Let $x = x(y)$ be the solution of the differential equation $y^2 dx + (x – \frac{1}{y}) dy = 0$. If $x(1) = 1$, then $x(\frac{1}{2})$ is:
- (1) $\frac{1}{2} + e$
- (2) $\frac{3}{2} + e$
- (3) $3 – e$
- (4) $3 + e$
Solution:
Rearranging the equation: $y^2 \frac{dx}{dy} + x = \frac{1}{y}$.
$$ \frac{dx}{dy} + \frac{1}{y^2} x = \frac{1}{y^3} $$
This is a Linear Differential Equation. Integrating Factor (I.F.):
$$ I.F. = e^{\int \frac{1}{y^2} dy} = e^{-1/y} $$
Solution is given by:
$$ x \cdot e^{-1/y} = \int \frac{1}{y^3} e^{-1/y} dy $$
Let $-\frac{1}{y} = t \Rightarrow \frac{1}{y^2} dy = dt$. Then $\frac{1}{y^3} dy = -t dt$.
$$ \int -t e^t dt = -(t e^t – e^t) + C = e^t(1-t) + C $$
Substituting back:
$$ x e^{-1/y} = e^{-1/y} \left(1 + \frac{1}{y}\right) + C $$
$$ x = 1 + \frac{1}{y} + C e^{1/y} $$
Given $x(1) = 1 \Rightarrow 1 = 1 + 1 + C e^1 \Rightarrow Ce = -1 \Rightarrow C = -1/e$.
Now find $x(1/2)$:
$$ x = 1 + \frac{1}{1/2} + \left(-\frac{1}{e}\right) e^{1/(1/2)} $$
$$ x = 1 + 2 – \frac{1}{e} \cdot e^2 = 3 – e $$
Ans. (3)