The solution of the differential equation `x(dy)/(dx)+y = y^2` is

Options

A. y=1+cxy
B. y=log (cxy)
C. y+1 = cxy
D. y=c+xy

Answer

To solve the differential equation \( x \frac{dy}{dx} + y = y^2 \), we can follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the given equation in a more standard form:
\[ x \frac{dy}{dx} + y - y^2 = 0 \]
This can be rearranged as:
\[ x \frac{dy}{dx} = y^2 - y \]

### Step 2: Separate variables
Next, we can separate the variables. We can rewrite the equation as:
\[ \frac{dy}{y^2 - y} = \frac{dx}{x} \]

### Step 3: Simplify the left side
We can factor the left-hand side:
\[ y^2 - y = y(y - 1) \]
Thus, we have:
\[ \frac{dy}{y(y - 1)} = \frac{dx}{x} \]

### Step 4: Integrate both sides
Now we integrate both sides. The left side requires partial fraction decomposition:
\[ \frac{1}{y(y - 1)} = \frac{A}{y} + \frac{B}{y - 1} \]
Multiplying through by \( y(y - 1) \) gives:
\[ 1 = A(y - 1) + By \]
Setting \( y = 1 \) gives \( A = 1 \), and setting \( y = 0 \) gives \( B = -1 \). Thus:
\[ \frac{1}{y(y - 1)} = \frac{1}{y} - \frac{1}{y - 1} \]
Now we can integrate:
\[ \int \left( \frac{1}{y} - \frac{1}{y - 1} \right) dy = \int \frac{dx}{x} \]
This results in:
\[ \ln |y| - \ln |y - 1| = \ln |x| + C \]

### Step 5: Simplify the logarithmic equation
Using properties of logarithms, we can combine the left side:
\[ \ln \left| \frac{y}{y - 1} \right| = \ln |x| + C \]
Exponentiating both sides gives:
\[ \left| \frac{y}{y - 1} \right| = e^C |x| \]
Let \( k = e^C \), then:
\[ \frac{y}{y - 1} = kx \]

### Step 6: Solve for \( y \)
Now we can solve for \( y \):
\[ y = kx(y - 1) \]
This simplifies to:
\[ y = kxy - kx \]
Rearranging gives:
\[ y(1 - kx) = -kx \]
Thus:
\[ y = \frac{-kx}{1 - kx} \]

### Final Solution
The general solution of the differential equation is:
\[ y = \frac{kx}{1 + kx} \]
where \( k \) is a constant.

---