If `(dy)/(dx)=(y(x-y))/(x(x+y))`, then solution of the diff. equation is ……….

Answer

To solve the differential equation given by

\[ \frac{dy}{dx} = \frac{y(x-y)}{x(x+y)}, \]

we will use the substitution \( y = vx \), where \( v \) is a function of \( x \). This substitution helps us express \( y \) in terms of \( x \) and \( v \).

### Step-by-Step Solution:

1. **Substitute \( y = vx \)**:
\[ \frac{dy}{dx} = v + x\frac{dv}{dx}. \]
Here, \( \frac{dy}{dx} \) is expressed using the product rule.

2. **Substitute into the original equation**:
Replace \( y \) in the original equation:
\[ v + x\frac{dv}{dx} = \frac{vx(x - vx)}{x(x + vx)}. \]

3. **Simplify the right-hand side**:
The right-hand side simplifies as follows:
\[ \frac{vx(x - vx)}{x(x + vx)} = \frac{v(x - vx)}{x + vx} = \frac{v(x(1 - v))}{x(1 + v)} = \frac{v(1 - v)}{1 + v}. \]

4. **Set the equation**:
Now we have:
\[ v + x\frac{dv}{dx} = \frac{v(1 - v)}{1 + v}. \]

5. **Rearranging the equation**:
Rearranging gives:
\[ x\frac{dv}{dx} = \frac{v(1 - v)}{1 + v} - v. \]
Simplifying the right-hand side:
\[ x\frac{dv}{dx} = \frac{v(1 - v) - v(1 + v)}{1 + v} = \frac{v(1 - v - 1 - v)}{1 + v} = \frac{-v^2}{1 + v}. \]

6. **Separate variables**:
This can be rewritten as:
\[ \frac{1 + v}{-v^2} dv = \frac{1}{x} dx. \]

7. **Integrate both sides**:
Integrating both sides:
\[ \int \frac{1 + v}{-v^2} dv = \int \frac{1}{x} dx. \]
The left-hand side can be split:
\[ \int \left(-\frac{1}{v} - \frac{1}{v^2}\right) dv = \ln |x| + C. \]
This gives:
\[ -\ln |v| + \frac{1}{v} = \ln |x| + C. \]

8. **Back substitute \( v = \frac{y}{x} \)**:
Replace \( v \) back:
\[ -\ln \left|\frac{y}{x}\right| + \frac{x}{y} = \ln |x| + C. \]

9. **Final form**:
Rearranging gives the implicit solution of the differential equation.