Problem 9 $$ \left(x^{2}+y^{2}\right) d x-... [FREE SOLUTION]

Chapter 2: Problem 9

$$ \left(x^{2}+y^{2}\right) d x-x y d y=0 $$

Short Answer

Expert verified

Solution: $ \ln |x| + \frac{xy^2}{2} = C $.

Step by step solution

Identify the Differential Equation Format

The given differential equation is $ (x^2 + y^2) dx - xy dy = 0 $. By rearranging the terms, it can be written in the form $ M(x, y) dx + N(x, y) dy = 0 $, where $ M(x, y) = x^2 + y^2 $ and $ N(x, y) = -xy $.

Check for Exactness

A differential equation $ M(x, y) dx + N(x, y) dy = 0 $ is exact if $ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $. Calculate $ \frac{\partial M}{\partial y} = 2y $ and $ \frac{\partial N}{\partial x} = -y $. Since $ 2y eq -y $, the given equation is not exact.

Solve by Finding an Integrating Factor

To solve the non-exact differential equation, we need an integrating factor. Checking if the equation is solvable via integrating factor involving $ x $, this is noted because $ \frac{M_y - N_x}{N} = \frac{2y + y}{-xy} = \frac{3}{x} $. This indicates the integrating factor could be $ \mu(x) = x^3 $.

Apply Integrating Factor

Multiply each term by $ x^3 $, resulting in $ x^3(x^2 + y^2) dx - x^4y dy = 0 $, or $ (x^5 + x^3y^2) dx - x^4y dy = 0 $.

Check Exactness Again

Now, recalculate for exactness with the modified equation: $ M = x^5 + x^3y^2 $ and $ N = -x^4y $. Then, calculate $ \frac{\partial M}{\partial y} = 2x^3y $ and $ \frac{\partial N}{\partial x} = -4x^3y $. Since $ 2x^3y eq -4x^3y $, correct mistake by using $ \mu(x) = x^{-3} $ instead.

Correct Integrating Factor and Multiply

Multiply each term by $ x^{-3} $, giving $(x^{-1}y^2 + 1) dx - xy dx = 0 $, simplify to (1 + x^{-1}y^2) dx - xy dy = 0.

Integrate to Find Solution

Now, integrate $ 1 + x^{-1}y^2 $ with respect to $ x $, and $-xy $ with respect to $ y $. This simplifies to $ \ln |x| + xy^2/2 = C$, which is the final solution, where $ C $ is a constant.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integrating Factor

An integrating factor is a function that we use to turn a non-exact differential equation into an exact one. This allows us to solve the equation more easily. In this exercise, the original differential equation was non-exact, which meant that its terms didn鈥檛 neatly align for solving.
The process involves:

Finding a function, typically in the form of a variable-dependent expression, such as $ \mu(x) $ or $ \mu(y) $.
Multiplying each term of the differential equation by this integrating factor.
Re-checking the exactness of the modified equation.

For this problem, we initially guessed $ \mu(x) = x^3 $ based on specific relationships between the terms. However, recalculating led to the choice $ \mu(x) = x^{-3} $, turning the differential equation into an exact one. This illustrates the trial-and-error involved in finding the perfect integrating factor.

Partial Derivatives

Partial derivatives play a vital role when determining if a differential equation is exact. They are used to find the rate at which a function changes with respect to one of its variables while keeping the others constant. In our exercise:
We calculate partial derivatives of $ M(x, y) $ and $ N(x, y) $.

The partial derivative of $ M $ with respect to $ y $, denoted $ \frac{\partial M}{\partial y} $, was $ 2y $.
The partial derivative of $ N $ with respect to $ x $, $ \frac{\partial N}{\partial x} $, was $ -y $.

By comparing these partial derivatives, we determine whether the equation is initially exact. Exactness is achieved only when these derivatives are equal, which in this case, they were not.

Exactness of Differential Equations

Exactness of a differential equation indicates that it can be integrated directly to find a solution. A differential equation in the form $ M(x, y)dx + N(x, y)dy = 0 $ is exact if:
$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $.In simpler terms, this equality means that the changes in each direction (with respect to $ x $ and $ y $) align perfectly. In this problem, initially:

$ \frac{\partial M}{\partial y} = 2y $.
$ \frac{\partial N}{\partial x} = -y $.

The inequality hinted that the given differential equation was non-exact. The whole goal then becomes adjusting the equation to make these partial derivatives equal by applying an integrating factor.

Non-Exact Differential Equation Solving

When you encounter a non-exact differential equation, like in our exercise, you must first recognize this by comparing partial derivatives. The initial step, after realizing the non-exact nature, is to find a way to convert it into an exact problem. Here's how we approach such cases:

Identify the inequality in the partial derivatives.
Search for an integrating factor that makes the equation exact.
Re-evaluate the resulting equation to confirm exactness.

In the instance provided, an integrating factor $ \mu(x) = x^{-3} $ rectified the inexactness. After applying it, the resulting differential equation had equal partial derivatives, thus allowing us to integrate and find the solution. The solution requires correcting the equation format for direct integration and dealing with any computational errors during the process.

91影视

$$ \left(x^{2}+y^{2}\right) d x-x y d y=0 $$

Short Answer

Step by step solution

Identify the Differential Equation Format

Check for Exactness

Solve by Finding an Integrating Factor

Apply Integrating Factor

Check Exactness Again

Correct Integrating Factor and Multiply

Integrate to Find Solution

Key Concepts

Integrating Factor

Partial Derivatives

Exactness of Differential Equations

Non-Exact Differential Equation Solving

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Calculus

Mechanics Maths

Pure Maths

Probability and Statistics

Logic and Functions

Study anywhere. Anytime. Across all devices.