Consider the differential equation $\frac{dy}{dx}=ay-by^{2}$, a, $b0$ and $y(0)=y_{0}$. As $x\rightarrow\infty$, the solution $y(x)$ tends to :

Question

Consider the differential equation $\frac{dy}{dx}=ay-by^{2}$, a, $b>0$ and $y(0)=y_{0}$. As $x\rightarrow\infty$, the solution $y(x)$ tends to :

Accepted Answer

To determine the behavior of the solution $ y(x) $ as $ x \rightarrow \infty $ for the differential equation $\frac{dy}{dx} = ay - by^2$, where $ a, b > 0 $ and $ y(0) = y_0 $, we need to analyze the equilibrium points and the stability of the system.

### Explanation for Option 1: $\frac{a}{b}$

The differential equation $\frac{dy}{dx} = ay - by^2$ is a first-order, nonlinear ordinary differential equation. To find the equilibrium points, we set the right-hand side to zero:

$$ ay - by^2 = 0 $$

Factoring out $ y $, we get:

$$ y(ay - by) = y(a - by) = 0 $$

This gives us two equilibrium points: $ y = 0 $ and $ y = \frac{a}{b} $.

To determine the stability of these equilibrium points, we consider the derivative of the right-hand side with respect to $ y $:

$$ \frac{d}{dy}(ay - by^2) = a - 2by $$

1. **At $ y = 0 $:**
   $$ \frac{d}{dy}(ay - by^2) \bigg|_{y=0} = a $$
   Since $ a > 0 $, the derivative is positive, indicating that $ y = 0 $ is an unstable equilibrium point.

2. **At $ y = \frac{a}{b} $:**
   $$ \frac{d}{dy}(ay - by^2) \bigg|_{y=\frac{a}{b}} = a - 2b\left(\frac{a}{b}\right) = a - 2a = -a $$
   Since $ -a < 0 $, the derivative is negative, indicating that $ y = \frac{a}{b} $ is a stable equilibrium point.

As $ x \rightarrow \infty $, the solution $ y(x) $ will tend towards the stable equilibrium point, which is $ y = \frac{a}{b} $. Therefore, Option 1 is correct.

### Explanation for Why Other Options are Incorrect:

- **Option 2: $\frac{b}{a}$**
  This option is incorrect because $\frac{b}{a}$ is not an equilibrium point of the differential equation. The equilibrium points are determined by setting the derivative to zero, and $\frac{b}{a}$ does not satisfy this condition.

- **Option 3: $y_0$**
  This option is incorrect because $ y_0 $ is simply the initial condition and does not represent the behavior of the solution as $ x \rightarrow \infty $. The solution will tend towards the stable equilibrium point, not necessarily remain at the initial condition.

- **Option 4: 0**
  This option is incorrect because $ y = 0 $ is an unstable equilibrium point. The solution will not tend towards an unstable equilibrium as $ x \rightarrow \infty $ unless it starts exactly at that point, which is not guaranteed by the initial condition $ y(0) = y_0 $.

In conclusion, the correct behavior of the solution as $ x \rightarrow \infty $ is to approach the stable equilibrium point $ \frac{a}{b} $, making Option 1 the correct answer.