$\endgroup$ 2The Legendre equation $$(1-x^2)y''-2xy'+\alpha(\alpha+1)y=0$$ has a polynomial solution $P_n$ when $\alpha$ is a non-negative number $n$. Find $P_1$ satisfying $P_1(1)=1$ and then find the general solution of the D.E. with $\alpha=1$ $$y=Ax+B$$$$y'=A,y''=0$$$$B=0, y=Ax$$$$y=x$$$$W=e^{\int \frac{-2x}{1-x^2}}=x^2-1$$$$y_2=x\int \frac{x^2-1}{x^2}dx=x^2+1$$$$y=A_1x+A_2x(x+\frac{1}{x})$$ But where to from here?
1 Answer
$\begingroup$$$\alpha=1\qquad\to\qquad (1-x^2)y''-2xy'+2y=0 \tag 1$$ You found a particular solution $y=x$ which is correct.
Or, more general, a family of solutions : $\quad y=C\:x\quad$ where $C$ is a constant.
In order to find the general solution of the ODE, one can use the method of variation of parameter.
In the present case, remplace the parameter $C$ by an unknown function $u(x)$:
$y=u(x)\:x \quad\to\quad y'=xu'+u \quad\to\quad y''=xu''+2u'$
Putting them into $(1)$ leads to : $$x(1-x^2)u''+2(1-2x^2)u'=0$$ $$\frac{u''}{u'}=2\frac{2x^2-1}{x(1-x^2)}$$ $$\ln|u'|=2\int \frac{2x^2-1}{x(1-x^2)}dx =-\ln|1-x^2|-2\ln|x|+\text{constant}$$
$$u'=\frac{c_1}{x^2(1-x^2)}$$ $$u=c_1\int \frac{dx}{x^2(1-x^2)} = c_1\left(-\frac{1}{x}+\frac{1}{2}\ln|1+x|-\frac{1}{2}\ln|1-x| \right)+c_2$$ The general solution of $(1)$ is : $$y(x)=c_1\int \frac{dx}{x^2(1-x^2)} = c_1\left(-1+\frac{1}{2}x\ln|1+x|-\frac{1}{2}x\ln|1-x| \right)+c_2x$$
$\endgroup$
