if `e^y +xy = e ` then the value of `(d^2 y)/( dx^2)` for x=0 is

Options

A. `1//e`
B. `1//e^2`
C. `1//e^3`
D. None of these

Answer

To find the value of \(\frac{d^2y}{dx^2}\) at \(x = 0\) for the equation \(e^y + xy = e\), we will follow these steps:

### Step 1: Find \(y\) when \(x = 0\)
Substituting \(x = 0\) into the equation:
\[ e^y + 0 \cdot y = e \implies e^y = e \]
Taking the natural logarithm of both sides:
\[ y = 1 \]

### Step 2: Differentiate the equation implicitly
Differentiating both sides of the equation \(e^y + xy = e\) with respect to \(x\):
\[ \frac{d}{dx}(e^y) + \frac{d}{dx}(xy) = \frac{d}{dx}(e) \]
Using the chain rule and product rule:
\[ e^y \frac{dy}{dx} + \left(x \frac{dy}{dx} + y\right) = 0 \]
This simplifies to:
\[ e^y \frac{dy}{dx} + x \frac{dy}{dx} + y = 0 \]
Rearranging gives:
\[ (e^y + x) \frac{dy}{dx} + y = 0 \]

### Step 3: Solve for \(\frac{dy}{dx}\) at \(x = 0\)
Substituting \(x = 0\) and \(y = 1\):
\[ (e^1 + 0) \frac{dy}{dx} + 1 = 0 \implies (e) \frac{dy}{dx} + 1 = 0 \]
Thus,
\[ \frac{dy}{dx} = -\frac{1}{e} \]

### Step 4: Differentiate again to find \(\frac{d^2y}{dx^2}\)
Differentiating the equation \(e^y \frac{dy}{dx} + (x \frac{dy}{dx} + y) = 0\) again:
\[ \frac{d}{dx}(e^y \frac{dy}{dx}) + \frac{d}{dx}(x \frac{dy}{dx}) + \frac{dy}{dx} = 0 \]
Using the product rule:
\[ \left(e^y \frac{d^2y}{dx^2} + e^y \left(\frac{dy}{dx}\right)^2\right) + \left(\frac{dy}{dx} + x \frac{d^2y}{dx^2}\right) + \frac{d^2y}{dx^2} = 0 \]
Combining terms gives:
\[ e^y \frac{d^2y}{dx^2} + e^y \left(\frac{dy}{dx}\right)^2 + \frac{dy}{dx} + x \frac{d^2y}{dx^2} + \frac{d^2y}{dx^2} = 0 \]
Factoring out \(\frac{d^2y}{dx^2}\):
\[ \left(e^y + x + 1\right) \frac{d^2y}{dx^2} + e^y \left(\frac{dy}{dx}\right)^2 + \frac{dy}{dx} = 0 \]

### Step 5: Substitute \(x = 0\) and \(y = 1\) to find \(\frac{d^2y}{dx^2}\)
Substituting \(x = 0\), \(y = 1\), and \(\frac{dy}{dx} = -\frac{1}{e}\):
\[ \left(e^1 + 0 + 1\right) \frac{d^2y}{dx^2} + e^1 \left(-\frac{1}{e}\right)^2 - \frac{1}{e} = 0 \]
This simplifies to:
\[ (2e) \frac{d^2y}{dx^2} + \frac{1}{e} - \frac{1}{e} = 0 \]
Thus:
\[ (2e) \frac{d^2y}{dx^2} = 0 \implies \frac{d^2y}{dx^2} = 0 \]

### Conclusion
The value of \(\frac{d^2y}{dx^2}\) at \(x = 0\) is:
\[ \frac{d^2y}{dx^2} = 0 \]