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Question Number 103411 by abdomsup last updated on 14/Jul/20

$$solve{x}^2{y}^{{}^{″}}+{xy}^{{}^{\prime }}+y=0$$

Answered by bemath last updated on 14/Jul/20

$$sety=x^r$$$$y^{\prime }={rx}^{r-1};y^{″}=r(r-1)x^{r-2}$$$$\Rightarrow x^2(r^2-r)x^{r-2}+x\left({rx}^{r-1}\right)+x^r=0$$$$x^r(r^2-r+r+1)=0$$$$\Rightarrow r^2+1=0,r=\pm i$$$$y=x^{\pm i}$$

Answered by OlafThorendsen last updated on 15/Jul/20

$$y=x^m$$$$y^{\prime }={mx}^{m-1}$$$$y^{″}=m(m-1)x^{m-2}$$$$\Rightarrow m(m-1)+m+1=0$$$$m^2+1=0$$$$m=\pm i$$$$\Rightarrow y={\mathrm{C}}_1{x}^i+{\mathrm{C}}_2{x}^{-i}$$$$y={\mathrm{C}}_1{e}^{i\ln x}+{\mathrm{C}}_2{e}^{-i\ln x}$$$$y=({\mathrm{C}}_1+{\mathrm{C}}_2)\cos \left(\ln x\right)+({\mathrm{C}}_1-{\mathrm{C}}_2)i\sin \left(\ln x\right)$$$${\mathrm{C}}_1-{\mathrm{C}}_2=0\Rightarrow {\mathrm{C}}_1={\mathrm{C}}_2$$$$\mathrm{Finally}y=\mathrm{Kcos}\left(\ln x\right)$$

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