(x^2-y^2)y'=2xy

4 min read Jun 17, 2024

Solving the Differential Equation (x^2 - y^2)y' = 2xy

This article will explore the solution to the differential equation (x^2 - y^2)y' = 2xy. This equation is a first-order, nonlinear differential equation. We'll use a combination of techniques to arrive at the solution.

1. Recognizing the Type of Equation

The given equation is a homogeneous differential equation. This is because we can rewrite it in the form:

y' = f(x,y) = (2xy) / (x^2 - y^2)

where f(tx, ty) = t * f(x,y) for any scalar t. This property is characteristic of homogeneous differential equations.

2. Substitution and Simplification

To solve homogeneous equations, we introduce a substitution:

v = y/x

This leads to:

y = vx

Differentiating both sides with respect to x, we get:

y' = v + x * dv/dx

Now, substitute these expressions for y and y' in the original differential equation:

(x^2 - (vx)^2) * (v + x * dv/dx) = 2x * (vx)

Simplifying the equation, we get:

x^2 * (1 - v^2) * (v + x * dv/dx) = 2x^2 * v

Dividing both sides by x^2 (assuming x ≠ 0) and simplifying further:

(1 - v^2) * (v + x * dv/dx) = 2v

3. Separating Variables

Now, we can separate the variables and integrate. Rearranging the equation:

x * dv/dx = (2v - v + v^3) / (1 - v^2)

x * dv/dx = (v + v^3) / (1 - v^2)

Separating variables:

(1 - v^2) / (v + v^3) dv = dx / x

4. Integration

Integrating both sides, we get:

∫ (1 - v^2) / (v + v^3) dv = ∫ dx / x

The left side can be integrated using partial fractions:

∫ (1/v - 2v / (1 + v^2)) dv = ln|x| + C

Integrating, we obtain:

ln|v| - ln|1 + v^2| = ln|x| + C

5. Solving for v and Substituting Back

Combining the logarithms and simplifying:

ln|v / (1 + v^2)| = ln|x| + C

Taking the exponential of both sides:

|v / (1 + v^2)| = e^C * |x|

We can denote e^C as a new constant, K:

|v / (1 + v^2)| = K|x|

Now, substitute v = y/x back into the equation:

|y / (x^2 + y^2)| = K|x|

This is the general solution to the differential equation (x^2 - y^2)y' = 2xy.

6. General Solution

The general solution in implicit form is given by:

y / (x^2 + y^2) = ±Kx

where K is an arbitrary constant.

7. Conclusion

The solution to the differential equation (x^2 - y^2)y' = 2xy involves recognizing the homogeneous nature of the equation, performing a substitution, separating variables, integrating, and finally substituting back to obtain the general solution in implicit form. This solution describes the family of curves that satisfy the original differential equation.