Question

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

Answer 1

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

Answer 2

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^+`.