The value of λ+k=?

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Answer 1 · 2018-02-26T21:16:02

#lambda+k = -24#

Considering

#y = e^(log y)# we have

#y^(1/5) = e^((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

#logy/5 = "arcosh"(x)# or

#y = e^(5 "arcosh"(x))#

After substituting into the differential equation we have

#(5 (lambda-1) x / sqrt[x^2-1] + k+25)e^(5 "arcosh"( x)) =0#

So the conditions for solution are

#lambda = 1#
#k+25=0#

and then

#lambda+k = -24#