https://file.aiquickdraw.com/mathbot/file/1726988558522x0lz5yya.jpg

To verify that \( u(x,t) = e^{xy} \cos(xt) \) is a solution of the differential equation \( u_{xx} = 25u_t \), we need to calculate the second derivative of \( u(x,t) \) with respect to \( x \) (i.e., \( u_{xx} \)) and the first derivative of \( u(x,t) \) with respect to \( t \) (i.e., \( u_t \)), then check if they satisfy the equation \( u_{xx} = 25u_t \).

Let's proceed step by step.

### Step 1: Compute \( u_t \) (Partial derivative of \( u \) with respect to \( t \))
Given:
\[
u(x,t) = e^{xy} \cos(xt)
\]
We differentiate \( u(x,t) \) with respect to \( t \):
\[
u_t = e^{xy} \frac{\partial}{\partial t} (\cos(xt)) = e^{xy} (-x \sin(xt)) = -x e^{xy} \sin(xt)
\]
So:
\[
u_t = -x e^{xy} \sin(xt)
\]

### Step 2: Compute \( u_x \) (Partial derivative of \( u \) with respect to \( x \))
Now differentiate \( u(x,t) \) with respect to \( x \):
\[
u_x = \frac{\partial}{\partial x} \left( e^{xy} \cos(xt) \right)
\]
Using the product rule:
\[
u_x = \left( y e^{xy} \cos(xt) \right) + \left( e^{xy} \cdot (-t \sin(xt)) \right)
\]
So:
\[
u_x = y e^{xy} \cos(xt) - t e^{xy} \sin(xt)
\]

### Step 3: Compute \( u_{xx} \) (Second partial derivative of \( u \) with respect to \( x \))
Differentiate \( u_x \) with respect to \( x \) again:
\[
u_{xx} = \frac{\partial}{\partial x} \left( y e^{xy} \cos(xt) - t e^{xy} \sin(xt) \right)
\]
Applying the product rule to each term:
\[
u_{xx} = \left( y^2 e^{xy} \cos(xt) \right) + \left( -ty e^{xy} \sin(xt) \right) + \left( -ty e^{xy} \sin(xt) \right) + \left( -t^2 e^{xy} \cos(xt) \right)
\]
Simplifying:
\[
u_{xx} = y^2 e^{xy} \cos(xt) - 2ty e^{xy} \sin(xt) - t^2 e^{xy} \cos(xt)
\]
Thus:
\[
u_{xx} = e^{xy} \left( (y^2 - t^2) \cos(xt) - 2ty \sin(xt) \right)
\]

### Step 4: Verify the equation \( u_{xx} = 25 u_t \)
We already computed:
\[
u_t = -x e^{xy} \sin(xt)
\]
So:
\[
25u_t = 25(-x e^{xy} \sin(xt)) = -25x e^{xy} \sin(xt)
\]

Now compare \( u_{xx} \) and \( 25u_t \).

From \( u_{xx} \), we see that there are terms involving \( \cos(xt) \) and \( \sin(xt) \), while \( 25u_t \) involves only the \( \sin(xt) \) term. For the equation to hold, we require the following condition:
\[
(y^2 - t^2) \cos(xt) - 2ty \sin(xt) = -25x \sin(xt)
\]

This must be true for all \( x \), \( t \), and \( y \). Thus, equating the \( \sin(xt) \) terms and the \( \cos(xt) \) terms gives the conditions:
1. \( -2ty = -25x \)
2. \( y^2 - t^2 = 0 \), which implies \( y = \pm t \).

Thus, under the conditions \( y = t \) and \( x = \frac{2t}{25} \), the given function satisfies the differential equation \( u_{xx} = 25 u_t \).

Would you like more details on any of the steps? Here are some related questions to consider:

1. How would the solution change if \( u(x,t) \) were a sine function instead of cosine?
2. Can you explain why the product rule was necessary in this solution?
3. What is the general solution for the given partial differential equation?
4. How does the structure of \( e^{xy} \) affect the form of the solution?
5. Can we generalize the method to verify solutions of other second-order PDEs?

**Tip**: When verifying solutions to PDEs, it’s important to carefully follow the chain rule and product rule for derivatives, especially when dealing with functions of multiple variables.

Verify that u(x,t) = e^{xy} \cos xt is the solution of the differential equation u_{xx} = 25u_t.

Verify that the function u(x,t) = e^{xy} cos xt satisfies the partial differential equation u_{xx} = 25u_t. This problem involves partial derivatives, chain rule, and product rule, suitable for advanced calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation