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

4x^2 y′′ +12xy′ + 3y = 0

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

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

cauchy − euler   4r^2 +(−4+12)r+3 = 0  4r^2 +8r+3 = 0  (2r+1)(2r+3) = 0⇒ { ((r=−(1/2))),((r=−(3/2))) :}  solution y= x^r  ⇒  { ((y=x^(−(1/2)) )),((y=x^(−(3/2)) )) :}

$$\mathrm{cauchy}\:−\:\mathrm{euler}\: \\ $$$$\mathrm{4r}^{\mathrm{2}} +\left(−\mathrm{4}+\mathrm{12}\right)\mathrm{r}+\mathrm{3}\:=\:\mathrm{0} \\ $$$$\mathrm{4r}^{\mathrm{2}} +\mathrm{8r}+\mathrm{3}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2r}+\mathrm{1}\right)\left(\mathrm{2r}+\mathrm{3}\right)\:=\:\mathrm{0}\Rightarrow\begin{cases}{\mathrm{r}=−\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{r}=−\frac{\mathrm{3}}{\mathrm{2}}}\end{cases} \\ $$$$\mathrm{solution}\:\mathrm{y}=\:\mathrm{x}^{\mathrm{r}} \:\Rightarrow\:\begin{cases}{\mathrm{y}=\mathrm{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} }\\{\mathrm{y}=\mathrm{x}^{−\frac{\mathrm{3}}{\mathrm{2}}} }\end{cases} \\ $$

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