Let $Y=Y\left(X\right)$ be a curve lying in the first quadrant such that the area enclosed by the line $Y-y=Y^{\text{'}}\left(x\right)(X-x)$ and the co-ordinate axes, where $(x\mathit{,}y)$ is any point on the curve, is always $\frac{-y^2}{2Y^{\text{'}}\left(x\right)}+1,Y^{\text{'}}\left(x\right)\ne 0$. If $Y\left(1\right)=1$, then $12 Y\left(2\right)$ equals ________.

Plotting the line $Y-y=Y^{\text{'}}\left(x\right)(X-x)$ we get,

 

Now, from the above diagram,

The area of the triangle formed will be,

$A=\frac{1}{2}\left(\frac{-y}{Y^{\text{'}}\left(x\right)}+x\right)(y-x{Y}^{\text{'}}\left(x\right))$

Now, according to the given condition we get,

$\frac{1}{2}\left(\frac{-y}{Y^{\text{'}}\left(x\right)}+x\right)(y-x{Y}^{\text{'}}\left(x\right))=\frac{-y^2}{2Y^{\text{'}}\left(x\right)}+1$

$\Rightarrow \left(-y+x{Y}^{\text{'}}\left(x\right)\right)\left(y-x{Y}^{\text{'}}\left(x\right)\right)=-y^2+2Y^{\text{'}}\left(x\right)$

$\Rightarrow -y^2+xy{Y}^{\text{'}}\left(x\right)+xy{Y}^{\text{'}}\left(x\right)-x^2{\left[Y^{\text{'}}\left(x\right)\right]}^2=-y^2+2Y^{\text{'}}\left(x\right)$

$\Rightarrow 2xy-x^2{Y}^{\text{'}}\left(x\right)=2$

$\Rightarrow \frac{dy}{dx}=\frac{2xy-2}{x^2}$

$\Rightarrow \frac{dy}{dx}-\frac{2}{x}y=\frac{-2}{x^2}$

It is a linear differential equation,

Now, finding $\mathrm{IF}=e^{-2\ln x}=\frac{1}{x^2}$

Now, the solution is given by,

$y·\frac{1}{x^2}=\frac{2}{3}{x}^{-3}+c$

Given at $x=1, y=1$

$\Rightarrow 1=\frac{2}{3}+c\Rightarrow c=\frac{1}{3}$

Hence, the equation of the curve will be,

$Y=\frac{2}{3}·\frac{1}{X}+\frac{1}{3}{X}^2$

$\Rightarrow 12Y\left(2\right)=\frac{5}{3}\times 12=20$