Equation of curve passing through `(1,1)` & satisfyng the differential equation `(xy^(2)+y^(2)+x^(2)y+2xy)dx+(2xy+x^(2))dy=0` is

Options

A. `xy(x+y)=2e^(1-x)`
B. `xy(x^(2)+y^(2))=2e^(x-1)`
C. `x^(2)y^(2)(x+y)=2`
D. `x^(2)y^(2)(x^(2)+y^(2))=2`

Explanation

To solve the given problem step by step, we need to find the equation of the curve that passes through the point (1,1) and satisfies the differential equation:

\[ (xy^2 + y^2 + x^2y + 2xy)dx + (2xy + x^2)dy = 0 \]

### Step 1: Rewrite the Differential Equation
We start by rewriting the differential equation in a more manageable form:

\[ (xy^2 + y^2 + x^2y + 2xy)dx + (2xy + x^2)dy = 0 \]

### Step 2: Group Terms
We can group the terms in the equation:

\[ (xy^2 + y^2 + x^2y + 2xy)dx + (2xy + x^2)dy = 0 \]

This can be rearranged as:

\[ (xy^2 + y^2 + x^2y + 2xy)dx + (2xy + x^2)dy = 0 \]

### Step 3: Multiply by \( e^x \)
To simplify the equation, we multiply through by \( e^x \):

\[ e^x(xy^2 + y^2 + x^2y + 2xy)dx + e^x(2xy + x^2)dy = 0 \]

### Step 4: Recognize the Total Differential
We can recognize that the left-hand side can be expressed as the total differential of a function \( F(x, y) \):

\[ dF = e^x(xy^2 + x^2y)dx + e^x(2xy + x^2)dy = 0 \]

### Step 5: Integrate Both Sides
Integrating both sides gives us:

\[ F(x, y) = e^x(xy^2 + x^2y) + C \]

### Step 6: Apply Initial Condition
We know the curve passes through the point (1,1). We substitute \( x = 1 \) and \( y = 1 \) into the equation to find \( C \):

\[ F(1, 1) = e^1(1 \cdot 1^2 + 1^2 \cdot 1) + C = 2e + C = 0 \]

Thus, \( C = -2e \).

### Step 7: Final Equation
The final equation of the curve is:

\[ e^x(xy^2 + x^2y) = 2e \]

Dividing both sides by \( e^x \):

\[ xy^2 + x^2y = 2e^{1-x} \]

### Step 8: Rearranging
We can factor out \( xy \):

\[ xy(x + y) = 2e^{1-x} \]

### Final Answer
Thus, the equation of the curve is:

\[ xy(x + y) = 2e^{1-x} \]