If the general solution of the differential equation `y'=y/x+phi(x/y)`, for some function `phi` is given by `y ln|cx|=x`, where c is an arbitray constant, then `phi(2)` is equal to (here `y'=(dy)/(dx)`)

Options

A. -4
B. `-1/4`
C. `1/4`
D. 4

Answer

To solve the problem, we start with the given general solution of the differential equation:

\[ y \ln |cx| = x \]

where \( c \) is an arbitrary constant. We need to find the value of \( \phi(2) \).

### Step 1: Differentiate the general solution
We differentiate both sides of the equation with respect to \( x \):

\[ \frac{d}{dx}(y \ln |cx|) = \frac{d}{dx}(x) \]

Using the product rule on the left side, we have:

\[ y' \ln |cx| + y \cdot \frac{1}{x} \cdot c = 1 \]

This simplifies to:

\[ y' \ln |cx| + \frac{y}{x} = 1 \]

### Step 2: Solve for \( y' \)
Rearranging the equation to isolate \( y' \):

\[ y' \ln |cx| = 1 - \frac{y}{x} \]

Thus, we can express \( y' \) as:

\[ y' = \frac{1 - \frac{y}{x}}{\ln |cx|} \]

### Step 3: Substitute \( \frac{x}{y} = 2 \)
Since we need to find \( \phi(2) \), we set \( \frac{x}{y} = 2 \), which implies \( y = \frac{x}{2} \).

### Step 4: Substitute \( y \) in the expression for \( y' \)
Substituting \( y = \frac{x}{2} \) into the expression for \( y' \):

\[ y' = \frac{1 - \frac{\frac{x}{2}}{x}}{\ln |cx|} = \frac{1 - \frac{1}{2}}{\ln |cx|} = \frac{\frac{1}{2}}{\ln |cx|} = \frac{1}{2 \ln |cx|} \]

### Step 5: Use the original differential equation
From the original differential equation \( y' = \frac{y}{x} + \phi\left(\frac{x}{y}\right) \):

Substituting \( y = \frac{x}{2} \):

\[ y' = \frac{\frac{x}{2}}{x} + \phi(2) = \frac{1}{2} + \phi(2) \]

### Step 6: Set the two expressions for \( y' \) equal
Now we equate the two expressions for \( y' \):

\[ \frac{1}{2 \ln |cx|} = \frac{1}{2} + \phi(2) \]

### Step 7: Solve for \( \phi(2) \)
Rearranging gives:

\[ \phi(2) = \frac{1}{2 \ln |cx|} - \frac{1}{2} \]

### Step 8: Evaluate \( \phi(2) \)
To find \( \phi(2) \), we need to evaluate \( \ln |cx| \) at \( x = 2y \):

Since \( y = \frac{x}{2} \), we can substitute \( x = 2 \):

\[ \phi(2) = \frac{1}{2 \ln |c \cdot 2|} - \frac{1}{2} \]

### Conclusion
After substituting and simplifying, we find that:

\[ \phi(2) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \]

Thus, the final answer is:

\[ \phi(2) = -\frac{1}{4} \]