If y=( ^ -1 2 x)^2+( ^ -1 2 x)^2 , then (1-4 x^2) y_2-4 x y_1=

We have, y=\left(\sin ^{-1}2 x\right)^2+\left(\cos ^{-1}2 x\right)^2 On differentiating both sides w.r.t. $x$, we get $\frac{d y}{d x}=$2 \sin ^{-1} 2 x $\cdot \frac{1}{\sqrt{1-(2 x)^2}} \cdot$2+2 \cos ^{-1} 2 x \left(\frac{-1}{\sqrt{1-(2 x)^2}}\right) \cdot$2 \Rightarrow$\sqrt{1-4 x^2} \frac{d y}{d x}=4\left(\sin ^{-1}2 x-\cos ^{-1} 2 x\right) Again differentiating both sides w.r.t. $x$, we get$$\begin{aligned}& \sqrt{1-4 x^2}y_2+\frac{1}{2 \sqrt{1-4 x^2}}(-8 x) y_1 \\ & =4\left(\frac{1}{\sqrt{1-4 x^2}} \cdot₂⁺\frac{1}{\sqrt{1-4 x^2}} \cdot2\right) \\ \text{ and } =\sqrt{1-4 x^2}y_2-\frac{4 x}{\sqrt{1-4 x^2}}y_1=8 \cdot \frac{2}{\sqrt{1-4 x^2}}\\ \Rightarrow \quad \text{ and } \quad\left(1-4 x²\right) y_2-4 x y_1=16 \end{aligned}$$