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

2 Answers
Apr 14, 2018

Induction Proof - Hypothesis

We seek to prove that:

If # y=e^(mtan^(-1)x) #, then:
# #
# (1+x^2)y_(n+1)+(2nx-m)y_n+n(n-1)y_(n-1)=0 # ..... [A]

Where #y_n # represents the #n^(th)# derivative

Induction Proof - Base case:

Differentiating wrt #x# (twice), we get:

# y_1 = e^(mtan^(-1)x) d/dx (mtan^(-1)x)#
# \ \ \ = (m \ e^(mtan^(-1)x) )/(1+x^2) #

# y_2 = m \ ( (1+x^2)(m \ (e^(mtan^(-1)x) )/(1+x^2)) - (2x)(e^(mtan^(-1)x)) ) / (1+x^2)^2#
# \ \ \ = (m(m-2x)e^(mtan^(-1)x) ) / (1+x^2)^2 #

And , when #n=1# we have:

# LHS = (1+x^2)y_(1)+(2x-m)y_1+1(0)y_(0)#
# \ \ \ \ \ \ \ \ = (1+x^2)((m(m-2x)e^(mtan^(-1)x) ) / (1+x^2)^2 )+(2x-m)((m \ e^(mtan^(-1)x) )/(1+x^2)) #
# \ \ \ \ \ \ \ \ = ((m(m-2x)+m(2x-m))/(1+x^2)) \ e^(mtan^(-1)x#
# \ \ \ \ \ \ \ \ = 0 #

And #LHS = RHS#, So the given result is true when #n=1#.

Induction Proof - General Case

Now, Let us assume that the given result [A] is true when #n=k#, for some #k in NN, k ge 1#, in which case for this particular value of #k# we have:

# (1+x^2)y_(k+1)+(2kx-m)y_k+k(k-1)y_(k-1)=0 # ..... [B]

Now, let us differentiate the expression [B] using the product rule:

# (1+x^2)y_(k+1)+(2kx-m)y_k+k(k-1)y_(k-1)=0 #

# :. (1+x^2)y_(k+2) + (2x)y_(k+1) + (2kx-m)y_(k+1) + (2k)y_k + k(k-1)y_(k) #

# :. (1+x^2)y_(k+2) + (2x + 2kx-m)y_(k+1) + (2k + k(k-1))y_(k) #

# :. (1+x^2)y_(k+2) + (2(k+1)x -m)y_(k+1) + k(k+1)y_(k) #

# :. (1+x^2)y_((k+1)+1) + (2(k+1)x -m)y_(k+1) + (k+1)((k+1)-1)y_((k+1)-1) #

Which is the given expression [A], with #n=k+1#

Induction Proof - Summary

So, we have shown that if the given result [A] is true for #n=k#, then it is also true for #n=k+1# where #k ge 1#. But we initially showed that the given result was true for #n=1# so it must also be true for #n=2, n=3, n=4, ... # and so on.

Induction Proof - Conclusion

Then, by the process of mathematical induction the given result [A] is true for #n in NN#

Hence we have:

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

Apr 14, 2018

See below

Explanation:

Assume it's true for n, then:

#P(n):= (1+x^2)y^((n+1))+(2nx-m)y^((n))+n(n-1)y^((n-1))=0#

Differentiate again:

#2xy^((n+1)) + (1+x^2)y^((n+2))+2ny^((n))+(2nx-m)y^((n+1))+n(n-1)y^((n))=0#

Collect terms:

#(1+x^2)y^((n+2))+ 2xy^((n+1))+(2nx-m)y^((n+1))+n(n-1)y^((n)) + 2ny^((n))=0#

#implies (1+x^2)y^((n+2))+ (2(n+1)x-m)y^((n+1))+(n+1)ny^((n))=0#

This is #P(n+1)#.

So if it's true for #n#, then it's true for #n+1, n+2, n+3, .....#.

We can now try #n = 1#. Is this true?

#(1+x^2)y''+(2x-m)y'=0 qquad triangle#

We need:

  • #D_x(e^(m tan^(-1)(x))) = (m e^(m tan^(-1)(x)))/(x^2 + 1) #

  • #D_x^2(e^(m tan^(-1)(x))) = (m (m - 2 x) e^(m tan^(-1)(x)))/(x^2 + 1)^2 #

Plug into #triangle#

#= (1+x^2)( (m (m - 2 x) e^(m tan^(-1)(x)))/(x^2 + 1)^2)+(2x-m)((m e^(m tan^(-1)(x)))/(x^2 + 1))#

#= (m e^(m tan^(-1) x))/(x^2 + 1)( m - 2 x + 2x-m) = 0#

So, by induction, it's true for #n= 1#, and therefore for #n in mathbb Z^+ #.