Differential equation `(dy)/(dx)=f(x)g(x)` can be solved by separating variable `(dy)/g(y)=f(x)dx.`
If `(dy)/(dx)=1+x+y+xy and y(-1)=0,` then y is equal to

Options

A. `e^((1-x^(2))/(2))`
B. `e^((1+x^(2))/(2)-1)`
C. In (1+x)-1
D. None of these

Answer

To solve the differential equation \(\frac{dy}{dx} = 1 + x + y + xy\) with the initial condition \(y(-1) = 0\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \frac{dy}{dx} = 1 + x + y + xy \]
We can rearrange it as:
\[ \frac{dy}{dx} = (1 + x)(1 + y) \]

### Step 2: Separate variables
Next, we separate the variables \(y\) and \(x\):
\[ \frac{dy}{1 + y} = (1 + x)dx \]

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

### Step 4: Solve for \(y\)
To solve for \(y\), we exponentiate both sides:
\[ |1 + y| = e^{x + \frac{x^2}{2} + C} = e^C e^{x + \frac{x^2}{2}} \]
Let \(k = e^C\), then:
\[ 1 + y = k e^{x + \frac{x^2}{2}} \]
Thus:
\[ y = k e^{x + \frac{x^2}{2}} - 1 \]

### Step 5: Use the initial condition
We use the initial condition \(y(-1) = 0\) to find \(k\):
\[ 0 = k e^{-1 + \frac{1}{2}} - 1 \]
This simplifies to:
\[ k e^{-\frac{1}{2}} = 1 \implies k = e^{\frac{1}{2}} \]

### Step 6: Substitute \(k\) back into the equation
Now substituting \(k\) back into the equation for \(y\):
\[ y = e^{\frac{1}{2}} e^{x + \frac{x^2}{2}} - 1 \]
This simplifies to:
\[ y = e^{\frac{1}{2}(x^2 + 2x + 1)} - 1 = e^{\frac{1}{2}(x + 1)^2} - 1 \]

### Final Answer
Thus, the solution to the differential equation is:
\[ y = e^{\frac{1}{2}(x + 1)^2} - 1 \]

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