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

Short Answer

Expert verified
The given equation is separable. The solution is y+2x=m x ln y=-2

Step by step solution

01

Identify the type of differential equation

The given differential equation is d(x+y)=0 Rewriting it gives us:dy/dx = f(x)dy=-x/(2x) which identifies it as a separable differential equation since it can be written as dy = g(x)dx, and the variables can be separated.
02

Separate the variables

Separate the variables by dividing both sides by (y+2x).dx/(y+2x)d(y+2x=1)ydx/dxy-(2in)dx/dx=dxd(x)+2dx=0
03

Integrate both sides

Integrate both sides of the equation: pm the function gives. $$g()$ m+Int dx = c( x*x.0-d8pm*dx)
04

Solve for the constant.

Rewrite the integrated terms to solve for the constant: y+2x-c8dx = ln (c+p) +c8.
05

Simplify the solution

Interestingly, it can be written asy y=-2xln(m*x)

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Key Concepts

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

First-Order Differential Equations
First-order differential equations involve the first derivative of a function but no higher-order derivatives.
In mathematical terms, they can be written in the form:
\[ \frac{dy}{dx} = f(x, y) \]
These equations describe how a function changes with respect to an independent variable.
Common types of first-order differential equations include separable, linear, and exact.
In the given exercise, the differential equation can be identified as separable because the variables can be separated on opposite sides of the equation.
Recognizing the type of differential equation helps in choosing the right method to solve it.
Integration
Integration is a fundamental concept in calculus used to find the antiderivative or the area under a curve.
In the context of solving differential equations, integration is applied after separating variables.
Once you have separated the variables, you integrate both sides with respect to their respective variables.
For instance, if we have:
\[ \frac{dy}{dx} = g(x)h(y) \]
We separate the variables as:
\[ \frac{1}{h(y)} dy = g(x) dx \]
Then, we integrate both sides:
\[ \begin{align*} \text{Integrate:} \ \frac{1}{h(y)} dy &= \text{Integral of } \frac{1}{h(y)} \text{ with respect to } y \ g(x) dx &= \text{Integral of } g(x) \text{ with respect to } x \end{align*} \]
This process gives us the solution that usually involves a constant of integration.
Clearly separating and then integrating is crucial in solving these problems.
Solving Differential Equations
Solving a differential equation involves finding a function that satisfies the given equation.
The steps usually include:
  • Identifying the type of differential equation.
  • Separating variables (if possible).
  • Integrating both sides.
  • Solving for the constant of integration.
  • Simplifying the solution.

In the given exercise:
  • The equation was identified as a separable differential equation.
  • Variables were separated: \[ (y+2x) dx - x dy = 0 \]
  • The equation was rearranged and integrated: \[ \frac{dy}{dx} = -\frac{x}{y+2x} \]
  • Both sides were integrated and simplified to find the general solution: \[ y + 2x = C \]
Solving differential equations often involves these systematic steps.
Each step needs to be carefully executed for accurate results.
Always verify your final solution by plugging it back into the original equation.

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Most popular questions from this chapter

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$x y^{\prime \prime}+y^{\prime}=4 x$$

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

By using Laplace transforms, solve the following differential equations subject to the given initial conditions. $$y^{\prime \prime}-2 y^{\prime}+y=2 \cos t, \quad y_{0}=5, y_{0}^{\prime}=-2$$

Find the family of curves satisfying the differential equation \((x+y) d y+(x-y) d x=0\) and also find their orthogonal trajectories.

Let \(D\) stand for \(d / d x,\) that is, \(D y=d y / d x ;\) then $$D^{2} y=D(D y)=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d^{2} y}{d x^{2}}, \quad D^{3} y=\frac{d^{3} y}{d x^{3}}, \text { etc. }$$ \(D\) (or an expression involving \(D\) ) is called a differential operator. Two operators are equal if they give the same results when they operate on \(y\). For example, $$D(D+x) y=\frac{d}{d x}\left(\frac{d y}{d x}+x y\right)=\frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=\left(D^{2}+x D+1\right) y$$ so we say that $$D(D+x)=D^{2}+x D+1$$ In a similar way show that: (a) \(\quad(D-a)(D-b)=(D-b)(D-a)=D^{2}-(b+a) D+a b\) for constant \(a\) and \(b\). (b) \(\quad D^{3}+1=(D+1)\left(D^{2}-D+1\right)\) (c) \(\quad D x=x D+1 .\) (Note that \(D\) and \(x\) do not commute, that is, \(D x \neq x D .\) ) (d) \(\quad(D-x)(D+x)=D^{2}-x^{2}+1,\) but \((D+x)(D-x)=D^{2}-x^{2}-1\) Comment: The operator equations in (c) and (d) are useful in quantum mechanics; see Chapter 12, Section 22.

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