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Question Number 130830 by Ar Brandon last updated on 29/Jan/21

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

Answered by bemath last updated on 29/Jan/21

$$\mathrm{Cauchy}-\mathrm{Euler}\mathrm{diff}\mathrm{eq}$$$$\mathrm{let}\mathrm{y}={\mathrm{x}}^{\mathrm{m}}\to \{\begin{array}{l}{\mathrm{y}}^{\prime }={\mathrm{mx}}^{\mathrm{m}-1}\\ {\mathrm{y}}^{″}=({\mathrm{m}}^2-\mathrm{m}){\mathrm{x}}^{\mathrm{m}-2}\end{array}$$$$\iff {\mathrm{m}}^2-2\mathrm{m}+1=0;\mathrm{m}=1;1$$$$\mathrm{y}={\mathrm{C}}_1\mathrm{x}+{\mathrm{C}}_2\mathrm{x}\ln \mathrm{x}=\mathrm{x}({\mathrm{C}}_1+{\mathrm{C}}_2\ln \mathrm{x})$$

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