The value of λ+k=?

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The value of λ+k=?

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Answer 1 · 2018-02-26T21:16:02

$λ + k = - 24$ $lambda+k = -24$

Explanation:

Considering

$y = e^{log y}$ $y = e^(log y)$ we have

$y^{\frac{1}{5}} = e^{\frac{log y}{5}}$ $y^(1/5) = e^((log y)/5)$

$x = \frac{y^{\frac{1}{5}} + y^{- \frac{1}{5}}}{2} = \frac{e^{\frac{log y}{5}} + e^{- \frac{log y}{5}}}{2} = cosh (\frac{log y}{5})$ $x = (y^(1/5)+y^(-1/5))/2 = (e^((log y)/5)+e^(-(log y)/5))/2 = cosh(log y/5)$

and then

$\frac{log y}{5} = arcosh (x)$ $logy/5 = "arcosh"(x)$ or

$y = e^{5 arcosh (x)}$ $y = e^(5 "arcosh"(x))$

After substituting into the differential equation we have

$(5 (λ - 1) \frac{x}{\sqrt{x^{2} - 1}} + k + 25) e^{5 arcosh (x)} = 0$ $(5 (lambda-1) x / sqrt[x^2-1] + k+25)e^(5 "arcosh"( x)) =0$

So the conditions for solution are

$λ = 1$ $lambda = 1$
$k + 25 = 0$ $k+25=0$

and then

$λ + k = - 24$ $lambda+k = -24$

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The value of λ+k=?

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