If `(dx)/(dy)=(e^(y)-x)`, where y(0)=0, then y is expressed explicitly as

Options

A. `(1)/(2)ln(1+x^(2))`
B. `ln(1+x^(2))`
C. `ln(x-sqrt(1+x^(2)))`
D. `ln(x+sqrt(1-x^(2)))`

Explanation

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

### Step 1: Rewrite the Differential Equation
We start with the given equation:
\[ \frac{dx}{dy} = e^y - x \]
Rearranging this gives:
\[ \frac{dx}{dy} + x = e^y \]

### Step 2: Identify the Form of the Equation
This is a linear first-order differential equation of the form:
\[ \frac{dx}{dy} + P(y)x = Q(y) \]
where \(P(y) = 1\) and \(Q(y) = e^y\).

### Step 3: Calculate the Integrating Factor
The integrating factor \(I(y)\) is given by:
\[ I(y) = e^{\int P(y) \, dy} = e^{\int 1 \, dy} = e^y \]

### Step 4: Multiply the Equation by the Integrating Factor
Multiplying the entire differential equation by the integrating factor:
\[ e^y \frac{dx}{dy} + e^y x = e^{2y} \]

### Step 5: Rewrite the Left Side
The left side can be expressed as the derivative of a product:
\[ \frac{d}{dy}(e^y x) = e^{2y} \]

### Step 6: Integrate Both Sides
Integrating both sides with respect to \(y\):
\[ \int \frac{d}{dy}(e^y x) \, dy = \int e^{2y} \, dy \]
This gives:
\[ e^y x = \frac{e^{2y}}{2} + C \]
where \(C\) is the constant of integration.

### Step 7: Solve for \(x\)
Now, we can solve for \(x\):
\[ x = \frac{e^{2y}}{2e^y} + \frac{C}{e^y} = \frac{e^y}{2} + Ce^{-y} \]

### Step 8: Apply the Initial Condition
Using the initial condition \(y(0) = 0\), we substitute \(y = 0\) and \(x = 0\):
\[ 0 = \frac{e^0}{2} + C \cdot e^0 \implies 0 = \frac{1}{2} + C \implies C = -\frac{1}{2} \]

### Step 9: Substitute \(C\) Back into the Equation
Substituting \(C\) back into the equation for \(x\):
\[ x = \frac{e^y}{2} - \frac{1}{2} e^{-y} \]

### Step 10: Rearranging to Express \(y\) Explicitly
To express \(y\) explicitly, we can rearrange the equation:
\[ 2x = e^y - e^{-y} \]
This can be rewritten as:
\[ e^y - 2x - e^{-y} = 0 \]
Recognizing that \(e^y - e^{-y} = 2\sinh(y)\):
\[ 2\sinh(y) = 2x \implies \sinh(y) = x \]
Thus,
\[ y = \sinh^{-1}(x) \]

### Final Answer
The explicit expression for \(y\) in terms of \(x\) is:
\[ y = \sinh^{-1}(x) \]