Question

Let \( x = x(y) \) be the solution of the differential equation \( y^2 dx + \left( x - \frac{1}{y} \right) dy = 0 \). If \( x(1) = 1 \), then \( x\left( \frac{1}{2} \right) \) is :

• A. \( \frac{1}{2} + e \)
• B. \( \frac{3}{2} + e \)
• C. \( 3 - e \)
• D. \( 3 + e \)

Hint:

For differential equations, use integrating factors to simplify the problem and apply initial conditions to find the solution.

Answer 1

The Correct Option is C

Differential Equation Solution

The provided differential equation is:

\( y^2 dx + \left( x - \frac{1}{y} \right) dy = 0 \)

The solution \( x = x(y) \) satisfies \( x(1) = 1 \). The objective is to determine \( x\left( \frac{1}{2} \right) \).

Step 1: Equation Rearrangement

Divide the equation by \( y^2 \) for simplification:

\[ \frac{y^2 dx}{y^2} + \frac{\left( x - \frac{1}{y} \right) dy}{y^2} = 0 \]

This simplifies to:

\[ dx + \left( \frac{x}{y^2} - \frac{1}{y^3} \right) dy = 0 \]

Step 2: Equation Solving

Separate the variables:

\[ dx = -\left( \frac{x}{y^2} - \frac{1}{y^3} \right) dy \]

Rearrange to isolate \( x \) and \( y \) terms:

\[ \frac{dx}{x} = \left( \frac{1}{y^3} - \frac{1}{y^2} \right) dy \]

Step 3: Integration

Integrate both sides:

\[ \int \frac{1}{x} dx = \int \left( \frac{1}{y^3} - \frac{1}{y^2} \right) dy \]

Left-hand side integral:

\[ \ln |x| = \int \left( \frac{1}{y^3} - \frac{1}{y^2} \right) dy \]

Right-hand side integrals:

\[ \int \frac{1}{y^3} dy = -\frac{1}{2y^2}, \quad \int \frac{1}{y^2} dy = -\frac{1}{y} \]

The integrated equation is:

\[ \ln |x| = -\frac{1}{2y^2} + \frac{1}{y} + C \]

where \( C \) is the constant of integration.

Step 4: Apply Initial Condition

Given \( x(1) = 1 \). Substitute \( x = 1 \) and \( y = 1 \):

\[ \ln |1| = -\frac{1}{2(1)^2} + \frac{1}{1} + C \]

Simplify:

\[ 0 = -\frac{1}{2} + 1 + C \]

\[ C = -\frac{1}{2} \]

Step 5: Final Equation Form

Substitute \( C = -\frac{1}{2} \) back into the equation:

\[ \ln |x| = -\frac{1}{2y^2} + \frac{1}{y} - \frac{1}{2} \]

Step 6: Calculate \( x\left( \frac{1}{2} \right) \)

Substitute \( y = \frac{1}{2} \):

\[ \ln |x\left( \frac{1}{2} \right)| = -\frac{1}{2 \left( \frac{1}{2} \right)^2} + \frac{1}{\frac{1}{2}} - \frac{1}{2} \]

Simplify:

\[ \ln |x\left( \frac{1}{2} \right)| = -\frac{1}{2 \times \frac{1}{4}} + 2 - \frac{1}{2} \]

\[ \ln |x\left( \frac{1}{2} \right)| = -\frac{1}{\frac{1}{2}} + 2 - \frac{1}{2} \]

\[ \ln |x\left( \frac{1}{2} \right)| = -2 + 2 - \frac{1}{2} \]

\[ \ln |x\left( \frac{1}{2} \right)| = -\frac{1}{2} \]

Exponentiate both sides:

\[ x\left( \frac{1}{2} \right) = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}} \]

The value of \( x\left( \frac{1}{2} \right) \) is \( \frac{1}{\sqrt{e}} \).