Problem 7

Question

Let $M=x^{2}-y^{2}$ and $N=x^{2}-2 x y$ so that $M_{y}=-2 y$ and $N_{x}=2 x-2 y .$ The equation is not exact.

Step-by-Step Solution

Verified

Answer

The equation is not exact; find a suitable integrating factor.

1Step 1: Check Exactness

To determine if the differential equation is exact, we need to compare $ M_y $ and $ N_x $ as follows: $ M_y = -2y $ and $ N_x = 2x - 2y $. Since $ -2y eq 2x - 2y $, the equation is not exact. This confirms what is given in the problem statement.

2Step 2: Find Integrating Factor

Since the equation is not exact, we may use an integrating factor to make it exact. A function $ \mu(y) $ or $ \mu(x) $ can be used, such that after multiplying the original functions $ M $ and $ N $ by $ \mu $, the resulting equation becomes exact.

3Step 3: Identify Potential Integrating Factor

For the simplicity, let's consider a possible integrating factor which is a function of $ x $ or $ y $. Matrix manipulation may guide us to such factor, for this exercise let's explore potential symmetry or linearity in terms of $ x $ or $ y $.

4Step 4: Apply Integrating Factor to Test Function

An integrating factor based on symmetry could be $ \mu(y) = \frac{1}{y} $ if it simplifies the problem, or $ \mu(x) = \frac{1}{x} $. We test these by recalculating partial derivatives. With the proper choice of the integrating factor, the equation should become exact.

5Step 5: Re-check Exactness With Integrating Factor

After identifying and choosing an appropriate integrating factor, multiply both $ M $ and $ N $ by this factor, and re-calculate $ M_y $ and $ N_x $. Check if $ M_y $ equals $ N_x $. If these are equal, the pair will satisfy exactness conditions, thereby rendering our new equation exact.

Key Concepts

Exact EquationsIntegrating FactorPartial DerivativesExactness Condition

Exact Equations

In the study of differential equations, an **exact equation** is one where there is a perfect balance between two expressions, usually indicated by partial derivatives. An equation is considered exact if there exists a function whose derivatives match the components of the equation. For instance, in a differential equation of the form $ M(x,y) + N(x,y)\frac{dy}{dx} = 0 $, exactness can be determined by checking if the partial derivative of $ M $ with respect to $ y $ is equal to the partial derivative of $ N $ with respect to $ x $.
This means:

$ M_y = N_x $

If these conditions are satisfied, the equation is exact, meaning it can be derived from a potential function $ \phi(x,y) $. Understanding the concept of exact equations helps in identifying situations where direct integration is possible.

Integrating Factor

When a differential equation isn't exact, we can employ an **integrating factor** to transform it into an exact one. An integrating factor is a function, $ \mu(x) $ or $ \mu(y) $, that when multiplied to the entire differential equation, creates a situation where exactness can be achieved.
The process often involves these steps:

Determine if $ \mu(x) $ or $ \mu(y) $ would simplify the equation.
Multiply the original differential components by this assumed function.
Check to see if the equation becomes exact after multiplication.

This might involve some trial and error, as the integrating factor often depends on the specific characteristics of the differential equation. A successful application of an integrating factor allows for straightforward integration and solution finding.

Partial Derivatives

**Partial derivatives** are a cornerstone in solving differential equations, especially when dealing with functions of multiple variables, $ M(x,y) $ and $ N(x,y) $. A partial derivative with respect to $ x $ shows how a function changes as $ x $ changes while $ y $ is held constant, and vice versa for $ y $.
In the context of differential equations:

$ M_y $ represents the partial derivative of $ M $ with respect to $ y $.
$ N_x $ represents the partial derivative of $ N $ with respect to $ x $.

These derivatives are critical for checking exactness, as they allow us to compare $ M_y $ with $ N_x $. Understanding partial derivatives is crucial, as they help determine how changes in variables affect the overall equation.

Exactness Condition

The **exactness condition** is a specific criterion used to assess whether a differential equation is exact. To satisfy the exactness condition, the partial derivative of $ M(x,y) $ with respect to $ y $, $ M_y $, must equal the partial derivative of $ N(x,y) $ with respect to $ x $, $ N_x $.
This boils down to:

$ M_y = N_x $

If this relationship holds true, the differential equation is exact, and it implies that a potential function $ \phi(x, y) $ exists, which can be integrated to find a solution. When the exactness condition isn't met, techniques like using an integrating factor become necessary to manipulate the equation into an exact form, enabling solution by integration. This condition serves as a critical test when first evaluating any differential equation.

Problem 7

Problem 8

Other exercises in this chapter

Problem 7

Letting $y=u x$ we have $$\begin{aligned} (u x-x) d x-(u x+x)(u d x+x d u) &=0 \\ \left(u^{2}+1\right) d x+x(u+1) d u &=0 \\ \frac{d x}{x}+\frac{u+1}{u^{2}+1}

View solution

Problem 7

For $y^{\prime}+\frac{1}{x} y=\frac{1}{x^{2}}$ an integrating factor is $e^{\int(1 / x) d x}=x$ so that $\frac{d}{d x}[x y]=\frac{1}{x}$ and \(y=\frac{1}{

View solution

Problem 8

A model is $$\begin{aligned}&\frac{d x_{1}}{d t}=(4 \mathrm{gal} / \mathrm{min})(0 \mathrm{lb} / \mathrm{gal})-(4 \mathrm{gal} / \mathrm{min})\left(\frac{1}{200

View solution

Problem 8

From $y e^{y} d y=\left(e^{-x}+e^{-3 x}\right) d x$ we obtain $y e^{y}-e^{y}+e^{-x}+\frac{1}{3} e^{-3 x}=c$.

View solution