Z=y²+2f[(1/x)+logy] Form an pde by eliminating arbitary function. First time hearing this question no idea about this not even integration differentiation and trigonometry and other stuff let's start from basics

Let's break down the problem step by step, starting with some basic concepts. 

### What is a PDE?
A **Partial Differential Equation (PDE)** is a type of mathematical equation that involves partial derivatives of an unknown function with respect to multiple variables. The process of forming a PDE involves eliminating arbitrary functions or constants from a given equation.

### Given Problem
You have the equation:

\[
Z = y^2 + 2f\left(\frac{1}{x} + \log y\right)
\]

where \( f \) is an arbitrary function of the expression \( \frac{1}{x} + \log y \).

### Steps to Form the PDE

1. **Differentiate with respect to \( x \)**:
   - When differentiating \( Z \) with respect to \( x \), treat \( y \) as a constant. The derivative of \( f \) with respect to \( x \) will involve the chain rule.

2. **Differentiate with respect to \( y \)**:
   - Similarly, differentiate \( Z \) with respect to \( y \), treating \( x \) as a constant.

3. **Eliminate the Arbitrary Function \( f \)**:
   - Use the results from the derivatives to eliminate the arbitrary function \( f \) and its derivative from the equations.

### Let's Perform the Differentiations

#### 1. Differentiate \( Z \) with respect to \( x \):
\[
\frac{\partial Z}{\partial x} = 2f'\left(\frac{1}{x} + \log y\right) \cdot \left(-\frac{1}{x^2}\right)
\]

Here, \( f' \) denotes the derivative of \( f \) with respect to its argument.

#### 2. Differentiate \( Z \) with respect to \( y \):
\[
\frac{\partial Z}{\partial y} = 2y + 2f'\left(\frac{1}{x} + \log y\right) \cdot \frac{1}{y}
\]

### Eliminate \( f \)
To eliminate \( f' \), let's divide the partial derivative with respect to \( y \) by the partial derivative with respect to \( x \):

\[
\frac{\frac{\partial Z}{\partial y} - 2y}{\frac{\partial Z}{\partial x}} = \frac{\frac{2f'}{y}}{-\frac{2f'}{x^2}} = -x^2 \cdot \frac{1}{y}
\]

This gives us the PDE:

\[
\frac{\partial Z}{\partial y} + x^2 \cdot \frac{1}{y} \cdot \frac{\partial Z}{\partial x} = 2y
\]

### Summary
The PDE formed by eliminating the arbitrary function \( f \) is:

\[
\frac{\partial Z}{\partial y} + x^2 \cdot \frac{1}{y} \cdot \frac{\partial Z}{\partial x} = 2y
\]

This is the required Partial Differential Equation.

Would you like to dive deeper into any of these steps or need further clarification? Here are some related questions:

1. What is the chain rule in differentiation, and how is it used in this context?
2. How do you perform partial differentiation for functions of multiple variables?
3. What are the differences between ordinary and partial differential equations?
4. How do we interpret the physical meaning of a PDE in applied contexts?
5. Why is eliminating the arbitrary function important in forming a PDE?

**Tip**: Understanding the chain rule and partial derivatives is crucial for working with PDEs. These are fundamental concepts in calculus that allow us to deal with functions of multiple variables.

Learn how to form a Partial Differential Equation (PDE) by eliminating arbitrary functions from a given equation step-by-step. This problem involves advanced mathematical concepts such as Partial Differential Equations, Chain Rule, and Partial Derivatives.

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