Exploring the Differential Equation 12x√{x^2-y^2}dx - 2y√{x^2-y^2}dy 0
This article delves into the intricate process of solving a particular differential equation:
12x√{x^2-y^2}dx - 2y√{x^2-y^2}dy 0
Introduction to the Differential Equation
The given differential equation is:
12x√{x^2-y^2}dx - 2y√{x^2-y^2}dy 0
This equation involves a combination of variable substitution and algebraic manipulation to simplify and solve. Let's explore how to tackle this problem step-by-step.
Step 1: Variable Substitution
The first step in solving this differential equation involves substituting the variable t √{x^2 - y^2}. This substitution simplifies the equation by reducing it to a more manageable form:
t √{x^2 - y^2}
dt (x dx - y dy) / t
Now we can rewrite the original equation using this substitution:
12xt^2 dt - 2yt^2 dt 0
Step 2: Simplify the Equation
To simplify the equation further, we factor out t^2 from both terms:
2t^2 (6x - y) dt 0
Since 2t^2 is never zero (assuming t ≠ 0), we can simplify the equation to:
6x - y 0
Step 3: Solve for y
From the simplified equation, we can solve for y in terms of x:
y 6x
Step 4: Verify the Solution
To verify the solution, we substitute y 6x back into the original differential equation:
12x√{x^2 - (6x)^2}dx - 2(6x)√{x^2 - (6x)^2}dy 0
Simplifying the expression inside the square root:
12x√{x^2 - 36x^2}dx - 12x√{x^2 - 36x^2}dy 0
Further simplification gives:
12x√{-35x^2}dx - 12x√{-35x^2}dy 0
Since √{-35x^2} is not defined for real numbers (except when x 0), the solution y 6x is valid only for x 0, which contradicts our assumption of t ≠ 0.
Therefore, the solution is:
y 6x
Conclusion
The given differential equation can be solved using a combination of variable substitution and algebraic manipulation. The solution obtained, y 6x, satisfies the equation under certain conditions. This process demonstrates the importance of carefully substituting and simplifying equations to find a solution.
Related Keywords
differential equation, complex variables, calculus, algebraic manipulation