The general solution of the equation `(dy)/(dx)=1+x y` is

Options

A. `y=ce^(-x^(2)//2)`
B. `y=ce^(x^(2)//2)`
C. `y=(x+c),e^(-x^(2)//2)`
D. None of these

Explanation

To solve the differential equation \(\frac{dy}{dx} = 1 + xy\), we will follow the steps to find the general solution.

### Step 1: Rewrite the equation
We start with the given equation:
\[ \frac{dy}{dx} = 1 + xy \]
This can be rearranged to fit the standard form of a linear differential equation:
\[ \frac{dy}{dx} - xy = 1 \]

### Step 2: Identify \(p(x)\) and \(q(x)\)
From the standard form \(\frac{dy}{dx} + p(x)y = q(x)\), we identify:
\[ p(x) = -x \quad \text{and} \quad q(x) = 1 \]

### Step 3: Find the integrating factor
The integrating factor \(\mu(x)\) is given by:
\[ \mu(x) = e^{\int p(x) \, dx} = e^{\int -x \, dx} \]
Calculating the integral:
\[ \int -x \, dx = -\frac{x^2}{2} \]
Thus, the integrating factor becomes:
\[ \mu(x) = e^{-\frac{x^2}{2}} \]

### Step 4: Multiply the differential equation by the integrating factor
We multiply the entire differential equation by the integrating factor:
\[ e^{-\frac{x^2}{2}} \frac{dy}{dx} - e^{-\frac{x^2}{2}} xy = e^{-\frac{x^2}{2}} \]

### Step 5: Recognize the left-hand side as a derivative
The left side can be rewritten as:
\[ \frac{d}{dx}(y \cdot e^{-\frac{x^2}{2}}) = e^{-\frac{x^2}{2}} \]

### Step 6: Integrate both sides
Now, we integrate both sides with respect to \(x\):
\[ \int \frac{d}{dx}(y \cdot e^{-\frac{x^2}{2}}) \, dx = \int e^{-\frac{x^2}{2}} \, dx \]
This gives us:
\[ y \cdot e^{-\frac{x^2}{2}} = \int e^{-\frac{x^2}{2}} \, dx + C \]
where \(C\) is the constant of integration.

### Step 7: Solve for \(y\)
To isolate \(y\), we multiply both sides by \(e^{\frac{x^2}{2}}\):
\[ y = e^{\frac{x^2}{2}} \left( \int e^{-\frac{x^2}{2}} \, dx + C \right) \]

### Final Result
The general solution of the differential equation \(\frac{dy}{dx} = 1 + xy\) is:
\[ y = e^{\frac{x^2}{2}} \left( \int e^{-\frac{x^2}{2}} \, dx + C \right) \]