Consider the differential equation `y^2dx+(x-1/y)dy=0` if `y(1)=1` then `x` is

Options

A. `3-(1)/(y)+(e^(1//y))/(e)`
B. `1+(1)/(y)-(e^(1/y))/(e)`
C. `1-(1)/(y)+(e^(1//y))/(e)`
D. `4-(2)/(y)-(e^(1//y))/(e)`

Explanation

To solve the differential equation \( y^2 dx + \left( x - \frac{1}{y} \right) dy = 0 \) with the initial condition \( y(1) = 1 \), we will follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the given differential equation in a more manageable form. We can express it as:
\[ y^2 dx = -\left( x - \frac{1}{y} \right) dy \]
Dividing both sides by \( y^2 \) gives:
\[ dx = -\left( \frac{x}{y^2} - \frac{1}{y^3} \right) dy \]

### Step 2: Rearranging into standard form
Next, we rearrange the equation into the standard form of a first-order linear differential equation:
\[ \frac{dx}{dy} + \frac{x}{y^2} = \frac{1}{y^3} \]

### Step 3: Identify the integrating factor
The integrating factor \( \mu(y) \) is given by:
\[ \mu(y) = e^{\int \frac{1}{y^2} dy} = e^{-\frac{1}{y}} \]

### Step 4: Multiply through by the integrating factor
Now we multiply the entire equation by the integrating factor:
\[ e^{-\frac{1}{y}} \frac{dx}{dy} + e^{-\frac{1}{y}} \frac{x}{y^2} = e^{-\frac{1}{y}} \frac{1}{y^3} \]

### Step 5: Left-hand side as a derivative
The left-hand side can be expressed as the derivative of a product:
\[ \frac{d}{dy} \left( x e^{-\frac{1}{y}} \right) = e^{-\frac{1}{y}} \frac{1}{y^3} \]

### Step 6: Integrate both sides
Integrating both sides with respect to \( y \):
\[ \int \frac{d}{dy} \left( x e^{-\frac{1}{y}} \right) dy = \int e^{-\frac{1}{y}} \frac{1}{y^3} dy \]
This gives:
\[ x e^{-\frac{1}{y}} = \int e^{-\frac{1}{y}} \frac{1}{y^3} dy + C \]

### Step 7: Solve for \( x \)
To isolate \( x \), we multiply both sides by \( e^{\frac{1}{y}} \):
\[ x = e^{\frac{1}{y}} \left( \int e^{-\frac{1}{y}} \frac{1}{y^3} dy + C \right) \]

### Step 8: Use the initial condition
Given \( y(1) = 1 \), we substitute \( y = 1 \) into our equation to find \( C \):
\[ 1 = e^{1} \left( \int e^{-1} \frac{1}{1^3} dy + C \right) \]
Solving this will yield the constant \( C \).

### Step 9: Final expression for \( x \)
Substituting \( C \) back into the expression for \( x \) gives us the final solution.

### Step 10: Evaluate \( x \) at \( y = 1 \)
Finally, we evaluate \( x \) when \( y = 1 \) to find the value of \( x \).