Let y = y ( x ) be the solution of the differential equation x 3 ? d y + ( x y - 1 ) d x = 0 , x 0 , y ( 1 2 ) = 3 - e . 1 Then y ( 1 ) is equal to [2023]

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Solution

Q.
$Let y = y (x) be the solution of the differential equation x^{3} d y + (x y - 1) d x = 0, x > 0, y (\frac{1}{2}) = 3 - e . 1$ $Then y (1) is equal to$ [2023]

1 $2 - e$

2 $e$

3 $1$

4 $3$

Ans.
(3)

$Given: x^{3} d y + (x y - 1) d x = 0$

$\Rightarrow x^{3} d y = (1 - x y) d x$

$\Rightarrow \frac{d y}{d x} = \frac{1 - x y}{x^{3}} \Rightarrow \frac{d y}{d x} + \frac{y}{x^{2}} = \frac{1}{x^{3}},$ which is a linear differential equation.

$Now, (I.F.) = e^{\int P d x} = e^{\int \frac{1}{x^{2}} d x} = e^{- 1 / x}$

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

Putting $u = \frac{- 1}{x} \Rightarrow d u = \frac{1}{x^{2}} d x,$ we get

$y e^{- 1 / x} = \int - e^{u} u d u + c = - [u e^{u} - \int e^{u} d u] + c$

$= - [u e^{u} - e^{u}] + c = (1 - u) e^{u} + c$

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

Put $x = \frac{1}{2}$ and $y = 3 - e$

$3 e^{- 2} - e^{- 1} = 3 e^{- 2} + c \Rightarrow c = - e^{- 1}$

Equation (i) becomes, $y e^{- 1 / x} = e^{- 1 / x} + \frac{e^{- 1 / x}}{x} - e^{- 1}$

Put $x = 1$ to get $y (1)$ ,

$\Rightarrow y e^{- 1} = e^{- 1} + e^{- 1} - e^{- 1} = e^{- 1} \Rightarrow y = 1$

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