JEE MAIN - Mathematics (2023 - 31st January Evening Shift - No. 14)

Let $y=y(x)$ be the solution of the differential equation

$\left(3 y^{2}-5 x^{2}\right) y \mathrm{~d} x+2 x\left(x^{2}-y^{2}\right) \mathrm{d} y=0$

such that $y(1)=1$. Then $\left|(y(2))^{3}-12 y(2)\right|$ is equal to :
64
$16 \sqrt{2}$
32
$32 \sqrt{2}$

Explanation

$\left(3 y^{2}-5 x^{2}\right) y \cdot d x+2 x\left(x^{2}-y^{2}\right) d y=0$

$$ \Rightarrow \frac{d y}{d x}=\frac{y\left(5 x^{2}-3 y^{2}\right)}{2 x\left(x^{2}-y^{2}\right)} $$

Put $\mathrm{y}=\mathrm{mx}$

$$ \Rightarrow m+x \cdot \frac{d m}{d x}=\frac{m\left(5-3 m^{2}\right)}{2\left(1-m^{2}\right)} $$

$$ \begin{aligned} & x \cdot \frac{d m}{d x}=\frac{\left(5-3 m^{2}\right) m-2 m\left(1-m^{2}\right)}{2\left(1-m^{2}\right)} \\\\ & \Rightarrow \frac{\mathrm{dx}}{\mathrm{x}}=\frac{2\left(\mathrm{~m}^{2}-1\right)}{\mathrm{m}\left(\mathrm{m}^{2}-3\right)} \mathrm{dm} \\\\ & \Rightarrow \frac{d x}{x}=\left(\frac{2}{m}-\frac{\frac{4}{3}}{m}+\frac{\frac{4 m}{3}}{\mathrm{~m}^{2}-3}\right) d m \end{aligned} $$

$\Rightarrow \int \frac{d x}{x}=\int \frac{\left(\frac{2}{3}\right)}{m}+\int \frac{2}{3}\left(\frac{2 m}{m^{2}-3}\right) d m$

$\Rightarrow \ln |\mathrm{x}|=\frac{2}{3} \ln |\mathrm{m}|+\frac{2}{3} \ln \left|\mathrm{m}^{2}-3\right|+\mathrm{C}$

Or, $\ln |\mathrm{x}|=\frac{2}{3} \ln \left|\frac{\mathrm{y}}{\mathrm{x}}\right|+\frac{2}{3} \ln \left|\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-3\right|+\mathrm{C}$

Put $(\mathrm{x}=1, \mathrm{y}=1)$ : we get $\mathrm{c}=-\frac{2}{3} \ln (2)$

$\Rightarrow \ln |\mathrm{x}|=\frac{2}{3} \ln \left|\frac{\mathrm{y}}{\mathrm{x}}\right|+\frac{2}{3} \ln \left|\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-3\right|-\frac{2}{3} \ln (2)$

$\Rightarrow\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\left[\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{2}-3\right]=2 .\left(\mathrm{x}^{3 / 2}\right)$ ........(1)

Put $\mathrm{x}=2$ in equation (1), we get

$\Rightarrow \mathrm{y}\left(\mathrm{y}^{2}-12\right)=4 \times 2 \times 2 \times 2 \sqrt{2}$

$\Rightarrow \mathrm{y}^{3}-12 \mathrm{y}=32 \sqrt{2}$

$\Rightarrow\left|\mathrm{y}^{3}(2)-12 \mathrm{y}(2)\right|=32 \sqrt{2}$

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