Solving Differential Equations: The Approach to 2xydx (x^2 - 1)dy 0
Differential equations are a fundamental topic in mathematics, with applications spanning various fields such as physics, engineering, and economics. This article focuses on a specific type of differential equation, 2xydx (x^2 - 1)dy 0. We will explore various methods to solve this equation, starting with variable separation and moving through partial fractions, integration, and logarithmic forms.
Approach 1: Variable Separation
The given equation is:
2xydx (x^2 - 1)dy 0
To use the variable separation method, we can rearrange the equation:
2xydx (1 - x^2)dy
(2x/(1 - x^2))dx dy/y
Let's decompose the left-hand side using a substitution:
u x^2
du 2xdx
(du/(1 - u)) dy/y
Integrate both sides:
-ln|1 - u| ln|y| C
ln|1 - u| -ln|y| - C
ln|1 - u| ln|1/y| - C
1 - u Ce^(-ln|y|)
1 - u C/y
Substitute back u x^2:
1 - x^2 C/y
y C/(1 - x^2)
Approach 2: Integration
Another method is to directly integrate the given equation:
2xydx (x^2 - 1)dy 0
2xydx - (x^2 - 1)dy
(2x/(x^2 - 1))dx -dy/y
Integrate both sides:
∫(2x/(x^2 - 1))dx -∫(1/y) dy
ln|x^2 - 1| -ln|y| C
ln|x^2 - 1| ln|y| C
ln|y(x^2 - 1)| C
y(x^2 - 1) e^C
y Ce^{1/(x^2 - 1)}
Recall that Ce^{1/(x^2 - 1)} K (where C and K are arbitrary constants), so:
y K/(x^2 - 1)
Additional Solution Methods
Another way to solve the equation is using partial fractions and logarithmic forms:
2xydx (1 - x^2)dy
(2x/(1 - x^2))dx dy/y
Integrate both sides:
∫(2x/(1 - x^2))dx ln|y| - ln|c|
ln|1 - x^2| ln|y/c|
1 - x^2 cy
y (1 - x^2)/c
Let's consider another form:
y (1 - x^2)c
Conclusion
By using variable separation, integration, and partial fractions, we have derived the general solution to the differential equation 2xydx (x^2 - 1)dy 0. The solution is:
y K/(1 - x^2)
Where K is an arbitrary constant. This article provides a comprehensive overview of the methods used to solve such differential equations, which can be adapted to similar problems by following a similar process of variable separation, integration, and logarithmic forms.
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