Question

If `y=acos(lnx)+bsin(lnx)` then show that? `x^2y^((n+2)) + (2n+1)xy^((n+1))+(n^2+1)y^((n)) = 0`

(Question Restore: portions of this question have been edited or deleted!)

Answer 1

We have poor notation as `y_n` is being used as the `n^(th)` derivative `y^((n))`, and `logx` is being used to represent the natural logarithm, `lnx`. There is also an error in the result that we are required to prove, which should read as follows

We should therefore write this as given that

`y=acos(lnx)+bsin(lnx)` ... [A]

Then prove that:

`x^2y^((n+2)) + (2n+1)xy^((n+1))+(n^2+1)y^((n)) = 0` ... [B}

(Notice the omission of `x` in the last term, which is incorrect in the original question.

We will prove this result is true for natural numbers `n in NN` by Mathematical Induction.

First we consider the case `n=0`; Differentiating [A] once wrt `x` using the chain rule gives:

`y' = -asin(lnx)*1/x +bcos(lnx)*1/x` ..... [1]

Differentiating [1] a second time (also using the product rule) we get:

`y'' = (-asin(lnx))(-1/x^2) + (-acos(lnx)*1/x)(1/x) + (bcos(lnx))(-1/x^2) + (-bsin(lnx)*1/x)(1/x)`

`:. y'' = asin(lnx)*1/x^2 -acos(lnx)*1/x^2 - bcos(lnx)*1/x^2 -bsin(lnx)*1/x^2` ..... [2]

Multiplying [1] by `x`, and Multiplying [2] by `x^2` gives:

`\ \ \ xy' = -asin(lnx) +bcos(lnx)`
`x^2y'' = asin(lnx) -acos(lnx) - bcos(lnx) -bsin(lnx)`

Substituting into given DE [B] (with `n=0`) we get:

`x^2y''+xy'+y = asin(lnx) -acos(lnx) - bcos(lnx) -bsin(lnx) -asin(lnx) +bcos(lnx)+acos(lnx)+bsin(lnx)`

`:. x^2y''+xy'+y = 0`

So the given result [B] is true when `n=0`

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

`x^2y^((k+2)) + (2k+1)xy^((k+1))+(k^2+1)y^((k)) = 0` ..... [3]

Now we differentiate [3] wrt `x` to get:

`(x^2)(y^((k+3))) + (2x)(y^((k+2))) + ((2k+1)x)(y^((k+2))) + (2k+1)(y^((k+1))) + (k^2+1)y^((n+1)) = 0`

`:. x^2y^((k+3)) + 2xy^((k+2)) + (2k+1)xy^((k+2)) + (2k+1)y^((k+1)) + (k^2+1)y^((k+1)) = 0`

`:. x^2y^((k+3)) + (2 + 2k+1)xy^((k+2)) + (2k+1+k^2+1)y^((k+1)) = 0`

`:. x^2y^((k+3)) + (2k+3)xy^((k+2)) + (k^2+2k+2)y^((k+1)) = 0`

`:. x^2y^(((k+1)+2)) + (2(k+1)+1)xy^(((k+1)+1)) + ((k+1)^2+1)y^((k+1)) = 0`

Which is the given result with `n=k+1`

So, we have shown that if the given result is true for `n=k`, then it is also true for `n=k+1`. But we initially showed that the given result was true for `n=0` so it must also be true for `n=1, n=2, n=3, n=4, ...` and so on.

Hence, by the process of mathematical induction the given result is true for `n in NN` QED

Answer 2

For Proof, see Explanation.

Explanation:

We note that, here, `y_1,y_2,...,y_n` denote the `1^(st), 2^(nd), ..., n^(th)`

Derivatives of `y,` resp., and, `logx` does `lnx.`

`y=acos(lnx)+bsin(lnx)`

`rArr y_1=a(-sin(lnx))*1/x+b(cos(lnx))*1/x`

`rArr xy_1=-a(sin(lnx)+bcos(lnx)`

Rediff.ing both sides w.r.t. `x` with the help of Product and

Chain Rule, we get,

`xy_2+y_1=-acos(lnx)*1/x-bsin(lnx)*1/x`

`rArr x^2y_2+xy_1=-{acos(lnx)+bsin(lnx)}=-y, or,`

`x^2y_2+xy_1+y=0`

`:. (x^2y_2+xy_1+y)_n=(0)_n=0....(star).`

To prove the Result, we, now, take help of Leibnitz Theorem (LT),

which states :

`(uv)_n=uv_n+""_nC_1u_1v_(n-1)+""_nC_2u_2v_(n-2)+...+u_nv.`

Using LT for `x^2y_2`, with, `u=x^2, and, v=y_2,` we have,

`(x^2y_2)_n=x^2(y_2)_n+""_nC_1(x^2)'(y_2)_(n-1)+""_nC_2(x^2)''(y_2)_(n-2)+""_nC_3(x^2)'''(y_2)_(n-3)+...`

`=x^2y_(n+2)+n*(2x)*y_(n+1)+(n(n-1))/2*(2)*y_n+0+0+...`

`:.(x^2y_2)_n=x^2y_(n+2)+2nxy_(n+1)+(n^2-n)y_n....(1).`

Similarly, we have, `(xy_1)_n=xy_(n+1)+ny_n...(2).`

Using `(1) and (2)` in `(star)`, we get,

`{x^2y_(n+2)+2nxy_(n+1)+(n^2-n)y_n}+{xy_(n+1)+ny_n}+{y_n}=0,`

what is the same as,

`x^2y_(n+2)+x(2n+1)y_(n+1)+(n^2+1)y_n=0.`

Enjoy Maths.!