To solve the differential equation
\[ y' = \frac{xy}{x^2 - y^2}, \]
we can start by rewriting it in a more manageable form. Substituting \( y' = \frac{dy}{dx} \), the equation becomes:
\[ \frac{dy}{dx} = \frac{xy}{x^2 - y^2}. \]
Next, we can separate the variables by rearranging the terms. This leads us to:
\[ (x^2 - y^2) , dy = xy , dx. \]
Now, let's isolate the terms involving \( y \) on one side and \( x \) on the other side:
\[ (x^2 - y^2) , dy - xy , dx = 0. \]
However, separating variables directly is challenging here. Instead, let's switch our approach and look for an implicit solution. We can attempt a substitution or implicit function transformation.
We can instead use homogeneity of the equation to make it easier to process. The given equation can be analyzed by noticing that both the numerator and denominator can be related by substitution.
Instead of separating, we can rewrite the equation as follows to identify suitable substitutions:
\[ \frac{dy}{dx} = \frac{xy}{x^2 - y^2} \]
Letting \( v = \frac{y}{x} \) gives \( y = vx \), leading to \( \frac{dy}{dx} = v + x\frac{dv}{dx} \).
Substituting \( y \) in the original equation yields:
\[ v + x\frac{dv}{dx} = \frac{x(vx)}{x^2 - (vx)^2} = \frac{vx^2}{x^2(1-v^2)} = \frac{v}{1-v^2}. \]
Rearranging gives:
\[ x\frac{dv}{dx} = \frac{v}{1-v^2} - v = v\left(\frac{1}{1-v^2} - 1\right) = v\frac{1-1+v^2}{1-v^2} = \frac{v^3}{1-v^2}. \]
This results in the separable equation:
\[ \frac{1-v^2}{v^3} dv = \frac{1}{x} dx. \]
Integrating both sides gives:
\[ \int \frac{1-v^2}{v^3} , dv = \int \frac{1}{x} , dx. \]
The left side can be simplified and integrated:
\[ \int \left(\frac{1}{v^3} - \frac{1}{v}\right) , dv = \int v^{-3} , dv - \int v^{-1} , dv = -\frac{1}{2v^2} - \ln |v| + C. \]
The right side gives:
\[ \ln |x| + C_1. \]
Rearranging the final integrated result yields something like:
\[ -\frac{1}{2v^2} - \ln |v| = \ln |x| + C. \]
Substituting back \( v = \frac{y}{x} \) gives us an implicit solution involving the relationship between \( y \) and \( x \) based on our integration. You can simplify and rearrange that result to isolate \( y \) in terms of \( x \) or leave it implicit.
This provides the general approach required for the solution of the differential equation:
- Separation and substitution techniques
- Integration to obtain implicit relationships
Thus the final answer is implicit in terms of \( x \) and \( y \).
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