JEE Main 2021 — Mathematics Calculus

$${y}^{\frac{1}{4}}+\frac{1}{{y}^{\frac{1}{4}}}=2x$$y41​+y41​1​=2x

$${({y}^{\frac{1}{4}})}^{2}-2x{y}^{\frac{1}{4}}+1=0$$(y41​)2−2xy41​+1=0

{y}^{\frac{1}{4}}=x+\sqrt{{x}^{2}-1}&x-\sqrt{{x}^{2}-1}

So, $\frac{1}{4}\frac{1}{{y}^{\frac{3}{4}}}\frac{dy}{dx}=1+\frac{x}{\sqrt{{x}^{2}-1}}$41​y43​1​dxdy​=1+x2−1​x​

$$\frac{1}{4}\frac{1}{{y}^{\frac{3}{4}}}\frac{dy}{dx}=\frac{{y}^{\frac{1}{4}}}{\sqrt{{x}^{2}-1}}$$41​y43​1​dxdy​=x2−1​y41​​

$$\frac{dy}{dx}=\frac{4y}{\sqrt{{x}^{2}-1}}...(1)$$dxdy​=x2−1​4y​...(1)

Hence

$$\frac{{d}^{2}y}{d{x}^{2}}=4\frac{(\sqrt{{x}^{2}-1}){y}^{'}-\frac{xy}{\sqrt{{x}^{2}-1}}}{{x}^{2}-1}$$dx2d2y​=4x2−1(x2−1​)y′−x2−1​xy​​

$$({x}^{2}-1){y}^{"}=\frac{4({x}^{2}-1){y}^{'}-xy}{\sqrt{{x}^{2}-1}}$$(x2−1)y"=x2−1​4(x2−1)y′−xy​

$$({x}^{2}-1)y{'}^{}=4[\sqrt{{x}^{2}-1}{y}^{'}-\frac{xy}{\sqrt{{x}^{2}-1}}]$$(x2−1)y′=4[x2−1​y′−x2−1​xy​]

$$({x}^{2}-1){y}^{"}=4[4y-\frac{x{y}^{'}}{4}](1)$$(x2−1)y"=4[4y−4xy′​](1)

$$({x}^{2}-1){y}^{"}+xy'-16y=0$$(x2−1)y"+xy′−16y=0

So $|\alpha -\beta |=17$∣α−β∣=17