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  • If <img alt="\mathrm{y^{1 / 4}+y^{-1 / 4}=2 x, and \left(x^2-1\right) \frac{d^2 y}{d x^2}+\alpha x \frac{d y}{d x}+\beta y=0, }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7By%5

If \mathrm{y^{1 / 4}+y^{-1 / 4}=2 x, and \left(x^2-1\right) \frac{d^2 y}{d x^2}+\alpha x \frac{d y}{d x}+\beta y=0, }then \mathrm{|\alpha-\beta| } is equal to

Option: 1

17


Option: 2

16


Option: 3

15


Option: 4

11


Answers (1)

best_answer

 We have, \mathrm{y^{1 / 4}+y^{-1 / 4}=2 x }
\mathrm{\begin{aligned} & \Rightarrow y^{2 / 4}+1=2 x y^{1 / 4} \Rightarrow\left(y^{1 / 4}\right)^2-2 x y^{1 / 4}+1=0 \\ & \Rightarrow y^{1 / 4}=\frac{2 x \pm \sqrt{4 x^2-4}}{2} \Rightarrow y^{1 / 4}=x \pm \sqrt{x^2-1} \\ & \Rightarrow y=\left(x \pm \sqrt{x^2-1}\right)^4 \end{aligned} }
Differentiating both sides w.r.t. x, we get
\mathrm{\begin{aligned} & \frac{d y}{d x}=4\left(x \pm \sqrt{x^2-1}\right)^3\left\{1 \pm \frac{1 \times 2 x}{2 \sqrt{x^2-1}}\right\}=4 \frac{\left(x \pm \sqrt{x^2-1}\right)^4}{\sqrt{x^2-1}} \\ & \Rightarrow \frac{d y}{d x}=\frac{4 y}{\sqrt{x^2-1}} \\ & \Rightarrow \frac{d^2 y}{d x^2}=\frac{4 \sqrt{x^2-1} \cdot \frac{d y}{d x}-4 y \cdot \frac{1 \times 2 x}{2 \sqrt{x^2-1}}}{\left(x^2-1\right)} \\ & \Rightarrow\left(x^2-1\right) \frac{d^2 y}{d x^2}=16 y-x \frac{d y}{d x} \quad \text { [Using (i)] } \end{aligned} }
[Using (ii)]
\mathrm{\begin{aligned} & \Rightarrow \quad\left(x^2-1\right) \frac{d^2 y}{d x^2}+x \frac{d y}{d x}-16 y=0 \Rightarrow \alpha=1, \beta=-16 \\ & \therefore \quad|\alpha-\beta|=|1+16|=17 \end{aligned} }

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