Let y = y ( x ) be the solution curve of the differential equation d y d x = y x ( 1 + x y 2 ( 1 + log e x ) ) , x 0 , y ( 1 ) = 3 . Then, y 2 ( x ) 9 is equal to [2023]

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Solution

Q.
Let $y = y (x)$ be the solution curve of the differential equation $\frac{d y}{d x} = \frac{y}{x} (1 + x y^{2} (1 + \log_{e} x)), x > 0, y (1) = 3 .$ Then, $\frac{y^{2} (x)}{9}$ is equal to [2023]

1 $\frac{x^{2}}{5 - 2 x^{3} (2 + \log_{e} x^{3})}$

2 $\frac{x^{2}}{3 x^{3} (1 + \log_{e} x^{2}) - 2}$

3 $\frac{x^{2}}{7 - 3 x^{3} (2 + \log_{e} x^{2})}$

4 $\frac{x^{2}}{2 x^{3} (2 + \log_{e} x^{3}) - 3}$

Ans.
(1)

$Let y = v x; \frac{d y}{d x} = v + x \frac{d v}{d x}$

$v + x \frac{d v}{d x} = v + v^{3} x^{3} (1 + \log_{e} x)$

$\int v^{- 3} d v = \int (x^{2} + x^{2} \log_{e} x) d x$

$\frac{v^{- 2}}{- 2} = \frac{x^{3}}{3} + \log x \frac{x^{3}}{3} - \int \frac{x^{2}}{3} d x = \frac{x^{3}}{3} + \frac{x^{3}}{3} \log x - \frac{x^{3}}{9} + C$

$- \frac{x^{2}}{2 y^{2}} = \frac{2}{9} x^{3} + \frac{x^{3}}{3} \log x + c; y (1) = 3$

$\frac{- 1}{2 \times 9} = \frac{2}{9} + C \Rightarrow C = \frac{- 5}{18}$

$\frac{- x^{2}}{2 y^{2}} = \frac{2}{9} x^{3} + \frac{x^{3}}{3} \log x - \frac{5}{18} \Rightarrow \frac{9}{y^{2}} = \frac{4 x^{3} + 6 x^{3} \log x - 5}{- x^{2}}$

$\Rightarrow \frac{y^{2} (x)}{9} = \frac{x^{2}}{5 - 2 x^{3} (2 + \log x^{3})}$

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