We are given the differential equation:

$\frac{dy}{dx} = y(1 + y)$

We are also provided with the solution to verify:

$y_2 = \frac{Ce^x}{1 - Ce^x}$

Step 1: Differentiate the proposed solution

Let's differentiate $y_2$ with respect to $x$. Using the quotient rule:

$y_2 = \frac{Ce^x}{1 - Ce^x}$

Let $u = Ce^x$ and $v = 1 - Ce^x$, then:

$\frac{dy_2}{dx} = \frac{v \cdot \frac{d}{dx}(u) - u \cdot \frac{d}{dx}(v)}{v^2}$

Now, compute the derivatives of $u$ and $v$:

• $\frac{d}{dx}(u) = Ce^x$
• $\frac{d}{dx}(v) = -Ce^x$

Substitute these into the quotient rule formula:

$\frac{dy_2}{dx} = \frac{(1 - Ce^x)(Ce^x) - (Ce^x)(-Ce^x)}{(1 - Ce^x)^2}$

Simplifying the numerator:

$(1 - Ce^x)(Ce^x) = Ce^x - C^2e^{2x}$ $-(Ce^x)(-Ce^x) = C^2e^{2x}$

Thus:

$\frac{dy_2}{dx} = \frac{Ce^x - C^2e^{2x} + C^2e^{2x}}{(1 - Ce^x)^2} = \frac{Ce^x}{(1 - Ce^x)^2}$

Step 2: Substitute the solution into the original equation

We need to verify that $\frac{dy_2}{dx} = y_2(1 + y_2)$.

First, find $1 + y_2$:

$1 + y_2 = 1 + \frac{Ce^x}{1 - Ce^x} = \frac{1 - Ce^x + Ce^x}{1 - Ce^x} = \frac{1}{1 - Ce^x}$

Now multiply $y_2$ by $1 + y_2$:

$y_2(1 + y_2) = \frac{Ce^x}{1 - Ce^x} \cdot \frac{1}{1 - Ce^x} = \frac{Ce^x}{(1 - Ce^x)^2}$

This is exactly what we obtained for $\frac{dy_2}{dx}$. Therefore, the given function:

$y_2 = \frac{Ce^x}{1 - Ce^x}$

is indeed a solution to the differential equation $\frac{dy}{dx} = y(1 + y)$.

Do you have any questions or would you like further details on any part?

Here are 5 related questions to extend this concept:

1. How would the solution change if the equation was $\frac{dy}{dx} = y(1 - y)$?
2. What are the possible boundary conditions that could determine $C$?
3. How can this solution be interpreted in terms of population growth models?
4. Can we find the general solution for the equation using an integrating factor?
5. How do initial conditions affect the behavior of the solution?

Tip: Always verify a proposed solution by substituting it back into the original equation and simplifying both sides to see if they match.