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

The given equation is: \[ y^2 \frac{dx}{dy} + \left( x - \frac{1}{y} \right) = 0 \] This is a first-order linear differential equation. We can use the integrating factor \( I.F. = e^{\int \frac{1}{y^2} dy} = e^{\frac{1}{y}} \). Thus, we have: \[ x \cdot e^{\frac{1}{y}} = \int e^{\frac{1}{t}} t^{-3} dt + C \] Put $-\frac{1}{y} = t$ $+\frac{1}{y^2} dy = dt$ $x \cdot e^{\frac{1}{y}} = -\int t \cdot e^t dt$ $x \cdot e^{\frac{1}{y}} = -te^t + e^t + C$ $x \cdot e^{\frac{1}{y}} = \frac{+1}{y} e^{\frac{1}{y}} + e^{\frac{1}{y}} + C$ $x = 1, y = 1$ $\frac{1}{e} = \frac{1}{e} + \frac{1}{e} + C$

 
$\Rightarrow C = -\frac{1}{e}$ Put $y = \frac{1}{2}$ $\frac{x}{e^2} = \frac{2}{e^2} + \frac{1}{e^2} - \frac{1}{e}$ $x = 3 - e$

after Solving for \( C \) and applying the initial condition, we find: \[ x = 3 - e \]