find the general solution of the differential equation y' = sin x + 2y

To find the general solution of the differential equation \( y' = \sin x + 2y \), we will use the method of integrating factors. This is a first-order linear differential equation of the form:

\[ y' + P(x)y = Q(x) \]

where \( P(x) = -2 \) and \( Q(x) = \sin x \).

1. **Compute the integrating factor \( \mu(x) \)**:

   The integrating factor \( \mu(x) \) is given by:
   \[
   \mu(x) = e^{\int P(x) \, dx} = e^{\int -2 \, dx} = e^{-2x}
   \]

2. **Multiply the entire differential equation by the integrating factor**:

   \[
   e^{-2x} y' + e^{-2x} \cdot 2y = e^{-2x} \sin x
   \]

   This simplifies to:
   \[
   \left(e^{-2x} y\right)' = e^{-2x} \sin x
   \]

3. **Integrate both sides with respect to \( x \)**:

   \[
   e^{-2x} y = \int e^{-2x} \sin x \, dx
   \]

   To integrate \( \int e^{-2x} \sin x \, dx \), we use integration by parts. Let:
   \[
   u = \sin x \quad \text{and} \quad dv = e^{-2x} dx
   \]
   Then,
   \[
   du = \cos x \, dx \quad \text{and} \quad v = \int e^{-2x} \, dx = -\frac{1}{2} e^{-2x}
   \]

   Now, apply integration by parts:
   \[
   \int u \, dv = uv - \int v \, du
   \]
   Substituting \( u \) and \( v \):
   \[
   \int e^{-2x} \sin x \, dx = \sin x \left(-\frac{1}{2} e^{-2x}\right) - \int \left(-\frac{1}{2} e^{-2x}\right) \cos x \, dx
   \]

   Simplify:
   \[
   \int e^{-2x} \sin x \, dx = -\frac{1}{2} e^{-2x} \sin x + \frac{1}{2} \int e^{-2x} \cos x \, dx
   \]

   To solve \( \int e^{-2x} \cos x \, dx \), we apply integration by parts again. Let:
   \[
   u = \cos x \quad \text{and} \quad dv = e^{-2x} dx
   \]
   Then,
   \[
   du = -\sin x \, dx \quad \text{and} \quad v = -\frac{1}{2} e^{-2x}
   \]

   Apply integration by parts again:
   \[
   \int e^{-2x} \cos x \, dx = \cos x \left(-\frac{1}{2} e^{-2x}\right) - \int \left(-\frac{1}{2} e^{-2x}\right) (-\sin x) \, dx
   \]

   Simplify:
   \[
   \int e^{-2x} \cos x \, dx = -\frac{1}{2} e^{-2x} \cos x + \frac{1}{2} \int e^{-2x} \sin x \, dx
   \]

   Combining the results of both parts, we get:
   \[
   I = \int e^{-2x} \sin x \, dx = -\frac{1}{2} e^{-2x} \sin x - \frac{1}{2} \left(-\frac{1}{2} e^{-2x} \cos x + \frac{1}{2} \int e^{-2x} \sin x \, dx\right)
   \]

   Solve for \( I \):
   \[
   I = -\frac{1}{2} e^{-2x} \sin x + \frac{1}{4} e^{-2x} \cos x - \frac{1}{4} I
   \]
   \[
   \frac{5}{4} I = -\frac{1}{2} e^{-2x} \sin x + \frac{1}{4} e^{-2x} \cos x
   \]
   \[
   I = -\frac{2}{5} e^{-2x} \sin x + \frac{1}{5} e^{-2x} \cos x
   \]

4. **Include the constant of integration**:

   \[
   e^{-2x} y = -\frac{2}{5} e^{-2x} \sin x + \frac{1}{5} e^{-2x} \cos x + C
   \]

5. **Solve for \( y \)**:

   \[
   y = -\frac{2}{5} \sin x + \frac{1}{5} \cos x + Ce^{2x}
   \]

Therefore, the general solution to the differential equation \( y' = \sin x + 2y \) is:

\[ y = -\frac{2}{5} \sin x + \frac{1}{5} \cos x + Ce^{2x} \]

Would you like any details or have any questions? Here are some related questions you might want to explore:

1. How do you solve a differential equation using separation of variables?
2. What are homogeneous and non-homogeneous differential equations?
3. How do you find particular solutions to non-homogeneous differential equations?
4. What are the applications of first-order differential equations in real life?
5. How do you solve second-order linear differential equations with constant coefficients?

**Tip:** When solving differential equations, always ensure that you correctly identify the type of equation and use the appropriate method for the solution.

Explore the step-by-step process to find the general solution of the differential equation y' = sin x + 2y using integrating factors and integration by parts. Learn about first-order linear differential equations and their applications. Suitable for college-level students and learners interested in differential equations.

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