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Question Number 45413 by mondodotto@gmail.com last updated on 12/Oct/18

if y=((sin^(−1) x)/(√(1+x^2 ))) show that (dy/dx)(1+x^2 )+xy=1

$$\boldsymbol{\mathrm{if}}\:\boldsymbol{{y}}=\frac{\boldsymbol{\mathrm{sin}}^{−\mathrm{1}} \boldsymbol{{x}}}{\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }}\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}} \\ $$$$\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \right)+\boldsymbol{{xy}}=\mathrm{1} \\ $$

Commented by maxmathsup by imad last updated on 13/Oct/18

we have y(x)=((arcsinx)/(√(1+x^2 ))) ⇒y(x) =(1+x^2 )^(−(1/2)) acsinx ⇒ y^′ (x) =−x (1+x^2 )^(−(3/2)) arcsinx +(1+x^2 )^(−(1/2)) (1/(√(1−x^2 ))) ⇒ (1+x^2 )y^′ (x) +xy =−x(1+x^2 )^(−(1/2)) arcsinx +((√(1+x^2 ))/(√(1−x^2 ))) + x (1+x^2 )^(−(1/2)) arcsinx =((√(1+x^2 ))/(√(1−x^2 ))) finally thre is a error at the question...

$${we}\:{have}\:{y}\left({x}\right)=\frac{{arcsinx}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:\Rightarrow{y}\left({x}\right)\:=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {acsinx}\:\Rightarrow \\ $$$${y}^{'} \left({x}\right)\:=−{x}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \:{arcsinx}\:+\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\Rightarrow \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{'} \left({x}\right)\:+{xy}\:=−{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{arcsinx}\:+\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\:+\:{x}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{arcsinx} \\ $$$$=\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\:\:\:\:{finally}\:{thre}\:{is}\:{a}\:{error}\:{at}\:{the}\:{question}... \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

(dy/dx)+(x/(1+x^2 ))y=(1/(1+x^2 )) intregaying factor e^(∫(x/(1+x^2 ))dx) e^((1/2)∫((d(1+x^3 ))/(1+x^2 ))) e^((1/2)ln(1+x^2 )) e^(ln((√(1+x^2 )) )) =(√(1+x^2 )) (√(1+x^2 )) (dy/dx)+x.(y/(√(1+x^2 )))=(√(1+x^2 )) (d/dx)((√(1+x^2 )) ×y)=(√(1+x^2 )) d(y×(√(1+x^2 )) )=(√(1+x^2 )) dx ∫d(y×(√(1+x^2 )) )=∫(√(1+x^2 )) dx y×(√(1+x^2 )) =(x/2)(√(1+x^2 )) +(1/2)ln(x+(√(x^2 +a^)^2 ))

$$\frac{{dy}}{{dx}}+\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{y}=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${intregaying}\:{factor} \\ $$$${e}^{\int\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}} \\ $$$${e}^{\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$${e}^{\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} \\ $$$${e}^{{ln}\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)} =\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\: \\ $$$$\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\frac{{dy}}{{dx}}+{x}.\frac{{y}}{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} \:}}=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\: \\ $$$$\frac{{d}}{{dx}}\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:×{y}\right)=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\: \\ $$$${d}\left({y}×\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$$$\int{d}\left({y}×\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)=\int\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$$${y}×\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:=\frac{{x}}{\mathrm{2}}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\left.\right)^{\mathrm{2}} } }\:\right. \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

check the question...

$${check}\:{the}\:{question}... \\ $$

Commented by mondodotto@gmail.com last updated on 12/Oct/18

sir we need to prove

$$\mathrm{sir}\:\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{prove} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

y=((sin^(−1) x)/(√(1−x^2 ))) (dy/dx)=(((√(1−x^2 )) ×(1/((√(1−x^2 )) ))−sin^(−1) x×((−2x)/(2(√(1−x^2 )))))/((1−x^2 ))) (1−x^2 )(dy/dx)=1+xy

$${y}=\frac{{sin}^{−\mathrm{1}} {x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\frac{{dy}}{{dx}}=\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:×\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:}−{sin}^{−\mathrm{1}} {x}×\frac{−\mathrm{2}{x}}{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)} \\ $$$$\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\frac{{dy}}{{dx}}=\mathrm{1}+{xy} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

pls check...

$${pls}\:{check}... \\ $$

Commented by mondodotto@gmail.com last updated on 12/Oct/18

oky sir

$$\mathrm{oky}\:\mathrm{sir} \\ $$

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