Let dy/dx = y(y-2) Yo >0 a) use qualitative analysis to classify the equilibrium solutions b) solve the equation and use your solution to confirm the results from a)

Let's break down and solve the given differential equation step by step.

### Given:
\[
\frac{dy}{dx} = y(y - 2)
\]
where \( y_0 > 0 \) is the initial condition.

### Part (a): Qualitative Analysis (Classifying Equilibrium Solutions)

To classify the equilibrium solutions, we need to set \(\frac{dy}{dx} = 0\) and solve for \( y \).

\[
y(y - 2) = 0
\]

The solutions are:

\[
y = 0 \quad \text{or} \quad y = 2
\]

These are the equilibrium solutions. Now, let's classify their stability by analyzing the sign of \(\frac{dy}{dx}\) near these points.

1. **For \( y = 0 \)**:  
   - For \( y > 0 \), we have \( y(y - 2) > 0 \) (since both \( y \) and \( y - 2 \) are positive).
   - For \( y < 0 \), we have \( y(y - 2) < 0 \) (since \( y \) is negative and \( y - 2 \) is negative for \( y < 2 \)).

   Conclusion: For \( y = 0 \), solutions move away from \( y = 0 \) on either side. Hence, \( y = 0 \) is an **unstable equilibrium**.

2. **For \( y = 2 \)**:  
   - For \( y > 2 \), we have \( y(y - 2) > 0 \) (since both \( y > 2 \) and \( y - 2 > 0 \)).
   - For \( y < 2 \), we have \( y(y - 2) < 0 \) (since \( y > 0 \) and \( y - 2 < 0 \)).

   Conclusion: For \( y = 2 \), solutions tend to move towards \( y = 2 \). Hence, \( y = 2 \) is a **stable equilibrium**.

### Part (b): Solve the Differential Equation

This is a separable differential equation. Let's solve it by separating the variables.

\[
\frac{1}{y(y - 2)} dy = dx
\]

Now, we use partial fraction decomposition on the left-hand side:

\[
\frac{1}{y(y - 2)} = \frac{A}{y} + \frac{B}{y - 2}
\]

Multiplying through by \( y(y - 2) \) to clear the denominator:

\[
1 = A(y - 2) + By
\]

Expanding and collecting like terms:

\[
1 = (A + B)y - 2A
\]

This gives two equations:
1. \( A + B = 0 \)
2. \( -2A = 1 \)

Solving for \( A \) and \( B \):
\[
A = -\frac{1}{2}, \quad B = \frac{1}{2}
\]

Thus, we can rewrite the integral as:

\[
\frac{1}{y(y - 2)} = -\frac{1}{2y} + \frac{1}{2(y - 2)}
\]

Now, integrating both sides:

\[
-\frac{1}{2} \int \frac{1}{y} \, dy + \frac{1}{2} \int \frac{1}{y - 2} \, dy = \int dx
\]

This yields:

\[
-\frac{1}{2} \ln|y| + \frac{1}{2} \ln|y - 2| = x + C
\]

Simplifying the logarithms:

\[
\frac{1}{2} \ln\left|\frac{y - 2}{y}\right| = x + C
\]

Multiplying both sides by 2:

\[
\ln\left|\frac{y - 2}{y}\right| = 2(x + C)
\]

Exponentiating both sides:

\[
\left|\frac{y - 2}{y}\right| = e^{2(x + C)} = e^{2x}e^{2C}
\]

Let \( e^{2C} = C_1 \), so:

\[
\left|\frac{y - 2}{y}\right| = C_1 e^{2x}
\]

Removing the absolute value and solving for \( y \):

\[
\frac{y - 2}{y} = \pm C_1 e^{2x}
\]

Multiplying through by \( y \) and solving for \( y \):

\[
y - 2 = \pm C_1 y e^{2x}
\]
\[
y(1 \mp C_1 e^{2x}) = 2
\]
\[
y = \frac{2}{1 \mp C_1 e^{2x}}
\]

### Confirming Equilibrium Analysis:

- As \( x \to \infty \), if \( C_1 > 0 \), \( y \to 2 \), which confirms \( y = 2 \) is a stable equilibrium.
- If \( y(0) = 0 \), the solution \( y = 0 \) is an unstable equilibrium as solutions move away from it.

### Follow-Up Questions:

1. How does the behavior change if \( y_0 \) is negative?
2. What are the possible long-term behaviors for different values of \( C_1 \)?
3. How would a phase portrait of this system look like?
4. Can you provide a numerical example of this solution?
5. How would adding a constant term to the differential equation affect the equilibria?

#### Tip: Stability can often be inferred by checking the sign of the derivative near equilibrium points.

Let dy/dx = y(y-2), y0 > 0 a) use qualitative analysis to classify the equilibrium solutions b) solve the equation and use your solution to confirm the results from a).

Solve the differential equation dy/dx = y(y-2) and classify its equilibrium solutions using qualitative analysis. Explore key mathematical concepts like stability of equilibrium solutions and separation of variables. Suitable for undergraduate students studying differential equations.

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