"If ${y^{1/4}} + {y^{ - 1/4}} = 2x$, and $({x^2} - 1){{{d^2}y} \over {d{x^2}}} + \alpha x{{dy} \over {dx}} + \beta y = 0$, then | $\alpha$ $-$ $\beta$ | is equal to __________."

Question

Accepted Answer

"${y^{{1 \over 4}}} + {1 \over {{y^{{1 \over 4}}}}} = 2x$

$$ \Rightarrow {\left( {{y^{{1 \over 4}}}} \right)^2} - 2x{y^{\left( {{1 \over 4}} \right)}} + 1 = 0$$

$\Rightarrow {y^{{1 \over 4}}} = x + \sqrt {{x^2} - 1}$ or $x - \sqrt {{x^2} - 1}$

So, $${1 \over 4}{1 \over {{y^{{3 \over 4}}}}}{{dy} \over {dx}} = 1 + {x \over {\sqrt {{x^2} - 1} }}$$

$$ \Rightarrow {1 \over 4}{1 \over {{y^{{3 \over 4}}}}}{{dy} \over {dx}} = {{{y^{{1 \over 4}}}} \over {\sqrt {{x^2} - 1} }}$$

$\Rightarrow {{dy} \over {dx}} = {{4y} \over {\sqrt {{x^2} - 1} }}$ .... (1)

Hence, $${{{d^2}y} \over {d{x^2}}} = 4{{\left( {\sqrt {{x^2} - 1} } \right)y' - {{yx} \over {\sqrt {{x^2} - 1} }}} \over {{x^2} - 1}}$$

$\Rightarrow ({x^2} - 1)y'' = 4{{({x^2} - 1)y' - xy} \over {\sqrt {{x^2} - 1} }}$

$$ \Rightarrow ({x^2} - 1)y'' = 4\left( {\sqrt {{x^2} - 1} y' - {{xy} \over {\sqrt {{x^2} - 1} }}} \right)$$

$\Rightarrow ({x^2} - 1)y'' = 4\left( {4y - {{xy'} \over 4}} \right)$ (from I)

$\Rightarrow ({x^2} - 1)y'' + xy' - 16y = 0$

So, | $\alpha$ $-$ $\beta$ | = 17"

If ${y^{1/4}} + {y^{ - 1/4}} = 2x$, and

$({x^2} - 1){{{d^2}y} \over {d{x^2}}} + \alpha x{{dy} \over {dx}} + \beta y = 0$, then | $\alpha$ $-$ $\beta$ | is equal to __________.

Solution

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If ${y^{1/4}} + {y^{ - 1/4}} = 2x$, and $({x^2} - 1){{{d^2}y} \over {d{x^2}}} + \alpha x{{dy} \over {dx}} + \beta y = 0$, then | $\alpha$ $-$ $\beta$ | is equal to __________.

Solution

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More Differential Equations questions

Drill 25 more like these. Every day. Free.

If ${y^{1/4}} + {y^{ - 1/4}} = 2x$, and

$({x^2} - 1){{{d^2}y} \over {d{x^2}}} + \alpha x{{dy} \over {dx}} + \beta y = 0$, then | $\alpha$ $-$ $\beta$ | is equal to __________.