Chapter 9: Problem 42
Show that \(z=x^{2} y^{2} /\left(x^{2}+y^{2}\right)\) satisfies the differential equation $$ x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}=2 z $$
Short Answer
Expert verified
The equation is satisfied as shown by the simplification steps.
Step by step solution
01
Compute Partial Derivative of z with respect to x
Given \( z = \frac{x^2 y^2}{x^2 + y^2} \), use the quotient rule to find \( \frac{\partial z}{\partial x} \). The quotient rule is \( \frac{f}{g} = \frac{f'g - fg'}{g^2} \). Here, \( f = x^2 y^2 \) and \( g = x^2 + y^2 \).The derivative of \( f \) with respect to \( x \) is \( 2xy^2 \). The derivative of \( g \) with respect to \( x \) is \( 2x \). Substitute into the quotient rule:\[ \frac{\partial z}{\partial x} = \frac{(2xy^2)(x^2 + y^2) - (x^2y^2)(2x)}{(x^2 + y^2)^2} = \frac{2x^3y^2 + 2xy^4 - 2x^3y^2}{(x^2 + y^2)^2} = \frac{2xy^4}{(x^2 + y^2)^2}. \]
02
Compute Partial Derivative of z with respect to y
Find \( \frac{\partial z}{\partial y} \) using the quotient rule. With \( f = x^2 y^2 \) and \( g = x^2 + y^2 \), the derivative of \( f \) with respect to \( y \) is \( 2x^2 y \), and the derivative of \( g \) with respect to \( y \) is \( 2y \). Applying the quotient rule:\[ \frac{\partial z}{\partial y} = \frac{(2x^2 y)(x^2 + y^2) - (x^2 y^2)(2y)}{(x^2 + y^2)^2} = \frac{2x^2 y^3 + 2x^4 y - 2x^2 y^3}{(x^2 + y^2)^2} = \frac{2x^4 y}{(x^2 + y^2)^2}. \]
03
Evaluate x(∂z/∂x) + y(∂z/∂y)
Substitute the partial derivatives back into the expression \( x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} \):\[ \begin{align*}x \frac{\partial z}{\partial x} & = x \cdot \frac{2xy^4}{(x^2 + y^2)^2} = \frac{2x^2y^4}{(x^2 + y^2)^2}, \y \frac{\partial z}{\partial y} & = y \cdot \frac{2x^4y}{(x^2 + y^2)^2} = \frac{2x^4y^2}{(x^2 + y^2)^2}. \end{align*}\]Add these results together:\[ \frac{2x^2y^4}{(x^2 + y^2)^2} + \frac{2x^4y^2}{(x^2 + y^2)^2} = \frac{2x^2y^4 + 2x^4y^2}{(x^2 + y^2)^2} = \frac{2x^2y^2(x^2 + y^2)}{(x^2 + y^2)^2} = \frac{2x^2y^2}{x^2 + y^2}. \]
04
Confirm Final Result
The expression \( x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} \) simplifies to:\[ \frac{2x^2y^2}{x^2 + y^2} = 2z. \]This confirms that the given differential equation \( x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = 2z \) is satisfied.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
The quotient rule is a tool used in calculus to differentiate functions that are the ratio of two differentiable functions. In this scenario, we are working with a function in the form of a fraction, with the numerator and denominator being functions of variables. The quotient rule is expressed as:
- If you have a function \( f(x) = \frac{u(x)}{v(x)} \), then the derivative \( f'(x) \) is \( \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \).
Partial Derivative
Partial derivatives are an essential concept in the study of multivariable calculus. When a function depends on more than one variable, partial derivatives are used to explore the rate of change with respect to one of those variables while keeping the others constant.
- It is denoted by the symbol \( \frac{\partial}{\partial x} \) or \( \partial_x \) for a variable \( x \).
Differential Equation Solution
A differential equation is an equation involving functional derivatives, which represent rates of change. Solving a differential equation provides a function or set of functions that satisfies the relationship dictated by the equation. In this example, our task was to demonstrate that a given function satisfies a specific differential equation. By substituting the partial derivatives calculated with the quotient rule back into the equation \( x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = 2z \), we were able to verify the solution. The equation simplifies to confirm that our original function \( z = \frac{x^2 y^2}{x^2 + y^2} \) indeed satisfies this relationship.
- This involves not just computing derivatives but combining them to match the pattern needed on both sides of the equation to verify the solution.