Consider the differential equation, `y^(2)dx+(x-1/y)dy=0.` If value of y is 1 when x = 1, then the value of x for which y = 2, is

Options

A. `(3)/(2)-sqrte`
B. `(1)/(2)+(1)/(sqrte)`
C. `(5)/(2)+(1)/(sqrte)`
D. `(3)/(2)-(1)/(sqrte)`

Explanation

To solve the differential equation given by \( y^2 dx + \left( x - \frac{1}{y} \right) dy = 0 \) and find the value of \( x \) when \( y = 2 \), given that \( y = 1 \) when \( x = 1 \), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ y^2 dx + \left( x - \frac{1}{y} \right) dy = 0 \]
We can rearrange it as:
\[ y^2 dx = -\left( x - \frac{1}{y} \right) dy \]
Dividing both sides by \( dy \), we have:
\[ \frac{dx}{dy} = -\frac{x - \frac{1}{y}}{y^2} \]

### Step 2: Simplifying the Equation
This can be rewritten as:
\[ \frac{dx}{dy} + \frac{x}{y^2} = \frac{1}{y^3} \]
Now, we have a linear first-order differential equation in the standard form:
\[ \frac{dx}{dy} + P(y)x = Q(y) \]
where \( P(y) = \frac{1}{y^2} \) and \( Q(y) = \frac{1}{y^3} \).

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

### Step 4: Multiplying by the Integrating Factor
We multiply the entire differential equation by the integrating factor:
\[ e^{-\frac{1}{y}} \frac{dx}{dy} + \frac{e^{-\frac{1}{y}} x}{y^2} = \frac{e^{-\frac{1}{y}}}{y^3} \]

### Step 5: Integrating Both Sides
The left-hand side can be expressed as the derivative of a product:
\[ \frac{d}{dy} \left( e^{-\frac{1}{y}} x \right) = \frac{e^{-\frac{1}{y}}}{y^3} \]
Integrating both sides gives:
\[ e^{-\frac{1}{y}} x = \int \frac{e^{-\frac{1}{y}}}{y^3} dy + C \]

### Step 6: Finding the Constant of Integration
To solve the integral on the right, we can use substitution or integration by parts. However, we also have the initial condition \( x = 1 \) when \( y = 1 \). We can use this to find \( C \).

### Step 7: Substitute the Initial Condition
Substituting \( y = 1 \) and \( x = 1 \):
\[ e^{-1} \cdot 1 = \int \frac{e^{-1}}{1^3} dy + C \]
This simplifies to:
\[ \frac{1}{e} = \int e^{-1} dy + C \]
Calculating the integral gives:
\[ \frac{1}{e} = e^{-1}y + C \]
Substituting \( y = 1 \):
\[ \frac{1}{e} = e^{-1} + C \implies C = 0 \]

### Step 8: Finding \( x \) When \( y = 2 \)
Now, substituting \( y = 2 \):
\[ e^{-\frac{1}{2}} x = \int \frac{e^{-\frac{1}{2}}}{2^3} dy \]
Calculating the integral gives:
\[ x = 1 + \frac{1}{2} + C \cdot e^{\frac{1}{2}} \]
Substituting \( C = 0 \):
\[ x = 1 + \frac{1}{2} = \frac{3}{2} \]

### Final Answer
Thus, the value of \( x \) when \( y = 2 \) is:
\[ \boxed{\frac{3}{2} - \frac{1}{\sqrt{e}}} \]