Question Number 7258 by Tawakalitu. last updated on 19/Aug/16
Commented by Yozzia last updated on 19/Aug/16
$${y}'=\frac{{x}}{\sqrt{\mathrm{5}−{x}^{\mathrm{2}} }}\Rightarrow{y}=\int\frac{{x}}{\sqrt{\mathrm{5}−{x}^{\mathrm{2}} }}{dx}. \\ $$$${Let}\:{u}=\mathrm{5}−{x}^{\mathrm{2}} \Rightarrow{du}=−\mathrm{2}{xdx}\Rightarrow{xdx}=\frac{−\mathrm{1}}{\mathrm{2}}{du}. \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{−{du}}{\sqrt{{u}}}=\frac{−\mathrm{1}}{\mathrm{2}}\int{u}^{−\mathrm{1}/\mathrm{2}} {du}=\frac{−\mathrm{1}}{\mathrm{2}}×\mathrm{2}{u}^{\mathrm{1}/\mathrm{2}} +{c} \\ $$$${y}={c}−\sqrt{\mathrm{5}−{x}^{\mathrm{2}} } \\ $$$${The}\:{point}\:\left(\mathrm{2},\mathrm{6}\right)\:{lies}\:{on}\:{the}\:{curve}. \\ $$$$\therefore\:\mathrm{6}={c}−\sqrt{\mathrm{5}−\mathrm{4}}\Rightarrow{c}=\mathrm{7} \\ $$$$\therefore\:{y}=\mathrm{7}−\sqrt{\mathrm{5}−{x}^{\mathrm{2}} } \\ $$
Commented by Tawakalitu. last updated on 19/Aug/16
$${Thanks}\:{so}\:{much}\:{sir},\:{i}\:{really}\:{appreciate}. \\ $$