Problem 29

Question

Determine which of the five types of differential equations we have studied the given equation falls into (see Table $1.12 .1),$ and use an appropriate technique to find the general solution. $$y[\ln (y / x)+1] d x-x d y=0$$

Step-by-Step Solution

Verified

Answer

The given differential equation is not one of the five types we have studied directly. However, by manipulating the equation and introducing a new variable $v = \frac{y}{x}$, we transformed the equation into a separable form. The general solution is then found to be: $\ln\left(\frac{y}{x}y\right) = \frac{1}{3}x^3 + C$.

1Step 1: Writing the given differential equation in the standard form

First, we rewrite the given differential equation in the standard form, which is $M(x, y)dx + N(x, y)dy = 0$. This makes it easier to analyze the equation and decide which type it is. $$ y[\ln(y/x)+1] dx - x dy = 0 $$ We can rewrite the given equation as: $$ (\ln(\frac{y}{x})+1) y dx - x dy = 0 $$ The standard form is $M(x, y)dx + N(x, y)dy = 0$, so we can say $M(x,y) = (\ln(\frac{y}{x})+1) y$, and $N(x,y) = -x$.

2Step 2: Determine the type of differential equation

Now, let's analyze to see which type of differential equation we have. 1. Linear: It is not a linear differential equation since it does not have the form $y' + p(x)y = q(x)$. 2. Separable: It is also not separable, as we cannot rewrite the equation so that all x's and y's appear on separate sides of the equation. 3. Bernoulli: The equation is not in the form $y' + p(x)y = q(x)y^n$, so it is not a Bernoulli equation. 4. Homogeneous: The given equation is not homogeneous because the terms in the equation do not involve a homogeneous function of the same order. Now let's check if the given equation is exact.

3Step 3: Check if the equation is exact

An equation is exact if: $$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$ Let's find the partial derivatives. First, find the partial derivative of M with respect to y: $$ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y} [(\ln(\frac{y}{x})+1) y] = \frac{1}{y} + 1 $$ Next, find the partial derivative of N with respect to x: $$ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(-x) = -1 $$ Since these two partial derivatives are not equal, the given differential equation is not exact. Since the given differential equation does not fall into any of the five types of differential equations we have studied, we may need to manipulate the equation before finding a suitable method to solve it.

4Step 4: Manipulate the equation

Let's rewrite the equation and then attempt to make it separable: $$ (\ln(\frac{y}{x})+1)yd x - x dy = 0 $$ Let's divide both sides of the equation by $xy$: $$ (\ln(\frac{y}{x})+1) \frac{y}{x} d x - dy = 0 $$ Notice that we can rewrite the $y/x$ term as $v$ if we let $v = \frac{y}{x}$, and we have the equation in the form: $$ (\ln(v)+1) v d x - d y = 0 $$ Now we can say, $$ v = \frac{y}{x} $$ Differentiating both sides with respect to x, assuming y(x) is a function of x: $$ \frac{d v}{dx} = \frac{ \frac{dy}{dx}x - y }{x^2} $$ Now let's rewrite the given equation by substituting $dv/dx$ for $dy$: $$ (\ln(v)+1) v d x - x^2 \frac{d v}{dx} = 0 $$ Rearrange the terms of the equation: $$ (\ln(v)+1) v d x - x^2 \frac{d v}{dx} = 0 $$ Notice that now the equation seems separable. The partial derivatives are equal, so the equation is exact. At this point, we solve the equation in its final form: $$ (\ln(v)+1) v d x = x^2 \frac{d v}{d x} $$

5Step 5: Solving the separable equation

Now, we can separate the variables to integrate: $$ \frac{(\ln(v)+1) }{v} d v = x^2 d x $$ Integrating both sides: $$ \int \frac{(\ln(v)+1) }{v} d v = \int x^2 d x $$ $$ \ln(v) + \int \frac{1}{v} d v = \frac{1}{3}x^3 + C $$ Now substitute back in $v = \frac{y}{x}$: $$ \ln\left(\frac{y}{x}\right) + \ln(y) = \frac{1}{3}x^3 + C $$ The general solution of the given differential equation is: $$ \ln\left(\frac{y}{x}y\right) = \frac{1}{3}x^3 + C $$

Key Concepts

Differential Equation TypesSeparable Differential EquationsExact Differential Equations

Differential Equation Types

Understanding the types of differential equations is essential for solving them effectively. Differential equations are mathematical equations that relate functions with their derivatives. They are used to describe various phenomena in engineering, physics, economics, and beyond.

There are several types of differential equations, including but not limited to:

Ordinary Differential Equations (ODEs): Involve functions of a single variable and their derivatives.
Partial Differential Equations (PDEs): Involve functions of multiple variables and their partial derivatives.
Linear Differential Equations: Have solutions that form a vector space.
Nonlinear Differential Equations: Solutions do not form a vector space and are more complex.
Homogeneous Differential Equations: Both sides of the equation can be expressed as a function of the derivative alone.
Inhomogeneous or Nonhomogeneous: Have a non-zero term that is independent of the function and its derivatives.
Exact Differential Equations: Have the form where there exists a function whose partial derivatives correspond to the terms in the equation.
Separable Differential Equations: Can be expressed as the product of two functions, each depending only on a single variable.

The strategy for solving a differential equation often depends on identifying its type correctly. Incorrect classification may lead one to choose an inefficient or incorrect method of solution.

Separable Differential Equations

Separable differential equations are a class of differential equations in which the variables can be separated on opposite sides of the equation. Essentially, all terms containing the independent variable x are moved to one side, while all terms containing the dependent variable y are moved to the other.

A general separable equation takes on the form: \

$ g(y)dy = f(x)dx $

One can integrate both sides to find the general solution: \

$ \int g(y)dy = \int f(x)dx $

This technique simplifies the equation into two separate integrals which are usually easier to solve.

However, not all equations are easily separated due to the complexity of their terms. Sometimes, additional manipulation, such as substituting variables or factoring, is required to transform the equation into a separable form. This was seen in the exercise provided, where substituting $ v = \frac{y}{x} $ allowed us to rewrite the given equation into a separable form. Thus, we were able to solve it by integrating both sides after separation.

Exact Differential Equations

Exact differential equations are a category where there exists a particular relationship between the terms involving the derivatives of a function. An exact differential equation is identifiable through the standard form \

$ M(x, y)dx + N(x, y)dy = 0 $

where $ M $ and $ N $ are functions of $ x $ and $ y $ respectively. For an equation to be exact, the following condition must be met: \

$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $

This implies that the cross partial derivatives must be equal. The rationale behind this condition lies in Clairaut's theorem on the equality of mixed partial derivatives for continuous functions.

When an equation is exact, it indicates that there is a function $ F(x, y) $ such that $ dF = Mdx + Ndy $. One can then typically find the solution by computing a potential function $ F $ for which $ M $ and $ N $ are the partial derivatives with respect to $ x $ and $ y $, respectively. In the exercise example, it was initially assumed the equation was not exact as the cross partial derivatives were not equal. However, after manipulation, it turned out to be a separable equation, showcasing the importance of thoroughly analyzing the differential equations we encounter.

Problem 29