Question Number 217245 by OmoloyeMichael last updated on 07/Mar/25
$${find}\:{the}\:{following}\:{differential}\:{equation}\: \\ $$$${by}\:{eliminating}\:{the}\:{arbritrary}\:{constant} \\ $$$$\left(\mathrm{1}\right){y}={Ae}^{{x}} +{Bcosx} \\ $$$$\left(\mathrm{2}\right)\:{xy}={Ae}^{{x}} +{Be}^{−{x}} +{x}^{\mathrm{2}} \\ $$$$ \\ $$
Answered by aleks041103 last updated on 07/Mar/25
$${for}\:{example}: \\ $$$$\left(\mathrm{1}\right)\:{y}={Ae}^{{x}} +{Bcosx} \\ $$$$\:{y}'={Ae}^{{x}} −{Bsinx} \\ $$$$\Rightarrow{y}−{y}'={Bcosx}+{Bsinx}={z} \\ $$$${z}''=−{Bcosx}−{Bsinx}=−{z} \\ $$$$\Rightarrow\left({y}−{y}'\right)''+\left({y}−{y}'\right)=\mathrm{0} \\ $$$$\Rightarrow{y}'''−{y}''+{y}'−{y}=\mathrm{0} \\ $$$$ \\ $$$$\left(\mathrm{2}\right){xy}={Ae}^{{x}} +{Be}^{−{x}} +{x}^{\mathrm{2}} \\ $$$$\Rightarrow{z}={xy}−{x}^{\mathrm{2}} ={Ae}^{{x}} +{Be}^{−{x}} \\ $$$${z}''={Ae}^{{x}} +{Be}^{−{x}} ={z} \\ $$$$\Rightarrow\left({xy}−{x}^{\mathrm{2}} \right)''−\left({xy}−{x}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\left({xy}\right)''−\mathrm{2}−{xy}+{x}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({y}+{xy}'\right)'−{xy}=\mathrm{2}−{x}^{\mathrm{2}} \\ $$$${y}'+{y}'+{xy}''−{xy}=\mathrm{2}−{x}^{\mathrm{2}} \\ $$$$\Rightarrow{xy}''+\mathrm{2}{y}'−{xy}=\mathrm{2}−{x}^{\mathrm{2}} \\ $$
Commented by aleks041103 last updated on 07/Mar/25
$${That}\:{is},\:{if}\:{I}\:{have}\:{understood}\:{the}\:{question} \\ $$$${correctly}… \\ $$