which is crucial for determining real-world implications based on model predictions.

Understanding constants during integration ensures accurate modeling in mathematical analyses.