The value is=?

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The value is=?

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Answer 1 · 2018-03-15T13:45:27

Answer is (4).

Explanation:

$y = {(x + \sqrt{x^{2} - 1})}^{15} + {(x - \sqrt{x^{2} - 1})}^{15}$ $y=(x+sqrt(x^2-1))^15+(x-sqrt(x^2-1))^15$

Hence $\frac{d y}{d x} = 15 {(x + \sqrt{x^{2} - 1})}^{14} \cdot (1 + \frac{2 x}{2 \sqrt{x^{2} - 1}}) + 15 {(x - \sqrt{x^{2} - 1})}^{14} \cdot (1 - \frac{2 x}{2 \sqrt{x^{2} - 1}})$ $(dy)/(dx)=15(x+sqrt(x^2-1))^14*(1+(2x)/(2sqrt(x^2-1)))+15(x-sqrt(x^2-1))^14*(1-(2x)/(2sqrt(x^2-1)))$

= $15 {(x + \sqrt{x^{2} - 1})}^{14} \cdot (1 + \frac{x}{\sqrt{x^{2} - 1}}) + 15 {(x - \sqrt{x^{2} - 1})}^{14} \cdot (1 - \frac{x}{\sqrt{x^{2} - 1}})$ $15(x+sqrt(x^2-1))^14*(1+x/sqrt(x^2-1))+15(x-sqrt(x^2-1))^14*(1-x/sqrt(x^2-1))$

= $15 {(x + \sqrt{x^{2} - 1})}^{14} \cdot (\frac{\sqrt{x^{2} - 1} + x}{\sqrt{x^{2} - 1}}) + 15 {(x - \sqrt{x^{2} - 1})}^{14} \cdot (\frac{\sqrt{x^{2} - 1} - x}{\sqrt{x^{2} - 1}})$ $15(x+sqrt(x^2-1))^14*((sqrt(x^2-1)+x)/sqrt(x^2-1))+15(x-sqrt(x^2-1))^14*((sqrt(x^2-1)-x)/sqrt(x^2-1))$

= $15 \frac{{(x + \sqrt{x^{2} - 1})}^{15}}{\sqrt{x^{2} - 1}} - 15 \frac{{(x - \sqrt{x^{2} - 1})}^{15}}{\sqrt{x^{2} - 1}}$ $15(x+sqrt(x^2-1))^15/sqrt(x^2-1)-15(x-sqrt(x^2-1))^15/sqrt(x^2-1)$

or $\frac{\sqrt{x^{2} - 1}}{15} \frac{d y}{d x} = {(x + \sqrt{x^{2} - 1})}^{15} - {(x - \sqrt{x^{2} - 1})}^{15}$ $sqrt(x^2-1)/15(dy)/(dx)=(x+sqrt(x^2-1))^15-(x-sqrt(x^2-1))^15$

and similarly we have

$\frac{\sqrt{x^{2} - 1}}{15} \frac{d^{2} y}{{d x}^{2}} + \frac{d y}{d x} \cdot \frac{2 x}{30 \sqrt{x^{2} - 1}} = 15 \frac{{(x + \sqrt{x^{2} - 1})}^{15}}{\sqrt{x^{2} - 1}} + 15 \frac{{(x - \sqrt{x^{2} - 1})}^{15}}{\sqrt{x^{2} - 1}}$ $sqrt(x^2-1)/15(d^2y)/(dx^2)+(dy)/(dx)*(2x)/(30sqrt(x^2-1))=15(x+sqrt(x^2-1))^15/sqrt(x^2-1)+15(x-sqrt(x^2-1))^15/sqrt(x^2-1)$

or $\frac{x^{2} - 1}{225} \frac{d^{2} y}{{d x}^{2}} + \frac{d y}{d x} \cdot \frac{x}{225} = {(x + \sqrt{x^{2} - 1})}^{15} + {(x - \sqrt{x^{2} - 1})}^{15} = y$ $(x^2-1)/225(d^2y)/(dx^2)+(dy)/(dx)*x/225=(x+sqrt(x^2-1))^15+(x-sqrt(x^2-1))^15=y$

or $(x^{2} - 1) \frac{d^{2} y}{{d x}^{2}} + x \frac{d y}{d x} = 225 y$ $(x^2-1)(d^2y)/(dx^2)+x(dy)/(dx)=225y$

Hence answer is (4).

Additional Information $-$ $-$ Observe that for $y = {(x + \sqrt{x^{2} - 1})}^{n} + {(x - \sqrt{x^{2} - 1})}^{n}$ $y=(x+sqrt(x^2-1))^n+(x-sqrt(x^2-1))^n$ , $(x^{2} - 1) \frac{d^{2} y}{{d x}^{2}} + x \frac{d y}{d x}$ $(x^2-1)(d^2y)/(dx^2)+x(dy)/(dx)$ would be $n^{2} y$ $n^2y$

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The value is=?

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