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(x/y)` is equal to (here `y'=(dy)/(dx)`)

Options

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

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 function \( \phi(x/y) \).

### Step 1: Differentiate the given equation

We will differentiate both sides of the equation with respect to \( x \). Using the product rule for differentiation, we have:

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

### Step 2: Apply the product rule

Using the product rule, where \( u = y \) and \( v = \ln |cx| \):

\[ \frac{dy}{dx} \ln |cx| + y \frac{d}{dx}(\ln |cx|) = 1 \]

Now, we need to differentiate \( \ln |cx| \):

\[ \frac{d}{dx}(\ln |cx|) = \frac{1}{cx} \cdot c = \frac{1}{x} \]

### Step 3: Substitute back into the equation

Substituting this back into our differentiated equation gives:

\[ \frac{dy}{dx} \ln |cx| + y \cdot \frac{1}{x} = 1 \]

### Step 4: Replace \( \frac{dy}{dx} \) with \( y' \)

Now, we can replace \( \frac{dy}{dx} \) with \( y' \):

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

### Step 5: Isolate \( y' \)

Rearranging the equation to isolate \( y' \):

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

### Step 6: Solve for \( y' \)

Now, we can express \( y' \):

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

### Step 7: Substitute \( \ln |cx| \)

From the original equation \( y \ln |cx| = x \), we can express \( \ln |cx| \):

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

Substituting this into the expression for \( y' \):

\[ y' = \frac{1 - \frac{y}{x}}{\frac{x}{y}} = \frac{y(1 - \frac{y}{x})}{x} = \frac{y - \frac{y^2}{x}}{x} \]

### Step 8: Rewrite \( y' \)

This simplifies to:

\[ y' = \frac{y}{x} - \frac{y^2}{x^2} \]

### Step 9: Compare with the given form

We know from the problem statement that:

\[ y' = \frac{y}{x} + \phi\left(\frac{x}{y}\right) \]

### Step 10: Identify \( \phi\left(\frac{x}{y}\right) \)

Comparing the two expressions for \( y' \):

\[ \frac{y}{x} - \frac{y^2}{x^2} = \frac{y}{x} + \phi\left(\frac{x}{y}\right) \]

This implies:

\[ \phi\left(\frac{x}{y}\right) = -\frac{y^2}{x^2} \]

Thus, the final answer is:

\[ \phi\left(\frac{x}{y}\right) = -\frac{y^2}{x^2} \]