If #y=e^(mtan^-1x)#, check whether the equation #(1+x^2)y_(n+1)+(2nx-m)+n(n+1)y_(n-2) = 0# ?

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Answer 1 · 2017-09-30T13:04:56

The specified function does not satisfy the given recurrence relationship.

We have:

# y_0 = e^(mtan^-1x) #

And we wish to show that:

#(1+x^2)y_(n+1)+(2nx-m)+n(n+1)y_(n-2) = 0#

Where #y_n# represents the #n^(th)# derivative wrt #x#.

Let us test the base case #n=2# of the recurrence relationship, which gives us:

#(1+x^2)y_(3)+(4x-m)+6y_(0) = 0#

Now, let us test a specific case of #m=0 => y=1#, Then, all derivatives are zero, and the base case recurrence relationship is as follows:L

# (1+x^2)y_(3)+(4x-m)+6y_(0) = (1+x^2)0+(4x-0)+6 = 0#
# " " = 4x+6 #
# " " cancel(-=) 0 #

Hence, the specified function does not satisfy the given recurrence relationship.