Math Problem Statement

(x+2y) dx+(2y-5y) dy=0

Solution

The given differential equation is:

(x+2y)dx+(2y5y)dy=0(x + 2y) \, dx + (2y - 5y) \, dy = 0

Simplifying the second term:

(2y5y)=3y(2y - 5y) = -3y

Now, the equation becomes:

(x+2y)dx3ydy=0(x + 2y) \, dx - 3y \, dy = 0

This is a first-order, non-linear differential equation. To solve it, we will try to separate the variables xx and yy.

Step 1: Separate the Variables

Rearranging the terms to separate xx and yy:

dx3y=dyx+2y\frac{dx}{-3y} = \frac{dy}{x + 2y}

This form is not separable in its current form, but we can try substitution to simplify the equation.

Step 2: Use a Substitution

Let us use the substitution v=xyv = \frac{x}{y}. Thus, x=vyx = vy and dx=vdy+ydvdx = v \, dy + y \, dv. Substituting these into the equation:

(vy+2y)(vdy+ydv)3ydy=0(vy + 2y) (v \, dy + y \, dv) - 3y \, dy = 0

After substitution, you can try solving for vv or check alternative methods like exact differential or integrating factor approaches.


Would you like a more detailed walkthrough of these steps or different methods for solving it?

Here are 5 related questions to deepen your understanding:

  1. What types of substitutions are common for solving differential equations?
  2. How do you identify an exact differential equation?
  3. What is an integrating factor, and when is it useful?
  4. Can this equation be solved by converting to a linear form?
  5. How do you verify the solution of a differential equation?

Tip: In many non-linear equations, substitution techniques such as v=xyv = \frac{x}{y} are extremely useful for simplifying expressions and reducing them to solvable forms.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Substitution Method
First-order Non-linear Equations

Formulas

(x + 2y) dx + (2y - 5y) dy = 0
v = x / y

Theorems

Substitution Techniques for Solving Non-linear Differential Equations

Suitable Grade Level

Undergraduate level (Calculus II or III)