Question

The differential equation $y\text{'}\text{'}-2xy\text{'}+2\alpha y=0$is known as Hermite’s equation of order a after the French mathematician Charles Hermite (1822–1901). Show that thegeneral solution of the equation is $y\left(x\right)=c_0{y}_1\left(x\right)+c_1{y}_2\left(x\right)$, where $$\begin{gathered} y_1\left(x\right)=1+\sum _{k=1}^{\infty }{(-1)}^k\frac{2^{k}n(n-2)L(n-2k+2)}{\left(2k\right)!}{x}^{2k} \\ {y}_2\left(x\right)=x+\sum _{k=1}^{\infty }{(-1)}^k\frac{2^k(n-1)(n-3)L(n-2k+1)}{(2k+1)!}{x}^{2k+1} \end{gathered}$$

are power series solutions centered at the ordinary point 0.

Short answer

(a)The polynomial solutions are,

For (Y₁(X))

$$\begin{gathered} n=0,y_{}\left(x\right)=1 \\ n=2,y_1\left(x\right)=1-2x^2 \\ n=4,y_1\left(x\right)=1-4x^2+\frac{4}{3}{x}^4 \end{gathered}$$

For Y₂(x)

$$\begin{gathered} n=1,y_2\left(x\right)=x \\ n=3,y_2\left(x\right)=x-\frac{2}{3}{x}^3 \\ n=5,y_2\left(x\right)=x-\frac{4}{3}{x}^3+\frac{4}{15}{x}^5 \end{gathered}$$

(b) Hence proved that the first six Hermite polynomials are,

Step-by-step solution

Define Chebyshev’s equation.

Let the given DE be $(1-x^2)y\text{'}\text{'}-xy\text{'}+{\alpha }^{2}y=0$ . The given differential equation has an ordinary point at X=0 . It is possible to state that, without a proof, if the differential equation has an ordinary point at, x=x₀ then it has a power series solution with two linear independent solution has the following form:

$y=\sum _{n=0}^{\infty }{c}_n(x-x_0{)}^n$

Determine the two linear independent solution

Let the form of the solution of the given differential equation be

$y=\sum _{n=0}^{\infty }{c}_n(x-x_0{)}^n$

The first and second derivation of the equation (1) is the linear independent solution, So,

$$\begin{gathered} y\text{'}=\sum _{n=0}^{\infty }{c}_{n}n{x}^{n-1}\left(a_0=0\right) \\ y\text{'}\text{'}=\sum _{n=0}^{\infty }{c}_{n}n(n-1)x^{n-2}\left(a_0=a_1=0\right) \end{gathered}$$

Shift the summation n=1 for, since the first term of the series is equal to zero, and n=2 for Y^''

$$\begin{gathered} y\text{'}=\sum _{n=1}^{\infty }{c}_{n}n{x}^{n-1} \\ y\text{'}\text{'}=\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2} \end{gathered}$$

Find the function value.

Substitute the value (1), (2), (3) in the given DE.

$$\begin{gathered} \left(1-x^2\right)\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2}-x\sum _{n=1}^{\infty }{c}_{n}n{x}^{n-1}+{\alpha }^2\sum _{n=0}^{\infty }{c}_n{x}^n \\ =\underset{a_2\to x^2}{\underbrace{\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^{n-2}}}-\underset{a_1\to x^1}{\underbrace{\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^n}}-\underset{a_0\to x^n}{\underbrace{\sum _{n=1}^{\infty }{c}_{n}n{x}^n}}+\underset{a_0\to x^0}{\underbrace{\sum _{n=0}^{\infty }{\alpha }^2{c}_n{x}^n}} \end{gathered}$$

Assure that the first terms, for the four-power series are raised to the same power, which is not the case. So,

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=[2c_2{x}^0+6c_3{x}^1+\underset{a_4\to x^2}{\underbrace{\sum _{n=4}^{\infty }{c}_{n}n(n-1)x^{n-2}}}]-\underset{a_2\to x^2}{\underbrace{\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^n}}-[c_1{x}^1+\underset{a_2\to x^2}{\underbrace{\left.\sum _{n=2}^{\infty }{c}_{n}n{x}^n\right]}} \\ +[{\alpha }^2{c}_0{x}^0+\alpha c_1{x}^1+\underset{a_2\to x^2}{\underbrace{\sum _{n=2}^{\infty }{\alpha }^2{c}_n{x}^n}}] \end{gathered}$$

Replace the summation index for the first power series from n to n+2.

$$\begin{gathered} f\left(y\text{'}\text{'},y\text{'},y\right)=2c_2{x}^0+6c_3{x}^1+\sum _{n+2=4}^{\infty }{c}_{n+2}(n+2)\left(\right(n+2)-1)x^{(n+2)-2}-\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^n-c_1{x}^1-\sum _{n=2}^{\infty }{c}_{n}n{x}^n \\ +{\alpha }^2{c}_0{x}^0+{\alpha }^2{c}_1{x}^1+\sum _{n=2}^{\infty }\alpha c_n{x}^n \\ =\left[2c_2+{\alpha }^2{c}_0\right]x^0+\left[6c_3-c_1+{\alpha }^2{c}_1\right]x^1+\sum _{n=2}^{\infty }{c}_{n+2}(n+2)(n+1)x^n-\sum _{n=2}^{\infty }{c}_{n}n(n-1)x^n \\ -\sum _{n=2}^{\infty }{c}_{n}n{x}^n+\sum _{n=2}^{\infty }\alpha c_n{x}^n \\ =\left[2c_2+{\alpha }^2{c}_0\right]x^0+\left[6c_3+c_1\left({\alpha }^2-1\right)\right]x^1+\sum _{n=2}^{\infty }\left[c_{n+2}(n+2)(n+1)-c_{n}n(n-1)-c_{n}n+{\alpha }^2{c}_n\right]x^n \\ =\left[2c_2+{\alpha }^2{c}_0\right]x^0+\left[6c_3+c_1\left({\alpha }^2-1\right)\right]x^1+\sum _{n=2}^{\infty }\left[c_{n+2}(n+2)(n+1)-c_n\left(n^2-\backslash \mathrm{not}x+\backslash \mathrm{not}x\right)+{\alpha }^2{c}_n\right]x^n \\ =\left[2c_2+{\alpha }^2{c}_0\right]x^0+\left[6c_3+c_1\left({\alpha }^2-1\right)\right]x^1+\sum _{n=2}^{\infty }\left[c_{n+2}(n+2)(n+1)-c_n\left(n^2-{\alpha }^2\right)\right]x^n \\ =0 \end{gathered}$$

 Step 4: Find the equation for the coefficients.

Find the coefficients for the series.

Expand the series.

Expand the series in (1) yields

$$\begin{gathered} y\left(x\right)=c_0+c_{1}x+c_2{x}^2+c_3{x}^3+c_4{x}^4+c_5{x}^5+\dots \\ =\left[c_0+c_2{x}^2+c_4{x}^4+\dots \right]+\left[c_{1}x+c_3{x}^3+c_5{x}^5+\dots \right] \\ =\left[c_0+\sum _{k=1}^{\infty }{c}_{2k}{x}^{2k}\right]+\left[c_{1}x+\sum _{k=1}^{\infty }{c}_{2k+1}{x}^{2k+1}\right] \\ =\left[c_0+\sum _{k=1}^{\infty }\frac{{(-1)}^k{2}^k\alpha \left(\alpha -2\right)...\left(\alpha -2k+2\right)}{\left(2k\right)!}{c}_0{x}^{2k}\right]+\left[c_{1}x+\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k{2}^k\left(\alpha -1\right)\left(\alpha -3\right)...\left(\alpha -2k+1\right)}{(2k+1)!}{c}_1{x}^{2k+1}\right]l \\ =c_0\left[1+\sum _{k=1}^{\infty }\frac{{(-1)}^k{2}^k\alpha \left(\alpha -2\right)...\left(\alpha -2k+2\right)}{\left(2k\right)!}{x}^{2k}\right]+c_1\left[x+\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k{2}^k\left(\alpha -1\right)\left(\alpha -3\right)...\left(\alpha -2k+1\right)}{(2k+1)!}{x}^{2k+1}\right] \\ =c_0{y}_1\left(x\right)+c_1{y}_2\left(x\right) \end{gathered}$$