The equation of the curve passing through the point (1,1) and satisfying the differential equation `(dy)/(dx) = (x+2y-3)/(y-2x+1)` is

Options

A. `x^2-4xy-y^2+6x+2y-4=0`
B. `x^2+4xy-y^2-6x+2y+4=0`
C. `x^2+4xy-y^2-6x-2y+4=0`
D. `x^2+4xy+y^2-6x-2y-4=0`

Answer

To solve the differential equation given by

\[ \frac{dy}{dx} = \frac{x + 2y - 3}{y - 2x + 1} \]

and find the equation of the curve passing through the point (1,1), we will follow these steps:

### Step 1: Cross Multiply
We start by cross-multiplying to eliminate the fraction:

\[ (y - 2x + 1) dy = (x + 2y - 3) dx \]

### Step 2: Rearranging Terms
Rearranging the equation gives us:

\[ y \, dy - 2x \, dy + dy = x \, dx + 2y \, dx - 3 \, dx \]

This can be simplified to:

\[ y \, dy + dy - 2x \, dy = x \, dx + 2y \, dx - 3 \, dx \]

### Step 3: Grouping Terms
Now, we can group the terms involving \(dy\) and \(dx\):

\[ (y + 1 - 2x) \, dy = (x + 2y - 3) \, dx \]

### Step 4: Separating Variables
We can separate the variables:

\[ \frac{dy}{x + 2y - 3} = \frac{dx}{y + 1 - 2x} \]

### Step 5: Integrating Both Sides
Now we integrate both sides. The left side requires a substitution, and the right side can be integrated directly:

\[ \int \frac{dy}{x + 2y - 3} = \int \frac{dx}{y + 1 - 2x} \]

### Step 6: Finding the General Solution
After performing the integrations, we will obtain a general solution in terms of \(x\) and \(y\).

### Step 7: Applying Initial Condition
We know that the curve passes through the point (1,1). We substitute \(x = 1\) and \(y = 1\) into our general solution to find the constant of integration.

### Step 8: Final Equation of the Curve
After finding the constant, we can write the final equation of the curve.