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Question Number 196846 by mokys last updated on 01/Sep/23

$$2{xy}^{″}+(1-4x)y^{\prime }+(2x-1)y=y$$

Answered by witcher3 last updated on 04/Sep/23

$${2\mathrm{xy}}^{″}+(1-4\mathrm{x}){\mathrm{y}}^{\prime }+(2\mathrm{x}-1)\mathrm{y}=0$$$$\mathrm{y}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}..\mathrm{solution}$$$$\mathrm{y}={\mathrm{ze}}^{\mathrm{x}}$$$$\Rightarrow {\mathrm{y}}^{\prime }=(\mathrm{z}+{\mathrm{z}}^{\prime }){\mathrm{e}}^{\mathrm{x}}$$$${\mathrm{y}}^{″}=({\mathrm{z}}^{″}+{2\mathrm{z}}^{\prime }+\mathrm{z}){\mathrm{e}}^{\mathrm{x}}$$$$\iff 2\mathrm{x}({\mathrm{z}}^{″}+{2\mathrm{z}}^{\prime }+\mathrm{z})+(1-4\mathrm{x})(\mathrm{z}+{\mathrm{z}}^{\prime })+(2\mathrm{x}-1)\mathrm{z}=0$$$${2\mathrm{xz}}^{″}+(4\mathrm{x}+1-4\mathrm{x}){\mathrm{z}}^{\prime }=0$$$${2\mathrm{xz}}^{″}+{\mathrm{z}}^{\prime }=0$$$$\frac{{\mathrm{z}}^{″}}{{\mathrm{z}}^{\prime }}=-\frac{1}{2\mathrm{x}}\Rightarrow {\mathrm{z}}^{\prime }=\ln \left(\frac{1}{\sqrt{\mid \mathrm{x}\mid }}\right)+\mathrm{c}$$$${\mathrm{z}}^{\prime }=\frac{\mathrm{k}}{\sqrt{\mid \mathrm{x}\mid }}\Rightarrow \mathrm{z}=2\mathrm{k}\sqrt{\mid \mathrm{x}\mid }+\mathrm{c}=$$$$\mathrm{y}={\mathrm{ce}}^{\mathrm{x}}+\mathrm{b}\sqrt{\mid \mathrm{x}\mid }{\mathrm{e}}^{\mathrm{x}}$$$$$$

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