` dy/dx + xy = xy^(2) " when "x=1 , y =4`

Answer

To solve the differential equation \( \frac{dy}{dx} + xy = xy^2 \) with the initial condition \( x = 1 \) and \( y = 4 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \frac{dy}{dx} + xy = xy^2 \]
We can rearrange it as:
\[ \frac{dy}{dx} = xy^2 - xy \]

### Step 2: Factor out \( xy \)
Factoring out \( xy \) from the right-hand side gives us:
\[ \frac{dy}{dx} = xy(y - 1) \]

### Step 3: Separate variables
We can separate the variables \( y \) and \( x \):
\[ \frac{dy}{y(y - 1)} = x \, dx \]

### 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:
\[ \int \left( \frac{1}{y} - \frac{1}{y - 1} \right) dy = \int x \, dx \]
This results in:
\[ \ln |y| - \ln |y - 1| = \frac{x^2}{2} + C \]

### Step 5: Simplify the left side
Using properties of logarithms, we can combine the left side:
\[ \ln \left| \frac{y}{y - 1} \right| = \frac{x^2}{2} + C \]

### Step 6: Exponentiate both sides
Exponentiating both sides gives:
\[ \frac{y}{y - 1} = e^{\frac{x^2}{2} + C} = Ke^{\frac{x^2}{2}} \quad \text{(where \( K = e^C \))} \]

### Step 7: Solve for \( y \)
Rearranging gives:
\[ y = K e^{\frac{x^2}{2}} (y - 1) \]
This simplifies to:
\[ y(1 + K e^{\frac{x^2}{2}}) = K e^{\frac{x^2}{2}} \]
Thus:
\[ y = \frac{K e^{\frac{x^2}{2}}}{1 + K e^{\frac{x^2}{2}}} \]

### Step 8: Use initial conditions to find \( K \)
We know \( y(1) = 4 \):
\[ 4 = \frac{K e^{\frac{1^2}{2}}}{1 + K e^{\frac{1^2}{2}}} \]
This simplifies to:
\[ 4(1 + K e^{\frac{1}{2}}) = K e^{\frac{1}{2}} \]
Rearranging gives:
\[ 4 = K e^{\frac{1}{2}} (1 - 4) \Rightarrow K e^{\frac{1}{2}} = -\frac{4}{3} \]
Thus:
\[ K = -\frac{4}{3} e^{-\frac{1}{2}} \]

### Step 9: Substitute \( K \) back into the equation
Substituting \( K \) back into our equation for \( y \):
\[ y = \frac{-\frac{4}{3} e^{-\frac{1}{2}} e^{\frac{x^2}{2}}}{1 - \frac{4}{3} e^{-\frac{1}{2}} e^{\frac{x^2}{2}}} \]

### Final Result
This gives us the solution for the differential equation with the given initial condition.