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Question Number 120058 by bramlexs22 last updated on 29/Oct/20 | ||
$$\:\frac{{d}^{\mathrm{2}} {y}}{{dx}}\:+{x}\:\frac{{dy}}{{dx}}\:−{y}=\mathrm{0} \\ $$ | ||
Answered by Olaf last updated on 29/Oct/20 | ||
$$ \\ $$$${y}''+{xy}'−{y}\:=\:\mathrm{0}\:\left(\mathrm{1}\right) \\ $$$${y}\:=\:{xu} \\ $$$${y}'\:=\:{xu}'+{u} \\ $$$${y}''\:=\:{xu}''+\mathrm{2}{u}' \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\::\:{xu}''+\mathrm{2}{u}'+{x}\left({xu}'+{u}\right)−{xu}\:=\:\mathrm{0} \\ $$$${xu}''+\mathrm{2}{u}'+{x}^{\mathrm{2}} {u}'\:=\:\mathrm{0} \\ $$$$\frac{{u}''}{{u}'}\:=\:−\frac{\mathrm{2}+{x}^{\mathrm{2}} }{{x}}\:=\:−\frac{\mathrm{2}}{{x}}−{x} \\ $$$$\mathrm{ln}\mid{u}'\mid\:=\:−\mathrm{2ln}\mid{x}\mid−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{C}_{\mathrm{1}} \\ $$$${u}'\:=\:\mathrm{C}_{\mathrm{2}} {e}^{\mathrm{ln}\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \:=\:\frac{\mathrm{C}_{\mathrm{2}} }{{x}^{\mathrm{2}} }{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$${u}\:=\:−\mathrm{C}_{\mathrm{2}} \left[\frac{{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} }{{x}}+\sqrt{\frac{\pi}{\mathrm{2}}}\mathrm{erf}\left(\frac{{x}}{\:\sqrt{\mathrm{2}}}\right)\right]+\mathrm{C}_{\mathrm{3}} \\ $$$$\left(\mathrm{By}\:\mathrm{parts}\right) \\ $$$$ \\ $$$${y}\:=\:{ux}\:=\:−\mathrm{C}_{\mathrm{2}} \left[{e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} +\sqrt{\frac{\pi}{\mathrm{2}}}{x}\mathrm{erf}\left(\frac{{x}}{\:\sqrt{\mathrm{2}}}\right)\right]+\mathrm{C}_{\mathrm{3}} {x} \\ $$ | ||