Question

:(a) Use the explicit solutions Y₁(X)and Y₂(X) of Legendre’sequation given in 32and the appropriate choice ofc₀and c₁to find the Legendre polynomials P₆(X) and P₇(X).

(b) Write the differential equations for whichP₆(X) andP₇(X)are particular solutions.

Short answer

(a) The Legendre polynomials are$y_1=-\frac{5}{16}+\frac{105}{16}{x}^2-\frac{315}{16}{x}^4+\frac{231}{16}{x}^6$and $y_2=-\frac{35}{16}x+\frac{315}{16}{x}^3-\frac{693}{16}{x}^5+\frac{429}{16}{x}^7$

(b) The differential equations are $\left(1-x^2\right)y\text{'}\text{'}-2xy\text{'}+42y=0$ and $\left(1-x^2\right)y\text{'}\text{'}-2xy\text{'}+56y=0$.

Step-by-step solution

Step 1:Define Legendre polynomials.

Let the Legendre equation be $(1-{\mathrm{x}}^2)\mathrm{y}\text{'}\text{'}-2\mathrm{xy}\text{'}+\mathrm{n}(\mathrm{n}+1)\mathrm{y}=0$. The two particular solutions are ${\mathrm{y}}_1={\mathrm{c}}_0\left[1-\frac{\mathrm{n}(\mathrm{n}+1)}{2!}{\mathrm{x}}^2+\frac{(\mathrm{n}-2)\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+3)}{4!}{\mathrm{x}}^4-\frac{(\mathrm{n}-4)(\mathrm{n}-2)\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+3)(\mathrm{n}+5)}{6!}{\mathrm{x}}^6+\dots \right]$and

${\mathrm{y}}_2={\mathrm{c}}_1\left[\mathrm{x}-\frac{(\mathrm{n}-1)(\mathrm{n}+2)}{3!}{\mathrm{x}}^3+\frac{(\mathrm{n}-3)(\mathrm{n}-1)(\mathrm{n}+2)(\mathrm{n}+4)}{5!}{\mathrm{x}}^5-\frac{(\mathrm{n}-5)(\mathrm{n}-3)(\mathrm{n}-1)(\mathrm{n}+2)(\mathrm{n}+4)(\mathrm{n}+6)}{7!}{\mathrm{x}}^7+\dots \right]$,

where the arbitrary constants are$c_0=\left\{\begin{array}{ll}1& \mathrm{if}\mathrm{n}=0\\ {(-1)}^{\mathrm{n}/2}\frac{1\times 3\dots (\mathrm{n}-1)}{2\times 4\dots \mathrm{n}}& \mathrm{if}\mathrm{n}=2,4,6,\dots \end{array}\right.$ and${\mathrm{c}}_1=\left\{\begin{array}{ll}1& \mathrm{if}\mathrm{n}=1\\ {(-1)}^{(\mathrm{n}-1)/2}\frac{1\times 3\dots \mathrm{n}}{2\times 4\dots (\mathrm{n}-1)}& \mathrm{if}\mathrm{n}=3,5,7,\dots \end{array}\right.$ .

Derive the Legendre polynomial of degree 6 .

The Legendre polynomial of degree $6$represents a particular solution $y_1$at $n=6$.

role="math" localid="1668519815116" $\begin{array}{c}{\mathrm{y}}_1={\left(-1\right)}^{6/2}\frac{1\times 3\times 5}{2\times 4\times 6}\left[1-\frac{6(6+1)}{2!}{\mathrm{x}}^2+\frac{(6-2)6(6+1)(6+3)}{4!}{\mathrm{x}}^4-\frac{(6-4)(6-2)6(6+1)(6+3)(6+5)}{6!}{\mathrm{x}}^6+0+0+\dots \right]\\ \\ =\frac{-5}{16}\left[1-21{\mathrm{x}}^2+63{\mathrm{x}}^4-\frac{231}{5}{\mathrm{x}}^6\right]\\ =-\frac{5}{16}+\frac{105}{16}{\mathrm{x}}^2-\frac{315}{16}{\mathrm{x}}^4+\frac{231}{16}{\mathrm{x}}^6\end{array}$

Derive the Legendre polynomial of degree .

The Legendre polynomial of degree $7$represents a particular solution $y_2$ at $n=7$.

$\begin{array}{c}{y}_2={\left(-1\right)}^{\left(7-1\right)/2}\frac{1\times 3\times 5\times 7}{2\times 4\times 6}\left[x-\frac{(7-1)(7+2)}{3!}{x}^3+\frac{(7-3)(7-1)(7+2)(7+4)}{5!}{x}^5-\frac{(7-5)(7-3)(7-1)(7+2)(7+4)(7+6)}{7!}{x}^7+0+0+\dots \right]\\ =\frac{-35}{16}\left[x-9x^3+\frac{99}{5}{x}^5-\frac{429}{35}{x}^7\right]\\ =-\frac{35}{16}x+\frac{315}{16}{x}^3-\frac{693}{16}{x}^5+\frac{429}{16}{x}^7\end{array}$

Derive the differential equation.

Let the differential equation be$(1-x^2)y\text{'}\text{'}-2xy\text{'}+n(n+1)y=0$.

From the above equation, the differential equation for which$P_6\left(x\right)$are particular solution is,

$\begin{array}{r}(1-{\mathrm{x}}^2)\mathrm{y}\text{'}\text{'}-2\mathrm{xy}\text{'}+6(6+1)\mathrm{y}=0\\ (1-{\mathrm{x}}^2)\mathrm{y}\text{'}\text{'}-2\mathrm{xy}\text{'}+42\mathrm{y}=0\end{array}$

From the above equation, the differential equation for which$P_7\left(x\right)$are particular solution is,

$\begin{array}{r}(1-x^2)y\text{'}\text{'}-2xy\text{'}+7(7+1)y=0\\ (1-x^2)y\text{'}\text{'}-2xy\text{'}+56y=0\end{array}$