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resolva algebricamente o problema da imagem

The image contains a mathematical problem, which appears to involve solving differential equations. Here’s a breakdown of the problem and the algebraic steps to solve it:

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**Problem Statement:**

Given the equation \(\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}\) on the interval \(0 \leq x \leq L\), find the solution \(u(x,t)\).

**Boundary Conditions:**
1. \( u(0,t) = 0 \)
2. \( u(L,t) = 0 \)

**Initial Condition:**
\[ u(x,0) = f(x) = \sin\left(\frac{n\pi x}{L}\right) \]

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**Step-by-Step Solution:**

1. **Separation of Variables:**
   Assume a solution of the form \( u(x,t) = X(x)T(t) \). Substituting into the partial differential equation (PDE):
   \[
   X(x) \frac{dT(t)}{dt} = T(t) \frac{d^2X(x)}{dx^2}
   \]
   Dividing both sides by \( X(x)T(t) \) gives:
   \[
   \frac{1}{T(t)} \frac{dT(t)}{dt} = \frac{1}{X(x)} \frac{d^2X(x)}{dx^2} = -\lambda
   \]
   where \(\lambda\) is a separation constant.

2. **Time-dependent Part:**
   Solve the equation:
   \[
   \frac{dT(t)}{dt} = -\lambda T(t)
   \]
   The solution is:
   \[
   T(t) = A e^{-\lambda t}
   \]

3. **Spatial Part:**
   Solve the equation:
   \[
   \frac{d^2X(x)}{dx^2} + \lambda X(x) = 0
   \]
   This is a second-order differential equation with the general solution:
   \[
   X(x) = B \sin(\sqrt{\lambda} x) + C \cos(\sqrt{\lambda} x)
   \]
   Applying the boundary conditions:
   - At \(x = 0\): \( u(0,t) = 0 \), hence \( C = 0 \).
   - At \(x = L\): \( u(L,t) = 0 \), hence \( B \sin(\sqrt{\lambda} L) = 0 \).

   For a non-trivial solution, \( \sin(\sqrt{\lambda} L) = 0 \), so:
   \[
   \sqrt{\lambda} L = n\pi \quad \text{for } n = 1, 2, 3, \dots
   \]
   Therefore, \( \lambda = \left(\frac{n\pi}{L}\right)^2 \).

   Thus:
   \[
   X(x) = B \sin\left(\frac{n\pi x}{L}\right)
   \]

4. **Combine the Solutions:**
   The general solution is:
   \[
   u(x,t) = \sum_{n=1}^{\infty} B_n \sin\left(\frac{n\pi x}{L}\right) e^{-\left(\frac{n\pi}{L}\right)^2 t}
   \]

5. **Apply the Initial Condition:**
   Use the initial condition \( u(x,0) = f(x) = \sin\left(\frac{n\pi x}{L}\right) \):
   \[
   u(x,0) = B_n \sin\left(\frac{n\pi x}{L}\right)
   \]
   Comparing with the initial condition:
   \[
   B_n = 1
   \]

   Therefore, the solution is:
   \[
   u(x,t) = \sin\left(\frac{n\pi x}{L}\right) e^{-\left(\frac{n\pi}{L}\right)^2 t}
   \]

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This is the solution to the problem. Let me know if you want more details on any part of the solution.

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**Related Questions:**

1. How do boundary conditions affect the solution of differential equations?
2. What is the significance of the separation constant \(\lambda\) in solving PDEs?
3. How would the solution change if the initial condition was different?
4. What happens to the solution as \( t \) approaches infinity?
5. How can the method of separation of variables be applied to other types of PDEs?

**Tip:** When solving PDEs with boundary conditions, always check the compatibility of the boundary conditions with your general solution. This ensures that you capture the correct physical behavior of the system.

Explore the algebraic solution to a partial differential equation (PDE) involving boundary and initial conditions. Learn how to solve PDEs using separation of variables method with detailed steps and examples. Suitable for advanced undergraduate students and researchers.

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