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Question Number 155983 by MathsFan last updated on 06/Oct/21

given y=(√((1−x)/(1+x))) , then (1−x^2 )(dy/dx)+ky=0. find k

$$\:{given}\:\:{y}=\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\:,\:{then} \\ $$$$\:\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\frac{{dy}}{{dx}}+\boldsymbol{{k}}{y}=\mathrm{0}.\:{find}\:\boldsymbol{{k}} \\ $$

Commented by cortano last updated on 06/Oct/21

k=1

$$\mathrm{k}=\mathrm{1} \\ $$

Commented by yeti123 last updated on 07/Oct/21

(1−x^2 )(dy/dx) + ky = 0 (1−x^2 )(d/dx)(√(((1−x)/(1+x)) )) + ky = 0 (1−x^2 )×((−1)/( (1+x^2 )(√((1−x)/(1+x))))) + ky = 0 (1−x^2 )×((−1)/((1+x^2 )y)) + ky = 0 −(1−x^2 ) + ky^2 (1 + x^2 ) = 0 k = ((1−x^2 )/(y^2 (1+x^2 )))

$$\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\frac{{dy}}{{dx}}\:+\:{ky}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\frac{{d}}{{dx}}\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:}\:+\:{ky}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{1}−{x}^{\mathrm{2}} \right)×\frac{−\mathrm{1}}{\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}}\:+\:{ky}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{1}−{x}^{\mathrm{2}} \right)×\frac{−\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}}\:+\:{ky}\:=\:\mathrm{0} \\ $$$$−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\:+\:{ky}^{\mathrm{2}} \left(\mathrm{1}\:+\:{x}^{\mathrm{2}} \right)\:=\:\mathrm{0} \\ $$$${k}\:=\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\: \\ $$

Answered by puissant last updated on 06/Oct/21

(1−x^2 )(dy/dx)+ky=0 ⇒ (1−x^2 )dy=−kydx ⇒ (dy/(ky))=−(dx/(1−x^2 )) ⇒ (1/k)∫(dy/y) = ∫(dx/(x^2 −1)) ⇒ (1/k)lny=−(1/2)ln(((1+x)/(1−x))) y=(√(((1−x)/(1+x)) ))⇒ (1/k)lny=(1/(2k))ln(((1−x)/(1+x))) ⇒ (1/(2k))ln(((1−x)/(1+x)))=−(1/2)ln(((1+x)/(1−x)))=(1/2)ln(((1−x)/(1+x))) ⇒ (1/(2k))=(1/2) ⇒ ∴∵ k = 1...■

$$\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\frac{{dy}}{{dx}}+{ky}=\mathrm{0} \\ $$$$\Rightarrow\:\left(\mathrm{1}−{x}^{\mathrm{2}} \right){dy}=−{kydx}\:\Rightarrow\:\frac{{dy}}{{ky}}=−\frac{{dx}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{{k}}\int\frac{{dy}}{{y}}\:=\:\int\frac{{dx}}{{x}^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow\:\frac{\mathrm{1}}{{k}}{lny}=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right) \\ $$$${y}=\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:}\Rightarrow\:\frac{\mathrm{1}}{{k}}{lny}=\frac{\mathrm{1}}{\mathrm{2}{k}}{ln}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right) \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}{k}}{ln}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right)=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right) \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}{k}}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\Rightarrow\:\:\therefore\because\:\:{k}\:=\:\mathrm{1}...\blacksquare \\ $$

Commented by MathsFan last updated on 07/Oct/21

merci

$${merci} \\ $$

Commented by puissant last updated on 07/Oct/21

D′accord j′espere que tous est claire..

$${D}'{accord}\:{j}'{espere}\:{que}\:{tous}\:{est}\:{claire}.. \\ $$

Commented by MathsFan last updated on 07/Oct/21

oui monsieur

$${oui}\:{monsieur} \\ $$

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