The function f is such that f''(x) = f(x). Prove by mathematical induction that, for every positive integer n, `(d²ⁿ⁻¹)/(dx²ⁿ⁻¹)(xf(x)) = xf'(x)+(2n-1)f(x)`.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2022). The mark or mark type is shown in brackets.

1. [B1] Base case `n=1`: `d/dx(xf(x)) = xf'(x) + f(x) = xf'(x) + (2(1)-1)f(x)` (Checks base case using product rule.)
2. [B1] Assume true for `n=k`: `(d^(2k-1))/(dx^(2k-1))(xf(x)) = xf'(x) + (2k-1)f(x)` (States inductive hypothesis.)
3. [M1 A1] Differentiate once to `(d^(2k))/(dx^(2k))(xf(x)) = xf''(x) + 2kf'(x)` (Differentiates once.)
4. [M1 A1] Differentiate again to `(d^(2k+1))/(dx^(2k+1))(xf(x)) = xf'''(x) + f''(x) + 2kf''(x) = xf'(x) + (2k+1)f(x)` (Differentiates again, using `f''=f` and `f'''=f'`.)
5. [A1] So, it is also true for `n=k+1`. Hence, by induction, true for all positive integers. (States conclusion.)