Question Number 118364 by mnjuly1970 last updated on 17/Oct/20
$$\:\:\:\:\:\:\:\:\:\:\:\:\:...{ordinary}\:{differential}\:{equation}... \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:{y}''\:−{xy}'+{y}=\mathrm{0} \\ $$$$\:\:\:\:{general}\:{solution}\:::=?? \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:.{m}.{n}.\mathrm{1970}. \\ $$$$ \\ $$
Answered by mindispower last updated on 17/Oct/20
$${we}\:{can}\:{see}\:{easly}\:{that}\:{y}={x}\:{is}\:{solution} \\ $$$${let}\:{y}={xz} \\ $$$$\Rightarrow{y}'={xz}'+{z} \\ $$$${y}''={xz}''+\mathrm{2}{z}' \\ $$$${xz}''+\mathrm{2}{z}'−{x}\left({xz}'+{z}\right)+{xz}=\mathrm{0} \\ $$$$\Rightarrow{xz}''+\left(\mathrm{2}−{x}^{\mathrm{2}} \right){z}'=\mathrm{0} \\ $$$$\Rightarrow\int\frac{{d}\left({z}'\right)}{{z}'}=\int\frac{{x}^{\mathrm{2}} −\mathrm{2}}{{x}}{dx}=\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{2}{ln}\left({x}\right)\right) \\ $$$$\Rightarrow{z}=\frac{{k}}{{x}^{\mathrm{2}} }{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$${by}\:{part}\:{z}=\left(−\frac{{k}}{{x}}{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \right)+\int{ke}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} {dx} \\ $$$${x}={iy}\sqrt{\mathrm{2}}\:{in}\:\mathrm{2}{nd} \\ $$$$={z}\left({x}\right)=−\frac{{k}}{\mathrm{2}}{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} +{k}\int{i}\sqrt{\mathrm{2}}{e}^{−{y}^{\mathrm{2}} } {dy} \\ $$$$=−\frac{{k}}{\mathrm{2}}{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} +{ik}\frac{\sqrt{\pi}}{\mathrm{2}}\:{erf}\left(\frac{{x}}{{i}\sqrt{\mathrm{2}}}\right) \\ $$$${y}\left({x}\right)={x}\left(−\frac{{k}}{\mathrm{2}}{e}^{\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} +{ik}\frac{\sqrt{\pi}}{\mathrm{2}}{erf}\left(\frac{{x}}{{i}\sqrt{\mathrm{2}}}\right)+{c}\right) \\ $$
Commented by mnjuly1970 last updated on 17/Oct/20
$${thank}\:{you}\:{so}\:{much}\:\:{mr}... \\ $$
Commented by mindispower last updated on 19/Oct/20
$${withe}\:{pleasur}\:{sir}\: \\ $$