**12.** 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: - (1) \( \frac{1}{2} + e \) - (2) \( \frac{3}{2} + e \) - (3) \( 3 - e \) - (4) \( 3 + e \)

1. Put in slope form

Divide by $dy$ (assuming $dy \neq 0$): [ y^2,\frac{dx}{dy} + \left( x - \frac{1}{y} \right) = 0 \quad\Rightarrow\quad \frac{dx}{dy} = -\frac{x}{y^2} + \frac{1}{y^3}. ]

2. Recognise a linear ODE

It has the form [ \frac{dx}{dy} + P(y),x = Q(y), \quad\text{where}\quad P(y)=\frac{1}{y^2},; Q(y)=\frac{1}{y^3}. ] That is the standard first-order linear type.

3. Integrating factor (IF)

The recipe says [ \text{IF} = e^{\int P(y),dy} = e^{\int \frac{1}{y^2},dy}= e^{-1/y}. ] Multiply everything by the IF: [ e^{-1/y},\frac{dx}{dy} + \frac{1}{y^2}e^{-1/y},x = \frac{1}{y^3}e^{-1/y}. ] Left side collapses into a neat derivative: [ \frac{d}{dy}\Bigl(x,e^{-1/y}\Bigr) = \frac{1}{y^3}e^{-1/y}. ]

4. Integrate both sides

[ x,e^{-1/y} = \int \frac{1}{y^3}e^{-1/y},dy + C. ] A substitution $t = \dfrac1y$ (so $dt = -\dfrac{1}{y^2}dy$) turns the integral into [ -\int t,e^{-t},dt = (t+1)e^{-t} + C = \Bigl(\tfrac1y+1\Bigr)e^{-1/y} + C. ] Thus [ x,e^{-1/y} = \left(1 + \frac{1}{y}\right)e^{-1/y} + C. ] Multiply by $e^{1/y}$ to isolate $x$: [ x(y) = 1 + \frac{1}{y} + C,e^{1/y}. ]

5. Use the initial point $(1,1)$

When $y = 1$, $x = 1$: [ 1 = 1 + 1 + C,e \quad\Rightarrow\quad C = -\frac{1}{e}. ] So the explicit solution is [ x(y) = 1 + \frac{1}{y} - e^{\frac{1}{y}-1}. ]

6. Evaluate at $y = \dfrac12$

Here $\dfrac{1}{y} = 2$: [ x!\left(\tfrac12\right) = 1 + 2 - e^{2-1} = 3 - e. ]

Hence the correct option is (3) $3 - e$.

What’s happening?
Imagine you have a tiny ant walking on a curve. The rule of the curve is hidden inside the little equation
$y^2\,dx + \left(x - \dfrac{1}{y}\right)dy = 0$.
It tells you how $x$ must change whenever $y$ changes.

Goal
We already know the ant is at the point $(x, y) = (1, 1)$. We want to know where the ant’s $x$-coordinate will be when $y$ shrinks to $\dfrac12$.

Plan

1. Rewrite the rule so we can see $\dfrac{dx}{dy}$ (how $x$ changes with $y$).
2. Notice it is a linear differential equation (very friendly!).
3. Use the “magic multiplier” called the integrating factor to solve it.
4. Plug in the known point to find the missing constant.
5. Finally, substitute $y = \dfrac12$ and read off $x$.

Step-by-step Solution

1. Rewrite in slope form $y^2\,dx + \Bigl(x - \frac{1}{y}\Bigr)dy = 0$ $\Rightarrow \frac{dx}{dy} = -\frac{x}{y^2} + \frac{1}{y^3}$

2. Identify P(y) and Q(y) $P(y)=\frac{1}{y^2}, \qquad Q(y)=\frac{1}{y^3}$

3. Integrating factor $\text{IF} = e^{\int P(y)dy} = e^{\int \frac{1}{y^2}dy} = e^{-1/y}$

4. Multiply and collapse $e^{-1/y}\,\frac{dx}{dy} + \frac{1}{y^2}e^{-1/y}\,x = \frac{1}{y^3}e^{-1/y}$ $\Rightarrow \frac{d}{dy}\Bigl(x e^{-1/y}\Bigr) = \frac{1}{y^3}e^{-1/y}$

5. Integrate $x e^{-1/y} = \int \frac{1}{y^3}e^{-1/y}dy + C$ Substitution $t = 1/y$:
   $x e^{-1/y} = -\int t e^{-t}dt + C$ $x e^{-1/y} = (t+1)e^{-t} + C = \Bigl(\frac{1}{y}+1\Bigr)e^{-1/y} + C$ $\boxed{\;x(y) = 1 + \frac{1}{y} + C e^{1/y}\;}$

6. Apply $x(1) = 1$ $1 = 1 + 1 + C e$ $\Rightarrow C = -\frac{1}{e}$ $x(y) = 1 + \frac{1}{y} - e^{\frac{1}{y}-1}$

7. Find $x\!(\tfrac12)$ $x\!\left(\tfrac12\right) = 1 + 2 - e^{2-1} = 3 - e$

[\boxed{;x!(\tfrac12) = 3 - e;}]

Chosen option: (3) $3 - e$