https://file.aiquickdraw.com/mathbot/file/1725376385931gkrbkb6s.jpg

The equation shown in the image is:

\[
\frac{dy}{dx} = -3 - e^x y^2
\]

This is a Bernoulli differential equation, which has the form:

\[
\frac{dy}{dx} + P(x)y = Q(x)y^n
\]

Here, the equation can be rearranged into a form that fits the Bernoulli equation structure. To solve it, we proceed as follows:

### Step 1: Rewrite the equation
First, we can rewrite the given equation as:

\[
\frac{dy}{dx} + e^x y^2 = -3
\]

Here, \( P(x) = 0 \) and \( Q(x) = e^x \), with \( n = 2 \).

### Step 2: Substitute \( v = y^{1-n} = y^{-1} \)
Let's define a new variable \( v = y^{-1} \), so \( y = v^{-1} \) and \( \frac{dy}{dx} = -v^{-2} \frac{dv}{dx} \).

### Step 3: Substitution into the equation
Substitute into the original equation:

\[
-\frac{1}{v^2} \frac{dv}{dx} + e^x \frac{1}{v^2} = -3
\]

Multiply through by \(-v^2\) to simplify:

\[
\frac{dv}{dx} = 3v^2 - e^x
\]

### Step 4: Solve the new linear equation
This equation is now linear in \( v \). The general approach to solve it is by using an integrating factor or a method appropriate for linear first-order differential equations.

However, let's move forward assuming you want an explicit integration process.

### Step 5: Find the integrating factor
For an equation of the form:

\[
\frac{dv}{dx} + P(x)v = Q(x)
\]

we need to find the integrating factor \( \mu(x) = e^{\int P(x) dx} \). In this case, \( P(x) = -3 \).

So:

\[
\mu(x) = e^{-3x}
\]

### Step 6: Multiply the equation by the integrating factor
Multiply the equation by the integrating factor \( e^{-3x} \):

\[
e^{-3x} \frac{dv}{dx} - 3e^{-3x} v = -e^{-2x}
\]

This simplifies to:

\[
\frac{d}{dx}(v e^{-3x}) = -e^{-2x}
\]

### Step 7: Integrate both sides
Integrate both sides with respect to \( x \):

\[
v e^{-3x} = \int -e^{-2x} dx
\]

The integral on the right-hand side is:

\[
\int -e^{-2x} dx = \frac{1}{2}e^{-2x} + C
\]

So:

\[
v e^{-3x} = \frac{1}{2}e^{-2x} + C
\]

### Step 8: Solve for \( v \)
Solve for \( v \):

\[
v = \frac{1}{2}e^{x} + Ce^{3x}
\]

### Step 9: Substitute back \( v = y^{-1} \)
Finally, substitute \( v = y^{-1} \):

\[
y = \frac{1}{\frac{1}{2}e^{x} + Ce^{3x}}
\]

Thus, the solution to the differential equation is:

\[
y(x) = \frac{2e^{-x}}{1 + 2Ce^{2x}}
\]

Would you like more details or explanations on any of the steps? 

Here are some related questions you might find interesting:
1. What is the general form of a Bernoulli differential equation?
2. How do you derive the integrating factor for a first-order linear differential equation?
3. Can all Bernoulli equations be transformed into linear differential equations?
4. What are the conditions for a differential equation to be exact?
5. How do we solve a first-order separable differential equation?

**Tip:** Always verify your final solution by substituting it back into the original differential equation to ensure it satisfies the equation.

Learn how to solve a Bernoulli differential equation involving exponential functions with a detailed step-by-step solution. This problem explores key concepts such as substitution, integrating factors, and the Bernoulli equation transformation, suitable for undergraduate students.

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