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Question Number 85706 by naka3546 last updated on 24/Mar/20

Find  the  general  solution  of        x^2  (√(y^2 +9))  dx + 5 (√(x^2 −3))  y dy  =  0

$${Find}\:\:{the}\:\:{general}\:\:{solution}\:\:{of} \\ $$$$\:\:\:\:\:\:{x}^{\mathrm{2}} \:\sqrt{{y}^{\mathrm{2}} +\mathrm{9}}\:\:{dx}\:+\:\mathrm{5}\:\sqrt{{x}^{\mathrm{2}} −\mathrm{3}}\:\:{y}\:{dy}\:\:=\:\:\mathrm{0} \\ $$

Answered by mind is power last updated on 24/Mar/20

⇔((ydy)/(√(y^2 +9)))=−((5(√(x^2 −3)))/x^2 )  ⇒(√(y^2 +9))=5∫−((√(x^2 −3))/x^2 )dx=((√(x^2 −3))/x)−∫(dx/(√(x^2 −3)))  =((√(x^2 −3))/x)−argsh((x/(√3)))+c  ⇒y=((((√(x^2 −3))/x)−argsh((x/(√3)))+c)^2 −9)^(1/2)

$$\Leftrightarrow\frac{{ydy}}{\sqrt{{y}^{\mathrm{2}} +\mathrm{9}}}=−\frac{\mathrm{5}\sqrt{{x}^{\mathrm{2}} −\mathrm{3}}}{{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\sqrt{{y}^{\mathrm{2}} +\mathrm{9}}=\mathrm{5}\int−\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{3}}}{{x}^{\mathrm{2}} }{dx}=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{3}}}{{x}}−\int\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} −\mathrm{3}}} \\ $$$$=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{3}}}{{x}}−{argsh}\left(\frac{{x}}{\sqrt{\mathrm{3}}}\right)+{c} \\ $$$$\Rightarrow{y}=\left(\left(\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{3}}}{{x}}−{argsh}\left(\frac{{x}}{\sqrt{\mathrm{3}}}\right)+{c}\right)^{\mathrm{2}} −\mathrm{9}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$

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