solve for variable separable y^' + y^2 sin x = 0. the answer must be 1/y + cos x = c

To solve the first-order ordinary differential equation \( y' + y^2 \sin x = 0 \) using the method of separation of variables, we proceed as follows:

Given:
\[
\frac{dy}{dx} + y^2 \sin x = 0
\]

1. **Rearrange the equation to separate the variables:**
   \[
   \frac{dy}{dx} = -y^2 \sin x
   \]
   \[
   \frac{dy}{y^2} = -\sin x \, dx
   \]

2. **Integrate both sides:**

   - The left side:
     \[
     \int \frac{dy}{y^2} = \int -y^{-2} \, dy = \int -y^{-2} \, dy
     \]
     The integral of \( -y^{-2} \) is \( y^{-1} \), so:
     \[
     \int -y^{-2} \, dy = \frac{1}{y}
     \]
   
   - The right side:
     \[
     \int -\sin x \, dx = \cos x
     \]

   Therefore, after integrating:
   \[
   \frac{1}{y} = \cos x + C
   \]

3. **Simplify the result:**

   Multiply both sides by \( y \) to express it in the standard form:
   \[
   \frac{1}{y} + \cos x = C
   \]

Thus, the solution is:
\[
\boxed{\frac{1}{y} + \cos x = C}
\]

Would you like further details or have any questions?

Here are five related questions you might explore:

1. What is the general method for solving variable separable differential equations?
2. How do you handle initial conditions in separable differential equations?
3. How can you verify a solution to a differential equation?
4. What are some examples of non-separable differential equations?
5. How does the constant of integration affect the solution to a differential equation?

**Tip:** When using separation of variables, always ensure that each side of the equation contains only one variable before integrating.

Learn how to solve the variable separable differential equation y' + y^2 sin x = 0 step-by-step with detailed explanations and integration techniques. Suitable for advanced learners interested in differential equations and calculus.

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