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• Notes

Let $w(x)$ be the Wronskian of the functions $\left\{1, x^{2}, f(x)\right\}$
Find $f(x)$ for which $w(x)=2 x^{2}$

$W=\left|\begin{array}{ccc}{1} & {x^{2}} & {f(x)} \\ {0} & {2 x} & {f^{\prime}} \\ {0} & {2} & {f^{\prime \prime}}\end{array}\right|$

$2 x f^{\prime \prime}-2 f^{\prime}=2 x^{2}$

$\rightarrow f^{\prime \prime}-\frac{2 f^{\prime}}{2 x}=\frac{2 x^{2}}{2 x}$

$\rightarrow f^{\prime \prime}-\frac{f^{\prime}}{x}=x$

$y=f^{\prime}$
$y^{\prime}=f^{\prime \prime}$

$y^{\prime}-\frac{1}{x} y=x$ linear in $y$

$\mu=e^{\int \frac{-1}{x} d x}=e^{-\ln (x)}=x^{-1}=\frac{1}{x}$

$\mu y=\int \mu \: Q(x) d x$

$\frac{1}{x} f^{\prime}(x)=\int \frac{1}{x} \cdot x d x + C$

$\longrightarrow \frac{1}{x} f^{\prime}(x)=x+c$

$f^{\prime}(x)=x^{2}+c x$

$\int f^{\prime}(x)=\int x^{2}+c x$

$\rightarrow f(x)=\frac{x^{3}}{3}+\frac{c x^{2}}{2}+k$

Without solving the equation, find the $w(y 1, y 2)$ of the DE:
$\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\alpha(\alpha+1) y=0$

Abel $\left(1-x^{2}\right) y^{\prime \prime}-2 x y^{\prime}+\alpha(\alpha+1) y=0$

$\div\left(1- x^{2}\right)$

$y^{\prime \prime}-\frac{2 x}{1-x^{2}} y^{\prime}+\frac{\alpha(\alpha+1)}{1-x^{2}} y=0$

$p(x)=\frac{-2 x}{1-x^{2}} \quad Q(x)=\frac{\alpha(\alpha+1)}{1-x^{2}}$

are cont, except at $x=\pm1$

then by Abel's Theorem :

$\rightarrow w\left(y_{1}, y_{2}\right)= c e^{-\int \rho(x) d x}=c e^{-\int \frac{-2 x}{1-x^{2}} d x}$

$=C e^{-\ln \left|1-x^{2}\right|}=\frac{c}{1-x^{2}}$