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

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