Question Number 115829 by mohammad17 last updated on 28/Sep/20
$${prove}\:{that}\:\left({y}−{c}\right)^{\mathrm{2}} ={cx}\:{its}\:{solution}\:{of}\:{the}\:{differention}\:{equation}\:\mathrm{4}{xy}^{''} +\mathrm{2}{xy}^{'} −{y}=\mathrm{0} \\ $$$$\left({m}.{o}\right) \\ $$
Answered by $@y@m last updated on 29/Sep/20
$${y}−{c}=\sqrt{{cx}}\: \\ $$$${y}=\sqrt{{c}\:}.\sqrt{{x}}\:+{c}\:...\left(\mathrm{1}\right) \\ $$$${y}'=\sqrt{{c}}.\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}\:}\:...\left(\mathrm{2}\right) \\ $$$${y}''=\frac{\sqrt{{c}}}{\mathrm{2}}.\frac{−\mathrm{1}}{\mathrm{2}{x}\sqrt{{x}}} \\ $$$${y}''=−\frac{\sqrt{{c}}}{\mathrm{4}}.\frac{\mathrm{1}}{{x}\sqrt{{x}}}\:...\left(\mathrm{3}\right) \\ $$$${Now}, \\ $$$$\mathrm{4}{xy}^{''} +\mathrm{2}{xy}^{'} −{y}=\mathrm{4}{x}×\left(−\frac{\sqrt{{c}}}{\mathrm{4}}.\frac{\mathrm{1}}{{x}\sqrt{{x}}}\:\right)+\mathrm{2}{x}×\frac{\sqrt{{c}}}{\mathrm{2}\sqrt{{x}}}−\sqrt{{cx}}−{c} \\ $$$$=\frac{−\sqrt{{c}}}{{x}\sqrt{{x}}}+\sqrt{{cx}}−.\sqrt{{cx}}−{c} \\ $$$$=\frac{−\sqrt{{c}}}{{x}\sqrt{{x}}}−{c} \\ $$$$\neq\mathrm{0} \\ $$$${Pl}.\:{check}\:{question} \\ $$$${OR},\:{my}\:{solution}. \\ $$$$\left.:\right) \\ $$
Commented by mohammad17 last updated on 29/Sep/20
$${thank}\:{you}\:{sir}\: \\ $$