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Question Number 113464 by bemath last updated on 13/Sep/20

$$(x^2\frac{d^{2}y}{{dx}^2}+x\frac{dy}{dx}+1).y=0$$

Answered by mathmax by abdo last updated on 13/Sep/20

$$\mathrm{y}=0\mathrm{is}\mathrm{soluton}\mathrm{if}\mathrm{y}\ne 0\mathrm{weget}{\mathrm{x}}^2{\mathrm{y}}^{{}^{″}}+\mathrm{x}{\mathrm{y}}^{{}^{\prime }}+1=0\mathrm{let}{\mathrm{y}}^{{}^{\prime }}=\mathrm{z}\Rightarrow$$$${\mathrm{x}}^2{\mathrm{z}}^{{}^{\prime }}+\mathrm{xz}+1=0\Rightarrow {\mathrm{z}}^{{}^{\prime }}+\frac{1}{\mathrm{x}}\mathrm{z}=-\frac{1}{{\mathrm{x}}^2}\left(\mathrm{e}\right)$$$$\mathrm{h}\to {\mathrm{z}}^{{}^{\prime }}+\frac{1}{\mathrm{x}}\mathrm{z}=0\Rightarrow \frac{{\mathrm{z}}^{{}^{\prime }}}{\mathrm{z}}=-\frac{1}{\mathrm{x}}\Rightarrow \ln \mid \mathrm{z}\mid =-\ln \mid \mathrm{x}\mid +\mathrm{c}\Rightarrow \mathrm{z}=\frac{\mathrm{k}}{\mathrm{x}}$$$$\mathrm{lagrange}\mathrm{method}\to {\mathrm{z}}^{{}^{\prime }}=\frac{{\mathrm{k}}^{{}^{\prime }}}{\mathrm{x}}-\frac{\mathrm{k}}{{\mathrm{x}}^2}$$$$\left(\mathrm{e}\right)\Rightarrow \frac{{\mathrm{k}}^{{}^{\prime }}}{\mathrm{x}}-\frac{\mathrm{k}}{{\mathrm{x}}^2}+\frac{\mathrm{k}}{{\mathrm{x}}^2}=-\frac{1}{{\mathrm{x}}^2}\Rightarrow {\mathrm{k}}^{{}^{\prime }}=-\frac{1}{\mathrm{x}}\Rightarrow \mathrm{k}=-\ln (\mid \mathrm{x}\mid )+\mathrm{c}\Rightarrow$$$$\mathrm{z}=-\frac{\ln \mid \mathrm{x}\mid }{\mathrm{x}}+\frac{\mathrm{c}}{\mathrm{x}}$$$${\mathrm{y}}^{{}^{\prime }}=\mathrm{z}\Rightarrow \mathrm{y}=\int \mathrm{z}\left(\mathrm{x}\right)\mathrm{dx}=-\int \frac{\ln \mid \mathrm{x}\mid }{\mathrm{x}}\mathrm{dx}+\mathrm{cln}\mid \mathrm{x}\mid +\lambda$$

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