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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 :

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

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

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