Let be the solution of the differential equation <img alt="\mathrm{\left(3 y^2-5 x^2\

Let $\mathrm{y=y(x)}$ be the solution of the differential equation $\mathrm{\left(3 y^2-5 x^2\right) y \mathrm{~d} x+2 x\left(x^2-y^2\right) \mathrm{d} y=0}$ such that $\mathrm{y(1)=1}$ . Then $\mathrm{\left|(y(2))^3-12 y(2)\right|}$ is equal to :

Option: 1
$16 \sqrt{2}$

Option: 2
$32 \sqrt{2}$

Option: 3
$32$

Option: 4
$64$

Answers (1)

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

$\mathrm{\frac{d y}{d x}=t+x \frac{d t}{d x}}$
$\mathrm{ t+x \frac{d t}{d x}=\frac{\left(5-3 t^2\right) t}{2\left(1-t^2\right)} }$
$\mathrm{x \frac{d t}{d x}=t\left[\frac{5-3 t^2-2+2 t^2}{2\left(1-t^2\right)}\right]}$
$\mathrm{ =\frac{t}{2}\left[\frac{3-t^2}{1-t^2}\right] }$
$\mathrm{ \frac{1}{3} \int \frac{3\left(1-t^2\right) d t}{3 t-t^3}=\int \frac{d x}{2 x} }$
$\mathrm{\frac{1}{3} \ln \left|3 t-t^3\right|=\frac{1}{2} \ln |x|+c }$
$\mathrm{2 \ell \ln \left|3 t-t^3\right|=3 \ln |x|+6 c }$
$\mathrm{ \left(3 t-t^3\right)^2=x^3 \lambda }$
$\mathrm{ \left(\frac{3 y}{x}-\frac{y^3}{x^3}\right)^2=x^3 \lambda }$
$\mathrm{ \left(3 y x^2-y^3\right)^2=x^9 \lambda }$
$\mathrm{ x=1 \Rightarrow y=1}$
$(3-1)^2=1 \times \lambda$
$\lambda=4$
$\mathrm{\left(3 y x^2-y^3\right)^2=4 x^9}$
Let $\mathrm{x=2}$
$\mathrm{ \left(3 y(2) \times 4-\left(y(2)^3\right)^2\right)=4(2)^9 }$
Taking square root both the sides,