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Question Number 96125 by john santu last updated on 30/May/20

$${4\mathrm{x}}^2{\mathrm{y}}^{″}+{12\mathrm{xy}}^{\prime }+3\mathrm{y}=0$$

Answered by john santu last updated on 30/May/20

$$\mathrm{cauchy}-\mathrm{euler}$$$${4\mathrm{r}}^2+(-4+12)\mathrm{r}+3=0$$$${4\mathrm{r}}^2+8\mathrm{r}+3=0$$$$(2\mathrm{r}+1)(2\mathrm{r}+3)=0\Rightarrow \{\begin{array}{l}\mathrm{r}=-\frac{1}{2}\\ \mathrm{r}=-\frac{3}{2}\end{array}$$$$\mathrm{solution}\mathrm{y}={\mathrm{x}}^{\mathrm{r}}\Rightarrow \{\begin{array}{l}\mathrm{y}={\mathrm{x}}^{-\frac{1}{2}}\\ \mathrm{y}={\mathrm{x}}^{-\frac{3}{2}}\end{array}$$

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