The equation of the curve through the point (1,1) and whose slope is `(2ay)/(x(y-a))` is

Options

A. `y^a . x^(2a) = e ^(y-1)`
B. ` y^2. x^(2x)= e^(y)`
C. `y^(2a) x^a = e^(y-1)`
D. none of these

Answer

To solve the given differential equation, we will follow these steps:

### Step 1: Set up the differential equation
The slope of the curve is given by:
\[ \frac{dy}{dx} = \frac{2ay}{x(y-a)} \]

### Step 2: Separate the variables
We can separate the variables \(y\) and \(x\):
\[ \frac{y - a}{y} dy = \frac{2a}{x} dx \]

### Step 3: Integrate both sides
Now we will integrate both sides:
\[ \int \left( \frac{y - a}{y} \right) dy = \int \frac{2a}{x} dx \]
The left side can be simplified as:
\[ \int \left( 1 - \frac{a}{y} \right) dy = \int 1 dy - a \int \frac{1}{y} dy \]
This gives:
\[ y - a \ln |y| = 2a \ln |x| + C \]

### Step 4: Rearranging the equation
Rearranging the equation gives us:
\[ y - a \ln |y| - 2a \ln |x| = C \]

### Step 5: Solve for \(C\) using the initial condition
We know the curve passes through the point (1, 1). Substituting \(x = 1\) and \(y = 1\):
\[ 1 - a \ln |1| - 2a \ln |1| = C \]
Since \(\ln |1| = 0\), we have:
\[ C = 1 \]

### Step 6: Substitute \(C\) back into the equation
Now substituting \(C = 1\) back into our equation:
\[ y - a \ln |y| - 2a \ln |x| = 1 \]

### Step 7: Rearranging the final equation
This can be rearranged to express \(y\) in terms of \(x\):
\[ y - a \ln |y| = 1 + 2a \ln |x| \]

### Final Equation
Thus, the equation of the curve is:
\[ y - a \ln |y| = 1 + 2a \ln |x| \]