The solution of `(dy)/(dx)=((x-1)^2+(y-2)^2tan^(- 1)((y-2)/(x-1)))/((x y-2x-y+2)tan^(- 1)((y-2)/(x-1)))` is equal to

Options

A. `{(x-1)^(2)+(y-1)^(2)}tan^(-1)((y-2)/(x-1))-2(x-1)(y-2)=2(x-1)^(2)logc(x-1)`
B. `{(x-1)^(2)+(y-1)^(2)}-2(x-1)(y-2)tan^(-1)((y-1)/(x-1))=2(x-1)^(2)logc`
C. `{(x-1)^(2)+(y-1)^(2)tan^(-1)((y-2)/(x-1))-2(x-1)(y-2)=logC(x-1)`
D. None of these

Answer

To solve the differential equation

\[ \frac{dy}{dx} = \frac{(x-1)^2 + (y-2)^2 \tan^{-1}\left(\frac{y-2}{x-1}\right)}{(xy - 2x - y + 2) \tan^{-1}\left(\frac{y-2}{x-1}\right)}, \]

we will follow these steps:

### Step 1: Simplify the equation
We start by rewriting the equation in a more manageable form. We can express the right-hand side in terms of new variables:

Let \( m = x - 1 \) and \( n = y - 2 \). Then, we have:
\[ \frac{dy}{dx} = \frac{m^2 + n^2 \tan^{-1}\left(\frac{n}{m}\right)}{(m + 1)(n + 2) - 2(m + 1) - (n + 2) + 2) \tan^{-1}\left(\frac{n}{m}\right)}. \]

### Step 2: Change of variables
Now, we can express \( \frac{dy}{dx} \) in terms of \( m \) and \( n \):
\[ \frac{dn}{dm} = \frac{m^2 + n^2 \tan^{-1}\left(\frac{n}{m}\right)}{(m + 1)(n + 2) - 2(m + 1) - (n + 2) + 2) \tan^{-1}\left(\frac{n}{m}\right)}. \]

### Step 3: Substitute and simplify
Next, we can substitute \( n = vm \) (where \( v = \frac{n}{m} \)) into the equation:
\[ \frac{dn}{dm} = v + m \frac{dv}{dm}. \]
This gives us:
\[ v + m \frac{dv}{dm} = \frac{m^2 + v^2 m^2 \tan^{-1}(v)}{m v \tan^{-1}(v)}. \]

### Step 4: Rearranging the equation
Rearranging the equation leads to:
\[ m \frac{dv}{dm} = \frac{1 + v^2 \tan^{-1}(v)}{v \tan^{-1}(v)} - v. \]

### Step 5: Integrate both sides
Now we can integrate both sides. We will use integration by parts for the left side:
\[ \int v \tan^{-1}(v) dv = \frac{1}{2} v^2 \tan^{-1}(v) - \frac{1}{2} \int \frac{v^2}{1 + v^2} dv. \]

### Step 6: Solve for the final expression
After performing the integration and substituting back for \( n \) and \( m \):
\[ \frac{1}{2} (y - 2)^2 \tan^{-1}\left(\frac{y - 2}{x - 1}\right) - \frac{1}{2} \left( \frac{y - 2}{x - 1} \right) = \ln|x - 1| + C. \]

### Final Result
Thus, the solution to the differential equation is:
\[ \frac{1}{2} (y - 2)^2 \tan^{-1}\left(\frac{y - 2}{x - 1}\right) - \frac{1}{2} \left( \frac{y - 2}{x - 1} \right) = \ln|x - 1| + C. \]