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The problem asks for the solution of the first-order differential equation:

\[
y' = -2xy^2
\]

with the initial condition \( y(0) = 5 \).

### Step-by-Step Solution:

#### 1. **Separate Variables:**

We can rewrite the equation in a separable form:

\[
\frac{dy}{dx} = -2xy^2
\]

Separate the variables \( y \) and \( x \):

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

#### 2. **Integrate Both Sides:**

Integrating both sides, we start with the left-hand side:

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

The integral on the left-hand side is:

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

The integral on the right-hand side is:

\[
\int -2x \, dx = -x^2
\]

So, we have:

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

#### 3. **Solve for \( y \):**

Multiplying through by \( -1 \), we get:

\[
\frac{1}{y} = x^2 - C
\]

Taking the reciprocal of both sides:

\[
y = \frac{1}{x^2 - C}
\]

#### 4. **Apply the Initial Condition \( y(0) = 5 \):**

Substitute \( x = 0 \) and \( y = 5 \) into the equation:

\[
5 = \frac{1}{0^2 - C} = \frac{1}{-C}
\]

Thus, \( -C = \frac{1}{5} \), which gives \( C = -\frac{1}{5} \).

#### 5. **Final Solution:**

Substitute \( C = -\frac{1}{5} \) into the general solution:

\[
y = \frac{1}{x^2 + \frac{1}{5}} = \frac{1}{x^2 + \frac{1}{5}}
\]

This is the solution to the differential equation.

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### Related Questions:
1. How do you verify a solution of a differential equation using initial conditions?
2. What techniques can be used to solve non-separable differential equations?
3. What is the general solution form for first-order separable equations?
4. How can we graphically interpret the solution \( y = \frac{1}{x^2 + \frac{1}{5}} \)?
5. What does the solution behavior near \( x = 0 \) suggest about the function's graph?

**Tip:** When solving differential equations, always check if you can separate variables early on, as it simplifies many problems.

Ache a solução da equação diferencial y' = -2xy^2, que atenda à condição inicial y(0) = 5 e assinale a alternativa que a contenha.

Find the solution to the first-order differential equation y' = -2xy^2 that satisfies the initial condition y(0) = 5. This solution involves separation of variables and integration, suitable for undergraduate students.

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