Math Problem Statement

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

Solution

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=y2+2f(1x+logy)Z = y^2 + 2f\left(\frac{1}{x} + \log y\right)

where ff is an arbitrary function of the expression 1x+logy\frac{1}{x} + \log y.

Steps to Form the PDE

  1. Differentiate with respect to xx:

    • When differentiating ZZ with respect to xx, treat yy as a constant. The derivative of ff with respect to xx will involve the chain rule.
  2. Differentiate with respect to yy:

    • Similarly, differentiate ZZ with respect to yy, treating xx as a constant.
  3. Eliminate the Arbitrary Function ff:

    • Use the results from the derivatives to eliminate the arbitrary function ff and its derivative from the equations.

Let's Perform the Differentiations

1. Differentiate ZZ with respect to xx:

Zx=2f(1x+logy)(1x2)\frac{\partial Z}{\partial x} = 2f'\left(\frac{1}{x} + \log y\right) \cdot \left(-\frac{1}{x^2}\right)

Here, ff' denotes the derivative of ff with respect to its argument.

2. Differentiate ZZ with respect to yy:

Zy=2y+2f(1x+logy)1y\frac{\partial Z}{\partial y} = 2y + 2f'\left(\frac{1}{x} + \log y\right) \cdot \frac{1}{y}

Eliminate ff

To eliminate ff', let's divide the partial derivative with respect to yy by the partial derivative with respect to xx:

Zy2yZx=2fy2fx2=x21y\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:

Zy+x21yZx=2y\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 ff is:

Zy+x21yZx=2y\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.

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Math Problem Analysis

Mathematical Concepts

Partial Differential Equations
Chain Rule
Partial Derivatives

Formulas

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Theorems

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Suitable Grade Level

Advanced Mathematics