verify that u(x,t)=x^2+2t, x^3+6xt and cos(2t-x)e^-x are each a solution of the heat equation u_t=u_xx with k=1

To verify that each of the given functions is a solution to the heat equation \( u_t = u_{xx} \) with \( k = 1 \), we need to perform the following steps:

1. **Find the partial derivative of \( u(x, t) \) with respect to \( t \) (i.e., \( u_t \)).**
2. **Find the second partial derivative of \( u(x, t) \) with respect to \( x \) (i.e., \( u_{xx} \)).**
3. **Check if \( u_t = u_{xx} \) holds true for each given function.**

Let's start with each function one by one.

### 1. \( u(x, t) = x^2 + 2t \)

- **First derivative with respect to \( t \):**
  \[
  u_t = \frac{\partial}{\partial t} (x^2 + 2t) = 2
  \]

- **Second derivative with respect to \( x \):**
  \[
  u_{xx} = \frac{\partial^2}{\partial x^2} (x^2 + 2t) = 2
  \]

- **Verification:**
  \[
  u_t = 2 \quad \text{and} \quad u_{xx} = 2
  \]
  Since \( u_t = u_{xx} \), the function \( u(x, t) = x^2 + 2t \) is a solution to the heat equation.

### 2. \( u(x, t) = x^3 + 6xt \)

- **First derivative with respect to \( t \):**
  \[
  u_t = \frac{\partial}{\partial t} (x^3 + 6xt) = 6x
  \]

- **Second derivative with respect to \( x \):**
  \[
  u_{xx} = \frac{\partial^2}{\partial x^2} (x^3 + 6xt) = 6x
  \]

- **Verification:**
  \[
  u_t = 6x \quad \text{and} \quad u_{xx} = 6x
  \]
  Since \( u_t = u_{xx} \), the function \( u(x, t) = x^3 + 6xt \) is also a solution to the heat equation.

### 3. \( u(x, t) = \cos(2t - x) e^{-x} \)

- **First derivative with respect to \( t \):**
  \[
  u_t = \frac{\partial}{\partial t} \left( \cos(2t - x) e^{-x} \right) = -2 \sin(2t - x) e^{-x}
  \]

- **Second derivative with respect to \( x \):**
  \[
  u_x = \frac{\partial}{\partial x} \left( \cos(2t - x) e^{-x} \right) = \sin(2t - x) e^{-x} - \cos(2t - x) e^{-x}
  \]
  \[
  u_{xx} = \frac{\partial}{\partial x} \left( \sin(2t - x) e^{-x} - \cos(2t - x) e^{-x} \right) = -2 \sin(2t - x) e^{-x}
  \]

- **Verification:**
  \[
  u_t = -2 \sin(2t - x) e^{-x} \quad \text{and} \quad u_{xx} = -2 \sin(2t - x) e^{-x}
  \]
  Since \( u_t = u_{xx} \), the function \( u(x, t) = \cos(2t - x) e^{-x} \) is also a solution to the heat equation.

### Conclusion:
All three functions \( u(x, t) = x^2 + 2t \), \( u(x, t) = x^3 + 6xt \), and \( u(x, t) = \cos(2t - x) e^{-x} \) satisfy the heat equation \( u_t = u_{xx} \) with \( k = 1 \).

---

Would you like to go into more detail on any of the steps? Here are some additional questions that can expand on the topic:

1. How would the verification process change if \( k \) were not equal to 1 in the heat equation?
2. Can you solve the heat equation for a different initial condition and boundary condition setup?
3. How would the solution differ if we added a source term to the heat equation?
4. What are the physical interpretations of the functions \( u(x, t) \) in the context of heat transfer?
5. How does the method of separation of variables apply to solving the heat equation?

**Tip:** When verifying solutions to differential equations, always ensure that you've correctly computed all required derivatives, and carefully compare both sides of the equation for equality.

Explore how to verify that the functions u(x,t)=x^2+2t, x^3+6xt, and cos(2t-x)e^-x are solutions to the heat equation u_t = u_xx with k=1. This step-by-step solution explains the use of partial derivatives and differential equations, ideal for undergraduate students studying partial differential equations.

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