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Question-66866

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Question Number 66866 by rajesh4661kumar@gmail.com last updated on 20/Aug/19
Commented by mathmax by abdo last updated on 20/Aug/19
first x+(√(1+x^2 ))>0 for all x so y=ln(x+(√(1+x^2 )))=argsh(x) ⇒ y^′ (x)=(1/( (√(1+x^2 )))) =(1+x^2 )^(−(1/2)) ⇒y^(′′) (x)=−(1/2)(2x)(1+x^2 )^(−(3/2)) =((−x)/((1+x^2 )^(3/2) )) ⇒(x^2 +1)y^(′′) +xy′=(x^2 +1)((−x)/((1+x^2 )^(3/2) )) +(x/( (√(1+x^2 )))) =((−x)/( (√(1+x^2 ))))+(x/( (√(1+x^2 )))) =0 so the Q must be (x^2 +1)y^(′′) +xy^′ =0
$${first}\:\:{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }>\mathrm{0}\:\:{for}\:{all}\:{x}\:{so}\:{y}={ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)={argsh}\left({x}\right)\:\Rightarrow \\ $$$${y}^{'} \left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\Rightarrow{y}^{''} \left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$=\frac{−{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\Rightarrow\left({x}^{\mathrm{2}} +\mathrm{1}\right){y}^{''} \:+{xy}'=\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\frac{−{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:+\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$=\frac{−{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}+\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:=\mathrm{0}\:\:\:{so}\:{the}\:{Q}\:{must}\:{be}\:\:\left({x}^{\mathrm{2}} +\mathrm{1}\right){y}^{''} \:+{xy}^{'} =\mathrm{0} \\ $$$$ \\ $$
Answered by Kunal12588 last updated on 20/Aug/19
y=log(x+(√(x^2 +1))) ⇒y′=(1/(x+(√(x^2 +1))))×(1+((2x)/(2(√(x^2 +1))))) ⇒y′=(1/(x+(√(x^2 +1))))×(((x+(√(x^2 +1)))/( (√(x^2 +1))))) ⇒y′=(1/( (√(x^2 +1)))) ⇒y′(√(x^2 +1))=1 ⇒(√(x^2 +1))×y′′+y′×((2x)/(2(√(x^2 +1))))=0 ⇒(x^2 +1)y′′+xy′=0
$${y}={log}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$\Rightarrow{y}'=\frac{\mathrm{1}}{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}×\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\right) \\ $$$$\Rightarrow{y}'=\frac{\mathrm{1}}{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}×\left(\frac{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\right) \\ $$$$\Rightarrow{y}'=\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\Rightarrow{y}'\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}=\mathrm{1} \\ $$$$\Rightarrow\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}×{y}''+{y}'×\frac{\mathrm{2}{x}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}=\mathrm{0} \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} +\mathrm{1}\right){y}''+{xy}'=\mathrm{0} \\ $$

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