For the differential equation :
`xy (dy)/(dx)=(x+2)(y+2)`,
find the solution of curve passing through the point (1, -1).

Answer

To solve the differential equation \( xy \frac{dy}{dx} = (x+2)(y+2) \) and find the solution of the curve passing through the point \( (1, -1) \), we will follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the given equation:
\[ xy \frac{dy}{dx} = (x+2)(y+2) \]
We can rearrange this equation to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{(x+2)(y+2)}{xy} \]

### Step 2: Separate Variables
Next, we separate the variables \( y \) and \( x \):
\[ \frac{dy}{(y+2)} = \frac{(x+2)}{x} dx \]

### Step 3: Integrate Both Sides
Now, we integrate both sides:
\[ \int \frac{dy}{(y+2)} = \int \left(1 + \frac{2}{x}\right) dx \]
The left side integrates to:
\[ \ln|y+2| + C_1 \]
The right side integrates to:
\[ x + 2 \ln|x| + C_2 \]
Thus, we have:
\[ \ln|y+2| = x + 2 \ln|x| + C \]
where \( C = C_2 - C_1 \).

### Step 4: Exponentiate to Solve for \( y \)
Exponentiating both sides gives:
\[ |y+2| = e^{x + 2 \ln|x| + C} = e^C \cdot x^2 \cdot e^x \]
Let \( K = e^C \), then:
\[ y + 2 = Kx^2 e^x \]
So,
\[ y = Kx^2 e^x - 2 \]

### Step 5: Use Initial Condition to Find \( K \)
Now we apply the initial condition \( (1, -1) \):
\[ -1 = K(1^2)e^1 - 2 \]
This simplifies to:
\[ -1 = Ke - 2 \]
Adding 2 to both sides gives:
\[ 1 = Ke \]
Thus,
\[ K = \frac{1}{e} \]

### Step 6: Write the Final Solution
Substituting \( K \) back into the equation for \( y \):
\[ y = \frac{1}{e} x^2 e^x - 2 \]
This simplifies to:
\[ y = x^2 - 2 \]

### Final Answer
The solution of the curve passing through the point \( (1, -1) \) is:
\[ y = x^2 - 2 \]

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