y_n means the nth differential of f(x)
f(x) = arcsinx
Prove by induction that (1-x^2)y_(n+2) - (2n+1)xy_(n+1) –y_n.n^2 = 0
Differentiating f(x),
y_1 = 1/(1 – x^2)^(1/2)
y_2 = x/(1 – x^2)^(3/2)
y_3 = 3x^2/(1 – x^2)^(5/2) + 1/(1 – x^2)^(3/2)
P(n) : (1-x^2)y_(n+2) - (2n+1)xy_(n+1) –y_n.n^2 = 0
The case n = 1 : (1-x^2)y_3 - 3xy_2 –y_n = 0
Substituting for the above differentials,
3x^2/(1 – x^2)^(3/2) + (1 – x^2)/(1 – x^2)^(3/2) – 3x^2/(1 – x^2)^(3/2) – (1 – x^2)/(1 – x^2)^(3/2) = 0
{3x^2 + (1 – x^2) – 3x^2 – (1 – x^2)}/(1 – x^2)^(3/2) = 0
{3x^2 + 1 – x^2 – 3x^2 – 1 + x^2)}/(1 – x^2)^(3/2) = 0
All the terms in the numerator cancel, so
0 = 0, which is true
Therefore the assertion P(n) holds for n = 1.
Assume that P(k) holds true for some k.
Then (1-x^2)y_(k+2) – (2k+1)xy_(k+1) – y_k.k^2 = 0
The case n = k+1 :
Differentiate P(k) to give
-2x.y_(k+2) + (1 – x^2).y_(k+3) – (2k + 1).y_(k+1) – (2k + 1)x.y_(k+2) – y_(k+1).k^2 = 0
(1-x^2)y_(k+3) - (2x+(2k+1)x)y_(k+2) – (2k+1+k^2)y_(k+1) = 0
(1-x^2)y_(k+3) - (2k+3)xy_(k+2) – y_(k+1).(k^2 + 2k + 1) = 0
(1-x^2)y_(k+3) - (2(k+1)+1)xy_(k+2) – y_(k+1).(k + 1)^2 = 0
Which shows that P(k+1) is true.
Hence, by mathematical induction, P(n) is true for all n >=1.