Let x = x(y) be the solution of the differential equation . If x(1) = 1, then is : [2025]

Question

Let x = x(y) be the solution of the differential equation $y^{2} d x + (x - \frac{1}{y}) d y = 0$ . If x(1) = 1, then $x (\frac{1}{2})$ is : [2025]

Smart Achievers · Accepted Answer

(4)

We have, $y^{2} d x + (x - \frac{1}{y}) d y = 0$

$\Rightarrow y^{2} d x = (\frac{1}{y} - x) d y \Rightarrow \frac{d x}{d y} + \frac{x}{y^{2}} = \frac{1}{y^{3}}$ ,

which is a linear D.E. with $P = \frac{1}{y^{2}}$ and $Q = \frac{1}{y^{3}}$

$I . F . = e^{\int P d y} = e^{\int \frac{1}{y^{2}} d y} = e^{\frac{- 1}{y}}$

$∴$ Solution is given by $x \cdot e^{- 1 / y} = \int \frac{1}{y^{3}} \cdot e^{- 1 / y} d y + c$

Putting $\frac{1}{y} = t \Rightarrow \frac{- 1}{y^{2}} d y = d t$

$\Rightarrow x \cdot e^{\frac{- 1}{y}} = - \int t \cdot e^{- t} d y + c \Rightarrow x \cdot e^{\frac{- 1}{y}} = - e^{- t} (- t - 1) + c$

$\Rightarrow x \cdot e^{\frac{- 1}{y}} = e^{\frac{- 1}{y}} (1 + \frac{1}{y}) + c \Rightarrow x = 1 + \frac{1}{y} + c e^{1 / y}$

Now, at x(1) = 1, i.e., when x = 1, y = 1

$\Rightarrow 1 = 1 + 1 + c e \Rightarrow c = \frac{- 1}{e}$

$∴$ Solution is, $x = 1 + \frac{1}{y} - \frac{1}{e} (e^{1 / y})$

$∴ x (\frac{1}{2}) = 1 + 2 - \frac{1}{e} \times e^{2} = 3 - e$

Solution

Q.
Let x = x(y) be the solution of the differential equation $y^{2} d x + (x - \frac{1}{y}) d y = 0$ . If x(1) = 1, then $x (\frac{1}{2})$ is : [2025]

1 3 + e

2 $\frac{3}{2} + e$

3 $\frac{1}{2} + e$

4 3 – e

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