Let x = x ( y ) be the solution of the differential equation 2 ( y + 2 ) log e ( y + 2 ) ? d x + ( x + 4 - 2 log e ( y + 2 ) ) ? d y = 0 , y - 1 with x ( e 4 - 2 ) = 1 .Then x ( e 9 - 2 ) is equal to [2023]

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Solution

Q.
Let $x = x (y)$ be the solution of the differential equation $2 (y + 2) \log_{e} (y + 2) d x + (x + 4 - 2 \log_{e} (y + 2)) d y = 0,$ $y > - 1$ with $x (e^{4} - 2) = 1$ .Then $x (e^{9} - 2)$ is equal to [2023]

1 $\frac{10}{3}$

2 $\frac{32}{9}$

3 $3$

4 $\frac{4}{9}$

Ans.
(2)

$2 (y + 2) \ln (y + 2) d x + (x + 4 - 2 \ln (y + 2)) d y = 0$

$\Rightarrow 2 \ln (y + 2) + (x + 4 - 2 \ln (y + 2)) \frac{1}{y + 2} \cdot \frac{d y}{d x} = 0$ ...(i)

Let $\ln (y + 2) = t$ $\Rightarrow$ $\frac{1}{y + 2} \cdot \frac{d y}{d x} = \frac{d t}{d x}$

Equation (i) becomes,

$2 t + (x + 4 - 2 t) \cdot \frac{d t}{d x} = 0 \Rightarrow (x + 4 - 2 t) \frac{d t}{d x} = - 2 t$

$\Rightarrow \frac{d x}{d t} = \frac{2 t - 4 - x}{2 t} \Rightarrow \frac{d x}{d t} + \frac{x}{2 t} = \frac{2 t - 4}{2 t}$

which is a linear differential equation in $x$ .

$I . F . = e^{\int \frac{1}{2 t} d t} = e^{\frac{1}{2} \ln t} = t^{1 / 2}$

Now, $x t^{1 / 2} = \int \frac{2 t - 4}{2 t} t^{1 / 2} d t + C$

$\Rightarrow x t^{1 / 2} = \int (t^{1 / 2} - \frac{2}{t^{1 / 2}}) d t + C$

$\Rightarrow x \cdot t^{1 / 2} = \frac{t^{3 / 2}}{3 / 2} - 2 \cdot \frac{t^{1 / 2}}{1 / 2} + C$

$\Rightarrow x \cdot t^{1 / 2} = \frac{2 t^{3 / 2}}{3} - 4 t^{1 / 2} + C \Rightarrow x = \frac{2}{3} \cdot t - 4 + C \cdot t^{- 1 / 2}$

Put $t = \ln (y + 2)$

$x = \frac{2}{3} \ln (y + 2) - 4 + C \cdot (\ln {(y + 2))}^{- 1 / 2}$ ...(ii)

Put $y = e^{4} - 2$ and $x = 1$

$1 = \frac{2}{3} \ln (e^{4} - 2 + 2) - 4 + C (\ln {(e^{4} - 2 + 2))}^{- 1 / 2}$

$\Rightarrow 1 = \frac{2}{3} \times 4 - 4 + C \times \frac{1}{2} \Rightarrow \frac{C}{2} = 5 - \frac{8}{3} = \frac{7}{3} \Rightarrow C = \frac{14}{3}$

Equation (ii) becomes,

$x = \frac{2}{3} \ln (y + 2) - 4 + \frac{14}{3} (\ln {(y + 2))}^{- 1 / 2}$

Put $y = e^{9} - 2$

$x (e^{9} - 2) = \frac{2}{3} \ln (e^{9} - 2 + 2) - 4 + \frac{14}{3} (\ln {(e^{9} - 2 + 2))}^{- 1 / 2}$

$\Rightarrow x (e^{9} - 2) = \frac{2}{3} \times 9 - 4 + \frac{14}{3} \times \frac{1}{3} = 2 + \frac{14}{9} = \frac{32}{9}$

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