Find the the solution of the differential equations 1. y'' + y = cot(x),

To solve the differential equation \( y'' + y = \cot(x) \), we can proceed as follows:

### Step 1: Solve the Homogeneous Equation
First, solve the associated homogeneous equation:

\[
y'' + y = 0
\]

The characteristic equation corresponding to this differential equation is:

\[
r^2 + 1 = 0
\]

This simplifies to:

\[
r^2 = -1 \quad \Rightarrow \quad r = \pm i
\]

The general solution to the homogeneous equation is:

\[
y_h(x) = c_1 \cos(x) + c_2 \sin(x)
\]

where \( c_1 \) and \( c_2 \) are arbitrary constants.

### Step 2: Find a Particular Solution
Next, we need to find a particular solution \( y_p(x) \) to the non-homogeneous equation:

\[
y_p'' + y_p = \cot(x)
\]

Given that the right-hand side \( \cot(x) \) is not a simple polynomial, exponential, or trigonometric function for which we have a direct method, we might use the method of undetermined coefficients or variation of parameters. However, since the function \( \cot(x) \) is not easily tackled with those methods, it's more practical to use the method of variation of parameters.

Let's use variation of parameters:

Assume the solution takes the form:

\[
y_p(x) = v_1(x) \cos(x) + v_2(x) \sin(x)
\]

where \( v_1(x) \) and \( v_2(x) \) are functions to be determined.

To find \( v_1(x) \) and \( v_2(x) \), we use the following system of equations derived from the original differential equation:

\[
v_1'(x) \cos(x) + v_2'(x) \sin(x) = 0
\]
\[
-v_1'(x) \sin(x) + v_2'(x) \cos(x) = \cot(x)
\]

From the first equation:

\[
v_1'(x) \cos(x) = -v_2'(x) \sin(x)
\]

Substituting this into the second equation gives:

\[
-v_1'(x) \sin(x) - \frac{v_1'(x) \cos^2(x)}{\sin(x)} = \cot(x)
\]

This simplifies to:

\[
v_1'(x) \left(-\sin(x) - \frac{\cos^2(x)}{\sin(x)} \right) = \cot(x)
\]

Simplifying further:

\[
v_1'(x) = -\csc(x)
\]

Integrating:

\[
v_1(x) = \ln|\csc(x) - \cot(x)|
\]

For \( v_2'(x) \):

\[
v_2'(x) = v_1'(x) \frac{\cos(x)}{\sin(x)} = -\cot(x) \cdot \csc(x)
\]

Integrating:

\[
v_2(x) = \ln|\sin(x)|
\]

Thus, the particular solution is:

\[
y_p(x) = \ln|\csc(x) - \cot(x)| \cdot \cos(x) + \ln|\sin(x)| \cdot \sin(x)
\]

### Step 3: General Solution
The general solution is:

\[
y(x) = y_h(x) + y_p(x)
\]

\[
y(x) = c_1 \cos(x) + c_2 \sin(x) + \ln|\csc(x) - \cot(x)| \cdot \cos(x) + \ln|\sin(x)| \cdot \sin(x)
\]

This is the solution to the differential equation.

Would you like more details on any of the steps?

### Follow-Up Questions
1. How is the characteristic equation derived from a homogeneous differential equation?
2. What is the method of undetermined coefficients?
3. Can you explain the method of variation of parameters in more detail?
4. What are the common types of particular solutions in non-homogeneous differential equations?
5. How do you solve integrals involving trigonometric functions like \( \csc(x) \)?

### Tip
When solving differential equations, always verify if a particular solution satisfies the non-homogeneous part by substituting it back into the original equation.

Find the solution of the differential equation y'' + y = cot(x).

This detailed solution explains how to solve the differential equation y'' + y = cot(x) using the method of variation of parameters. The problem involves key concepts in differential equations, including solving the characteristic equation and finding a particular solution. Suitable for undergraduate students.

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