Solution of differential equation `y-x(dy)/(dx)=y^(2)+(dy)/(dx),` when x=1,y=2, is

Options

A. (1+x)(1-y)+y=0
B. (1+x)(1-y)-y=0
C. (1+x)(1+y)+y=0
D. (1+x)(1+y)-y=0

Answer

To solve the differential equation \( y - x \frac{dy}{dx} = y^2 + \frac{dy}{dx} \) with the initial condition \( x = 1 \) and \( y = 2 \), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the given differential equation:
\[ y - x \frac{dy}{dx} = y^2 + \frac{dy}{dx} \]
Rearranging it gives:
\[ y - y^2 = x \frac{dy}{dx} + \frac{dy}{dx} \]
Factoring out \( \frac{dy}{dx} \) on the right side:
\[ y - y^2 = (x + 1) \frac{dy}{dx} \]

### Step 2: Separating Variables
Now we can separate the variables:
\[ \frac{dy}{y - y^2} = \frac{dx}{x + 1} \]

### Step 3: Simplifying the Left Side
The left side can be simplified:
\[ \frac{dy}{y(1 - y)} = \frac{dx}{x + 1} \]
We can use partial fractions to decompose \( \frac{1}{y(1-y)} \):
\[ \frac{1}{y(1-y)} = \frac{A}{y} + \frac{B}{1-y} \]
Multiplying through by \( y(1-y) \) and solving for \( A \) and \( B \):
\[ 1 = A(1-y) + By \]
Setting \( y = 0 \) gives \( A = 1 \). Setting \( y = 1 \) gives \( B = 1 \). Thus:
\[ \frac{1}{y(1-y)} = \frac{1}{y} + \frac{1}{1-y} \]

### Step 4: Integrating Both Sides
Now we integrate both sides:
\[ \int \left( \frac{1}{y} + \frac{1}{1-y} \right) dy = \int \frac{dx}{x + 1} \]
The left side integrates to:
\[ \ln |y| - \ln |1-y| = \ln \left| \frac{y}{1-y} \right| \]
The right side integrates to:
\[ \ln |x + 1| + C \]
Thus, we have:
\[ \ln \left| \frac{y}{1-y} \right| = \ln |x + 1| + C \]

### Step 5: Exponentiating Both Sides
Exponentiating both sides gives:
\[ \frac{y}{1-y} = K(x + 1) \]
where \( K = e^C \).

### Step 6: Applying Initial Conditions
Now we apply the initial conditions \( x = 1 \) and \( y = 2 \):
\[ \frac{2}{1-2} = K(1 + 1) \]
This simplifies to:
\[ \frac{2}{-1} = 2K \implies -2 = 2K \implies K = -1 \]

### Step 7: Final Equation
Substituting \( K \) back into the equation gives:
\[ \frac{y}{1-y} = -1(x + 1) \]
Multiplying both sides by \( 1-y \):
\[ y = -1(x + 1)(1 - y) \]
Expanding and rearranging yields:
\[ y + xy + y = -x - 1 \implies y(1 + x) = -x - 1 \implies y = \frac{-x - 1}{1 + x} \]

### Final Answer
Thus, the solution of the differential equation is:
\[ y = \frac{-x - 1}{1 + x} \]