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Question-48500

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Question Number 48500 by peter frank last updated on 24/Nov/18
Commented by Abdo msup. last updated on 24/Nov/18
y(x)=e^(arctan(x)) ⇒y^′ (x)=(1/(1+x^2 )) y(x) ⇒ y^(′′) (x) =−((2x)/((1+x^2 )^2 ))y(x) +(1/(1+x^2 ))y^′ (x) =((−2x)/((1+x^2 )^2 ))y(x) +(1/((1+x^2 )^2 ))y(x) =((−2x+1)/((1+x^2 )^2 ))y(x) ⇒ (1+x^2 )y^(′′) −(2x−1)y^′ =(((1+x^2 )(−2x+1))/((1+x^2 )^2 ))y(x)−(2x−1)(1/(1+x^2 ))y(x) =(((1+x^2 )(−2x+1))/((1+x^2 )^2 ))y(x) +(((1−2x)(1+x^2 ))/((1+x^2 )^2 ))y(x) =0 ⇒(1+x^2 )y^(′′) −(2x−1)y^′ =0 so there is a error at?the question!
$${y}\left({x}\right)={e}^{{arctan}\left({x}\right)} \:\Rightarrow{y}^{'} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:{y}\left({x}\right)\:\Rightarrow \\ $$$${y}^{''} \left({x}\right)\:=−\frac{\mathrm{2}{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{y}\left({x}\right)\:+\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{y}^{'} \left({x}\right) \\ $$$$=\frac{−\mathrm{2}{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{y}\left({x}\right)\:+\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{y}\left({x}\right)\:=\frac{−\mathrm{2}{x}+\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{y}\left({x}\right)\:\Rightarrow \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{''} \:−\left(\mathrm{2}{x}−\mathrm{1}\right){y}^{'} \\ $$$$=\frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(−\mathrm{2}{x}+\mathrm{1}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{y}\left({x}\right)−\left(\mathrm{2}{x}−\mathrm{1}\right)\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }{y}\left({x}\right) \\ $$$$=\frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(−\mathrm{2}{x}+\mathrm{1}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{y}\left({x}\right)\:+\frac{\left(\mathrm{1}−\mathrm{2}{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{y}\left({x}\right) \\ $$$$=\mathrm{0}\:\:\Rightarrow\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{''} −\left(\mathrm{2}{x}−\mathrm{1}\right){y}^{'} =\mathrm{0}\:{so}\:{there}\:{is}\:{a}\:{error}\:{at}?{the}\: \\ $$$${question}! \\ $$
Commented by Abdo msup. last updated on 24/Nov/18
let I =∫_0 ^(1/2) arcosx dx by parts I = [x arvosx]_0 ^(1/2) −∫_0 ^(1/2) ((−x)/( (√(1−x^2 ))))dx =(π/6) + ∫_0 ^(1/2) ((xdx)/( (√(1−x^2 )))) chsngement x=sinθgive ∫_0 ^(1/2) ((xdx)/( (√(1−x^2 )))) =∫_0 ^(π/6) ((sinθ)/(cosθ)) cosθ dθ =∫_0 ^(π/6) sinθdθ =[−cosθ]_0 ^(π/6) =1−((√3)/2) ⇒ I =(π/6) +1−((√3)/2)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:{arcosx}\:{dx}\:\:{by}\:{parts} \\ $$$${I}\:=\:\left[{x}\:{arvosx}\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:−\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{−{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$$=\frac{\pi}{\mathrm{6}}\:+\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\frac{{xdx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\:{chsngement}\:{x}={sin}\theta{give} \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\:\frac{{xdx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:\:\:\frac{{sin}\theta}{{cos}\theta}\:{cos}\theta\:{d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:{sin}\theta{d}\theta \\ $$$$=\left[−{cos}\theta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \:=\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow\:{I}\:=\frac{\pi}{\mathrm{6}}\:+\mathrm{1}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$

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