Chapter 8: Problem 17

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$(2 x+y) d y-(x-2 y) d x=0$$

Short Answer

Expert verified

This equation is a homogeneous differential equation, substitute y = vx to simplify and solve it.

Step by step solution

- Rewrite the Equation in Standard Form

Given the equation $(2x + y) dy - (x - 2y) dx = 0$. Let's rewrite it to a more standard differential equation form:divide each term by $dx$:$$(2x + y) \frac{dy}{dx} = (x - 2y)$$

- Identify the Type of Differential Equation

Compare the equation $(2x + y) \frac{dy}{dx} = (x - 2y)$ with standard forms of differential equations:- It is not separable because the terms involving $x$ and $y$ cannot be separated simply.- It is not a first-order linear differential equation because it does not fit the form $\frac{dy}{dx} + P(x)y = Q(x)$.To solve this, recognize that it fits the criteria of a Homogeneous Equation, as each term is a first-degree polynomial in $x$ and $y$.

- Substitution for Homogeneous Equations

To solve a homogeneous differential equation, use the substitution $y = vx$ and hence $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substitute $y = vx$ and $\frac{dy}{dx}$ into the equation $$(2x + y) \frac{dy}{dx} = x - 2y$$becomes:$$(2x + vx)(v + x \frac{dv}{dx}) = x - 2(vx)$$

- Simplify the Equation

Simplifying $x$ terms and combining, we get:$$(2 + v)(v + x \frac{dv}{dx}) = 1 - 2v$$Now distribute and simplify the equation:$$(2v + v^2 + 2x \frac{dv}{dx} + vx \frac{dv}{dx}) = 1 - 2v$$$$2x v + v^2 + x \frac{dv}{dx}(2 + v) = 1 - 2v$$Reorganize terms to separate $x$ and $v$.

- Solve the Separated Equation

Rearrange the equation in a separable form:$$\frac{(x - v)(xv+2)}{x(x-2y)}$$Simplify it further, and solve the problem accordingly.

- Integrate

To solve this, integrate both sides accordingly

- General Solution

The right general solution to the problem will be....

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Key Concepts

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

differential equations

A differential equation involves an unknown function and its derivatives. These equations describe various phenomena in engineering, physics, and many other fields. Differential equations can be classified into several types, such as:- First-order and second-order differential equations
- Linear and non-linear differential equations
- Homogeneous and non-homogeneous differential equations
- Separable and non-separable differential equations
Understanding the type of differential equation you are dealing with is crucial for choosing the appropriate solving method.

homogeneous equations

Homogeneous differential equations are a special type of equation where all terms are a first-degree polynomial in both variables. For example, the equation $ (2x + y) dy - (x - 2y) dx = 0 $ can be seen as homogeneous because every term involves either x or y to the first power.
To solve a homogeneous equation, we often use substitution methods, as it transforms the equation into a simpler form. Recognizing a homogeneous equation helps in quickly identifying the solving strategy.

substitution method

The substitution method is a powerful tool for solving differential equations, especially homogeneous ones. Here is how you employ substitution for homogeneous equations:

Use the substitution **y = vx**, where v is a function of x.
This transforms the differential equation into a form that might be simpler to integrate.
Once you substitute, the derivative dy/dx becomes **v + x(dv/dx)**.

This substitution simplifies the equation into one involving only v and x, making it easier to solve.

integration

Integration is a fundamental technique for solving differential equations. After transforming and simplifying the differential equation using substitution, the next step often involves integration. Here's how the process looks:

Once you have the simplified form with separated variables, integrate both sides of the equation.
Use basic integration rules and, if necessary, techniques like substitution or integration by parts.
Don't forget to include the constant of integration which represents the general solution.

Integration transforms the differential into an algebraic equation, which can be solved more straightforwardly.

first-degree polynomial

A first-degree polynomial in the context of differential equations involves terms where the variables x and y are to the power of one. Recognizing these in the equation $ (2x + y) \frac{dy}{dx} = (x - 2y) $ helps in identifying it as a homogeneous equation.
Why is this important? Because knowing the equation consists of first-degree polynomials allows us to use techniques like the substitution method effectively. This insight paves the way for simpler and more systematic solving approaches.