(a) When = n is a nonnegative integer, Hermites differential equation always possesses a polynomial solution...

(a) When α = n is a nonnegative integer, Hermite’s differential equation always possesses a polynomial solution of degree n. Use y₁(x), given in Problem 51, to find polynomial solutions for n = 0, n = 2, and n = 4.

Then use y₂(x) to find polynomial solutions for n = 1, n = 3, and n = 5.

(b) A Hermite polynomial H_n(x) is defined to be the nth degree polynomial solution of Hermite’s equation multiplied by an appropriate constant so that the coefficient of xⁿ in H_n(x) is 2_n. Use the polynomial solutions in part (a) to show that the first six Hermite polynomials are

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H(x) = 1 H(x) = 2x Н.(х) — 4х? — 2 H3(x) = 8x – 12x НА(х) 3D 16х4 — 48х? + 12 На (х) %3D 32х5 — 160х3 + 120х.