if `y+x(dy)/(dx)=x(phi(xy))/(phi'(xy))` then `phi(xy)` is equation to

Options

A. `ke^(x^(2)//2)`
B. `ke^(y^(2)//2)`
C. `ke^(xy//2)`
D. `ke^(xy)`

Explanation

To solve the given differential equation \( y + x \frac{dy}{dx} = x \frac{\phi(xy)}{\phi'(xy)} \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Substitution**: Let \( xy = v \). Then, we have:
\[ y = \frac{v}{x} \]
Differentiating both sides with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{x} \frac{dv}{dx} - \frac{v}{x^2} \]

2. **Rewrite the Original Equation**: Substitute \( y \) and \( \frac{dy}{dx} \) into the original equation:
\[ \frac{v}{x} + x \left( \frac{1}{x} \frac{dv}{dx} - \frac{v}{x^2} \right) = x \frac{\phi(v)}{\phi'(v)} \]
Simplifying this gives:
\[ \frac{v}{x} + \frac{dv}{dx} - \frac{v}{x} = x \frac{\phi(v)}{\phi'(v)} \]
Thus, we have:
\[ \frac{dv}{dx} = x \frac{\phi(v)}{\phi'(v)} \]

3. **Rearranging the Equation**: Rearranging gives:
\[ \frac{\phi'(v)}{\phi(v)} dv = x dx \]

4. **Integrating Both Sides**: Now we integrate both sides:
\[ \int \frac{\phi'(v)}{\phi(v)} dv = \int x dx \]
The left side integrates to \( \ln |\phi(v)| \) and the right side integrates to \( \frac{x^2}{2} + C \):
\[ \ln |\phi(v)| = \frac{x^2}{2} + C \]

5. **Exponentiating Both Sides**: To eliminate the logarithm, we exponentiate both sides:
\[ |\phi(v)| = e^{C} e^{\frac{x^2}{2}} \]
Let \( k = e^{C} \), then:
\[ \phi(v) = k e^{\frac{x^2}{2}} \]

6. **Substituting Back for \( v \)**: Recall that \( v = xy \):
\[ \phi(xy) = k e^{\frac{x^2}{2}} \]

### Final Result:
Thus, the function \( \phi(xy) \) is given by:
\[ \phi(xy) = k e^{\frac{x^2}{2}} \]