Thanks. can you sole the two separate ordinary differential equations (ODEs) then give me the final answer considering the no-homogeneous part as well?

Let's solve the two separate ordinary differential equations (ODEs) that result from the method of separation of variables, and then incorporate the non-homogeneous part to get the final solution.

### Recall the Separated Equations:

From the separation of variables, we assumed \( w(x,t) = X(x)T(t) \). Substituting this into the homogeneous PDE and dividing by \( X(x)T(t) \) gives us:

1. **Spatial Equation (for \( X(x) \))**:
   \[
   EI \frac{d^4 X(x)}{dx^4} - \lambda X(x) = 0
   \]
   
2. **Temporal Equation (for \( T(t) \))**:
   \[
   c \frac{dT(t)}{dt} + (k + \lambda) T(t) = 0
   \]

### Step 1: Solve the Spatial ODE
The spatial ODE is:
\[
EI \frac{d^4 X(x)}{dx^4} = \lambda X(x)
\]
This is a fourth-order linear homogeneous differential equation.

Let's consider a solution of the form:
\[
X(x) = e^{\beta x}
\]

Substituting into the ODE gives:
\[
EI \beta^4 e^{\beta x} = \lambda e^{\beta x}
\]

Simplifying, we obtain the characteristic equation:
\[
EI \beta^4 = \lambda
\]

This gives:
\[
\beta = \left(\frac{\lambda}{EI}\right)^{\frac{1}{4}}
\]

The general solution for \( X(x) \) is then:
\[
X(x) = C_1 e^{\beta x} + C_2 e^{-\beta x} + C_3 \cos(\beta x) + C_4 \sin(\beta x)
\]

where \( \beta = \left(\frac{\lambda}{EI}\right)^{\frac{1}{4}} \), and \( C_1, C_2, C_3, \) and \( C_4 \) are constants determined by boundary conditions.

### Step 2: Solve the Temporal ODE
The temporal ODE is:
\[
c \frac{dT(t)}{dt} + (k + \lambda) T(t) = 0
\]

This is a first-order linear differential equation. Its solution is:
\[
T(t) = A e^{-\frac{k + \lambda}{c} t}
\]

where \( A \) is a constant of integration.

### Step 3: Combine the Solutions
The general solution to the homogeneous PDE is:
\[
w(x,t) = \left( C_1 e^{\beta x} + C_2 e^{-\beta x} + C_3 \cos(\beta x) + C_4 \sin(\beta x) \right) \cdot A e^{-\frac{k + \lambda}{c} t}
\]

### Step 4: Incorporate the Non-Homogeneous Term
To account for the non-homogeneous term \( q_0 \), we assume a particular solution \( w_p(x,t) \) that satisfies:
\[
EI \frac{\partial^4 w_p(x,t)}{\partial x^4} + c \frac{\partial w_p(x,t)}{\partial t} + k w_p(x,t) - q_0 = 0
\]

Given that \( q_0 \) is a constant, a reasonable particular solution might be a constant \( w_p \) as well:
\[
w_p(x,t) = \frac{q_0}{k}
\]

### Final Solution
The final solution combining the homogeneous and particular solutions is:
\[
w(x,t) = \left( C_1 e^{\beta x} + C_2 e^{-\beta x} + C_3 \cos(\beta x) + C_4 \sin(\beta x) \right) \cdot A e^{-\frac{k + \lambda}{c} t} + \frac{q_0}{k}
\]

where \( \beta = \left(\frac{\lambda}{EI}\right)^{\frac{1}{4}} \), and \( C_1, C_2, C_3, C_4, \) and \( A \) are constants determined by boundary conditions.

---

Let me know if you need further clarification on any part, or if you'd like to discuss the implications of this solution!

### Related Questions:

1. How would the solution change if \( q_0 \) were a function of \( x \) or \( t \)?
2. What boundary conditions would you apply to determine the constants \( C_1 \), \( C_2 \), \( C_3 \), \( C_4 \), and \( A \)?
3. How does the term \( \frac{q_0}{k} \) influence the physical behavior of the system?
4. What does this solution tell you about the stability of the system over time?
5. How does changing \( EI \) or \( c \) affect the solution's frequency and damping?

**Tip:** Boundary conditions are crucial for determining the specific form of the solution in practical problems. Ensure they are well-understood and applied correctly.

Learn how to solve two separate ordinary differential equations (ODEs) using the method of separation of variables and incorporate the non-homogeneous part to obtain the final solution. This detailed solution covers mathematical concepts such as ODEs, separation of variables, and their applications in physics and engineering.

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