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Consider the differential equation $x^{2} y^{\prime \prime}+3 x y^{\prime}-3 y=\frac{12 x^{2}}{1+x^{2}}, x>0$
Find two linearly independent solutions of the form $x^{r}$ to the
associated homogeneous differential equation
Find the general solutions of the given equation.

(a) If $x^{r}$ is asolution to the assectated homo $-D \cdot E$

$y=x^{r}$ is a solution

$y^{\prime}=r x^{r-1}$

$y^{\prime \prime}=(r)(r-1) x^{r-2}$ Substitute in the Homo. D.E:

$x^{2}\left[(r)(r-1) x^{r-2}\right]+3 x\left[r x^{r-1}\right]-3 x^{r}=0$

$(r)(r-1) x^{r}+3 r x^{r}-3 x^{r}=0$

$\rightarrow x^{r}[r(r-1)+3 r-3]=0 \quad \div x^{r}$

$r^{2}-r+3 r-3=0 \rightarrow r^{2}+2 r-3=0$

$(r+3)(r-1)=0$

$r=-3$
$r=+1$

$y=x^{-3}, y=x^{1}$

$x^{2} y^{\prime \prime}+3 x y^{\prime}-3 y=\frac{12 x^{2}}{1+x^{2}} \quad \div x^{2}$

$y^{\prime \prime}+\left(\frac{3}{x}\right) y^{\prime}-\left(\frac{3}{x^{2}}\right) y=\frac{12}{1+x^{2}}$

\ $y_{c}=c_{1} x+c_{2} {x}^{-3}$

$y p=A_{1}(x) x+B(x) x^{-3}$

$A^{\prime} x+B^{\prime} {x}^{-3}=0 \longrightarrow (1)$

$A^{\prime}-3 B^{\prime} x^{-4}=\frac{12}{1+x^{2}} \longrightarrow (2)$

$-3 B^{\prime} {x}^{-4}=\frac{12}{1+x^{2}}$ multiply by $(-x)$

$3 {B^\prime} {x}^{-3}=\frac{-12 x}{1+x^{2}} \quad (1) + (2)$

$4 B^{\prime} x^{-3}=\frac{-12 x}{1+x^{2}} \quad \div 4 x^{-3}$

$B^{\prime}=\frac{-12 x}{1+x^{2}} \times \frac{1}{4 x^{-3}}=\frac{-3 x^{4}}{1+x^{2}}$ in $(1)$

$A^\prime x+B^\prime x^{-3}=0 \rightarrow A^\prime x+\left(\frac{-3 x^{4}}{1+x^{2}}\right) x^{-3}=0$

${A^\prime} x+\frac{-3 x}{1+x^{2}}=0 \Rightarrow A^{\prime}=\frac{+3 x}{1+x^{2}} * \frac{1}{x}=\frac{3}{1+x^{2}}$

$A^{\prime}=\frac{3}{1+x^{2}}$

$A=\int \frac{3}{1+x^{2}} d x=3 \cdot \tan ^{-1}(x)$

$B=\int \frac{-3 x^{4}}{1+x^{2}} d x$

$=\int\left[-3 x^{2}+3 \frac{-3}{x^{2}+1}\right] d x$

$B=\frac{-3 x^{3}}{3}+3 x-3 \tan ^{-1} x$

$y p=A x+B x^{-3}=3x \tan ^{-1}(x)+\left(-x^{3}+3 x-3 \tan ^{-1}(x)\right) x^{-3}$

$y_{g . s}=y_{p}+y_{c}$