Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 2: Problem 10
Classify each differential equation as separable, exact, linear, homogeneous, or Bernoulli. Some equations may be more than one kind. Do not solve the equations. $$2 x y y^{\prime}+y^{2}=2 x^{2}$$
Short Answer
Expert verified
The equation is a Bernoulli differential equation.
Step by step solution
01
Identify the Given Differential Equation
The given equation is \( 2xyy' + y^2 = 2x^2 \). We need to classify this differential equation using the properties and forms of different types of differential equations.
02
Check for Separability
A differential equation is separable if it can be expressed in the form \( g(y) \, dy = f(x) \, dx \). Rearranging the given equation, we have \( 2xy \frac{dy}{dx} + y^2 = 2x^2 \). However, isolating \( y' \) and attempting to express the terms involving \( x \) and \( y \) as products of functions of \( x \) and functions of \( y \) is not possible. Therefore, the equation is not separable.
03
Check for Exactness
An exact equation has the form \( M(x,y) \, dx + N(x,y) \, dy = 0 \) where the partial derivatives satisfy \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). Rewriting our equation gives \( -2x^2 \, dx + (2xy + y^2) \, dy = 0 \). Here, \( M = -2x^2 \) and \( N = 2xy + y^2 \). Since \( \frac{\partial M}{\partial y} = 0 \) and \( \frac{\partial N}{\partial x} = 2y \), they are not equal, so the equation is not exact.
04
Check for Linearity
The differential equation is linear if it can be written in the form \( \frac{dy}{dx} + P(x)y = Q(x) \). Comparing with the form \( y' + P(x)y = Q(x) \), the presence of \( y^2 \) (a non-linear term) in the equation makes it non-linear, as it cannot be rearranged to fit the linear form.
05
Check for Homogeneity
A first-order differential equation is homogeneous if it can be expressed as \( y' = F(\frac{y}{x}) \). Rearranging the original equation into this form isn't feasible, and using the test for homogeneous functions on \( M \) and \( N \) shows they do not exhibit the same degree, confirming the equation is not homogeneous.
06
Check for Bernoulli Form
Bernoulli's equation is of the form \( y' + P(x)y = Q(x)y^n \). Observing the equation \( 2xyy' + y^2 = 2x^2 \), rewriting as \( y' + Py = Qy^2 \) or \( 2xy \cdot \frac{dy}{dx} = 2x^2 - y^2 \), you see \( y^2 \) is present, which matches the Bernoulli form with \( n=2 \). Thus, this equation is a Bernoulli equation.
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.
Separable Differential Equations
Separable differential equations are a category of differential equations where the variables can be separated onto opposite sides of the equation. The standard form is \[ g(y) \, dy = f(x) \, dx \].
In other words, when you can rearrange the equation such that one function involving only y is on one side and one function involving only x is on the other, it is separable.
What makes these equations convenient is that you can solve them by integrating both sides separately.
In the given equation \( 2xyy' + y^2 = 2x^2 \), we find that it does not fit the separable form because we can't cleanly separate the y terms from the x terms. It would require us to express terms like \( 2xy \) or \( y^2 \) independently matched to functions that are purely y or x, which isn't possible here.
In other words, when you can rearrange the equation such that one function involving only y is on one side and one function involving only x is on the other, it is separable.
What makes these equations convenient is that you can solve them by integrating both sides separately.
- Identify functions of y and x distinctly.
- Solve by integration.
In the given equation \( 2xyy' + y^2 = 2x^2 \), we find that it does not fit the separable form because we can't cleanly separate the y terms from the x terms. It would require us to express terms like \( 2xy \) or \( y^2 \) independently matched to functions that are purely y or x, which isn't possible here.
Exact Differential Equations
An exact differential equation has a particular form where the equation can be expressed as \( M(x,y) \, dx + N(x,y) \, dy = 0 \) and the condition \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \) is satisfied.
This condition ensures that there exists a potential function \( \Phi(x, y) \) such that \( d\Phi = M dx + N dy \). Here's a simple process to test exactness:
In our case, the given equation, when rewritten as \( -2x^2 \, dx + (2xy + y^2) \, dy = 0 \), identifies \( M = -2x^2 \) and \( N = 2xy + y^2 \).Since \( \frac{\partial M}{\partial y} = 0 \) and \( \frac{\partial N}{\partial x} = 2y \), the equation is not exact, because the partial derivatives are not equal.
This condition ensures that there exists a potential function \( \Phi(x, y) \) such that \( d\Phi = M dx + N dy \). Here's a simple process to test exactness:
- Identify M and N from the equation.
- Calculate \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \).
- Check if they are equal.
In our case, the given equation, when rewritten as \( -2x^2 \, dx + (2xy + y^2) \, dy = 0 \), identifies \( M = -2x^2 \) and \( N = 2xy + y^2 \).Since \( \frac{\partial M}{\partial y} = 0 \) and \( \frac{\partial N}{\partial x} = 2y \), the equation is not exact, because the partial derivatives are not equal.
Linear Differential Equations
Linear differential equations are those which can be expressed in the form \( \frac{dy}{dx} + P(x)y = Q(x) \).
Here, \( P(x) \) and \( Q(x) \) are functions of x only. Importantly, in a linear equation, the dependent variable y and its derivative y' appear to the power of 1.
Attributes of linear differential equations include:
In the equation \( 2xyy' + y^2 = 2x^2 \), the presence of the term \( y^2 \) directly violates the linearity condition.Since it involves y raised to a power higher than one, the equation cannot be categorized as linear.
Here, \( P(x) \) and \( Q(x) \) are functions of x only. Importantly, in a linear equation, the dependent variable y and its derivative y' appear to the power of 1.
Attributes of linear differential equations include:
- No products or nonlinear functions of y and its derivatives.
- Standard operations involved are addition and scalar multiplication.
In the equation \( 2xyy' + y^2 = 2x^2 \), the presence of the term \( y^2 \) directly violates the linearity condition.Since it involves y raised to a power higher than one, the equation cannot be categorized as linear.
Bernoulli Differential Equations
Bernoulli differential equations provide a bridge between linear and nonlinear equations. They are of the form \( y' + P(x)y = Q(x)y^n \), with \( n eq 0,1 \).
This non-linear term \( y^n \) makes it challenging yet solvable through transformation.
How to approach Bernoulli equations:
In analyzing the equation \( 2xyy' + y^2 = 2x^2 \), it shows characteristics of a Bernoulli equation.
By recognizing the appearance of \( y^2 \) and rewriting, it conforms to the Bernoulli structure \( y' + P(x)y = Q(x)y^2 \) with \( n = 2 \).Therefore, we can classify it as a Bernoulli equation.
This non-linear term \( y^n \) makes it challenging yet solvable through transformation.
How to approach Bernoulli equations:
- Recognize the equation's form \( y' + P(x)y = Q(x)y^n \).
- If \( n eq 1 \), the equation is a Bernoulli equation.
- Introduce a substitution \( v = y^{1-n} \) to convert it to a linear form.
In analyzing the equation \( 2xyy' + y^2 = 2x^2 \), it shows characteristics of a Bernoulli equation.
By recognizing the appearance of \( y^2 \) and rewriting, it conforms to the Bernoulli structure \( y' + P(x)y = Q(x)y^2 \) with \( n = 2 \).Therefore, we can classify it as a Bernoulli equation.
