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Question Number 156931 by cortano last updated on 17/Oct/21

(1−x^2 )y′′−4xy′−(1+x^2 )y=0

$$\left(\mathrm{1}−{x}^{\mathrm{2}} \right){y}''−\mathrm{4}{xy}'−\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}=\mathrm{0} \\ $$

Commented by john_santu last updated on 17/Oct/21

y=((sin x)/(1−x^2 )) + c

$${y}=\frac{\mathrm{sin}\:{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\:+\:{c} \\ $$

Commented by aliyn last updated on 18/Oct/21

Solution: y^′ = ((4x ± (√(16x^2 +4 + 4 x^2 )))/(2(1−x^2 ))) y^′ = ((2x ± (√(5x^2 + 1)))/(1−x^2 )) dy = ((2x )/(1−x^2 )) ± ((√(5x^2 + 1))/(1−x^2 )) dx y = ln ∣ (1/( (√( 1−x^2 )))) ∣ ± ∫ ((√(5x^2 + 1))/(1−x^2 )) dx now : x = (1/( (√5))) tany → dx = (1/( (√5))) sec^2 y dy complete

$$ \\ $$$$\boldsymbol{{Solution}}: \\ $$$$ \\ $$$$\boldsymbol{{y}}^{'} \:=\:\frac{\mathrm{4}\boldsymbol{{x}}\:\pm\:\sqrt{\mathrm{16}\boldsymbol{{x}}^{\mathrm{2}} \:+\mathrm{4}\:+\:\mathrm{4}\:\boldsymbol{{x}}^{\mathrm{2}} }}{\mathrm{2}\left(\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} \right)} \\ $$$$ \\ $$$$\boldsymbol{{y}}^{'} \:=\:\frac{\mathrm{2}\boldsymbol{{x}}\:\pm\:\sqrt{\mathrm{5}\boldsymbol{{x}}^{\mathrm{2}} \:+\:\mathrm{1}}}{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} } \\ $$$$ \\ $$$$\boldsymbol{{dy}}\:=\:\frac{\mathrm{2}\boldsymbol{{x}}\:}{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} }\:\pm\:\frac{\sqrt{\mathrm{5}\boldsymbol{{x}}^{\mathrm{2}} \:+\:\mathrm{1}}}{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} }\:\boldsymbol{{dx}} \\ $$$$ \\ $$$$\boldsymbol{{y}}\:=\:\:\boldsymbol{{ln}}\:\mid\:\frac{\mathrm{1}}{\:\sqrt{\:\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} }}\:\mid\:\pm\:\int\:\:\frac{\sqrt{\mathrm{5}\boldsymbol{{x}}^{\mathrm{2}} \:+\:\mathrm{1}}}{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} }\:\boldsymbol{{dx}} \\ $$$$ \\ $$$$\boldsymbol{{now}}\::\:\boldsymbol{{x}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\boldsymbol{{tany}}\:\rightarrow\:\boldsymbol{{dx}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{y}}\:\boldsymbol{{dy}} \\ $$$$ \\ $$$$\boldsymbol{{complete}} \\ $$

Answered by mindispower last updated on 17/Oct/21

z=(1−x^2 )y z′=−2xy+(1−x^2 )y′ z′′=−2y−4xy′+(1−x^2 )y′′ z+z′′=−(1+x^2 )y−4xy′+(1−x^2 )y′′ z+z′′=0⇒z=acos(x)+bsin(x) y=((acos(x)+bsin(x))/(1−x^2 ))

$${z}=\left(\mathrm{1}−{x}^{\mathrm{2}} \right){y} \\ $$$${z}'=−\mathrm{2}{xy}+\left(\mathrm{1}−{x}^{\mathrm{2}} \right){y}' \\ $$$${z}''=−\mathrm{2}{y}−\mathrm{4}{xy}'+\left(\mathrm{1}−{x}^{\mathrm{2}} \right){y}'' \\ $$$${z}+{z}''=−\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}−\mathrm{4}{xy}'+\left(\mathrm{1}−{x}^{\mathrm{2}} \right){y}'' \\ $$$${z}+{z}''=\mathrm{0}\Rightarrow{z}={acos}\left({x}\right)+{bsin}\left({x}\right) \\ $$$${y}=\frac{{acos}\left({x}\right)+{bsin}\left({x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$ \\ $$

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