If `x(dy)/(dx) + (y^(2))/(x)=y`, then `(x)/(y)=log x +c`

Answer

To solve the differential equation \( x \frac{dy}{dx} + \frac{y^2}{x} = y \), we will follow these steps:

### Step 1: Rearranging the Equation
Start with the given equation:
\[ x \frac{dy}{dx} + \frac{y^2}{x} = y \]
Divide the entire equation by \( x \):
\[ \frac{dy}{dx} + \frac{y^2}{x^2} = \frac{y}{x} \]

**Hint:** Dividing by \( x \) simplifies the equation and makes it easier to work with.

### Step 2: Substituting Variables
Now, let’s use the substitution \( y = vx \), where \( v \) is a function of \( x \). Then, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]

**Hint:** The substitution \( y = vx \) helps to express \( y \) in terms of \( x \) and another variable \( v \).

### Step 3: Substitute into the Equation
Substituting \( y = vx \) and \( \frac{dy}{dx} = v + x \frac{dv}{dx} \) into the rearranged equation:
\[ v + x \frac{dv}{dx} + \frac{(vx)^2}{x^2} = \frac{vx}{x} \]
This simplifies to:
\[ v + x \frac{dv}{dx} + v^2 = v \]
Cancelling \( v \) from both sides gives:
\[ x \frac{dv}{dx} + v^2 = 0 \]

**Hint:** Simplifying the equation helps in isolating the derivative term.

### Step 4: Separating Variables
Rearranging gives:
\[ x \frac{dv}{dx} = -v^2 \]
Now, separate the variables:
\[ \frac{dv}{v^2} = -\frac{dx}{x} \]

**Hint:** Separating variables allows us to integrate both sides independently.

### Step 5: Integrating Both Sides
Integrate both sides:
\[ \int \frac{dv}{v^2} = \int -\frac{dx}{x} \]
This results in:
\[ -\frac{1}{v} = -\log x + C \]
Rearranging gives:
\[ \frac{1}{v} = \log x - C \]

**Hint:** Remember to include the constant of integration when integrating.

### Step 6: Substitute Back for \( v \)
Recall that \( v = \frac{y}{x} \). Substituting back gives:
\[ \frac{x}{y} = \log x - C \]
Rearranging gives:
\[ \frac{x}{y} = \log x + C \]

**Hint:** Substituting back to the original variables is crucial to express the solution in terms of \( x \) and \( y \).

### Final Result
Thus, we have shown that:
\[ \frac{x}{y} = \log x + C \]

**Conclusion:** The original statement is true.