Question Number 59926 by rahul 19 last updated on 16/May/19
Commented by rahul 19 last updated on 16/May/19
$$\mathrm{5},\mathrm{6}. \\ $$
Answered by tanmay last updated on 16/May/19
$${cosxdy}={ysinxdx}−{y}^{\mathrm{2}} {dx} \\ $$$$−{ysinxdx}+{cosxdy}=−{y}^{\mathrm{2}} {dx} \\ $$$$\frac{−{ysinxdx}+{cosxdy}}{\mathrm{1}}=−{y}^{\mathrm{2}} {dx} \\ $$$${d}\left({cosx}×{y}\right)=−{y}^{\mathrm{2}} {dx} \\ $$$$\frac{{d}\left({ycosx}\right)}{{y}^{\mathrm{2}} {cos}^{\mathrm{2}} {x}}=−\frac{{dx}}{{cos}^{\mathrm{2}} {x}} \\ $$$$\frac{−\mathrm{1}}{\left({ycosx}\right)}=−{tanx}−{c} \\ $$$${secx}={y}\left({tanx}+{c}\right) \\ $$$$ \\ $$$$ \\ $$
Answered by tanmay last updated on 16/May/19
$$\frac{{dy}}{{dx}}=\frac{{xy}^{\mathrm{5}} +\mathrm{2}{y}}{{x}} \\ $$$$\frac{{dy}}{{dx}}={y}^{\mathrm{5}} +\frac{\mathrm{2}{y}}{{x}} \\ $$$$\frac{{dy}}{{dx}}−\frac{\mathrm{2}{y}}{{x}}={y}^{\mathrm{5}} \\ $$$${y}^{−\mathrm{5}} \frac{{dy}}{{dx}}−\frac{\mathrm{2}{y}^{−\mathrm{4}} }{{x}}=\mathrm{1} \\ $$$${t}={y}^{−\mathrm{5}+\mathrm{1}} \\ $$$$\frac{{dt}}{{dx}}=−\mathrm{4}×{y}^{−\mathrm{5}} ×\frac{{dy}}{{dx}} \\ $$$$\frac{−\mathrm{1}}{\mathrm{4}}×\frac{{dt}}{{dx}}−\frac{\mathrm{2}{t}}{{x}}=\mathrm{1} \\ $$$$\frac{{dt}}{{dx}}+\frac{\mathrm{8}{t}}{{x}}=−\mathrm{4} \\ $$$${Intregating}\:{factor}\:{e}^{\int\frac{\mathrm{8}}{{x}}{dx}} ={e}^{\mathrm{8}{lnx}} ={x}^{\mathrm{8}} \\ $$$${x}^{\mathrm{8}} \frac{{dt}}{{dx}}+\mathrm{8}{x}^{\mathrm{7}} {t}=−\mathrm{4}{x}^{\mathrm{8}} \\ $$$$\frac{{d}}{{dx}}\left({x}^{\mathrm{8}} ×{t}\right)=−\mathrm{4}{x}^{\mathrm{8}} \\ $$$${d}\left({x}^{\mathrm{8}} {t}\right)=−\mathrm{4}{x}^{\mathrm{8}} {dx} \\ $$$$\int{d}\left({x}^{\mathrm{8}} {t}\right)=−\mathrm{4}\int{x}^{\mathrm{8}} {dx} \\ $$$${x}^{\mathrm{8}} {t}=\frac{−\mathrm{4}}{\mathrm{9}}×{x}^{\mathrm{9}} +{c} \\ $$$${x}^{\mathrm{8}} ×{y}^{−\mathrm{4}} =\frac{−\mathrm{4}}{\mathrm{9}}×{x}^{\mathrm{9}} +{c} \\ $$$$\mathrm{9}{x}^{\mathrm{8}} =−\mathrm{4}{x}^{\mathrm{9}} {y}^{\mathrm{4}} +{y}^{\mathrm{4}} ×\mathrm{9}{c} \\ $$$$\mathrm{9}{x}^{\mathrm{8}} +\mathrm{4}{x}^{\mathrm{9}} {y}^{\mathrm{4}} ={y}^{\mathrm{4}} ×\mathrm{9}{c} \\ $$
Commented by rahul 19 last updated on 16/May/19
$${thank}\:{you}\:{sir}! \\ $$
Commented by tanmay last updated on 16/May/19
$${most}\:{welcome}…{now}\:{i}\:{think}\:{you}\:{are}\:{counting} \\ $$$${days}\:{to}\:{wrestle}\:{with}\:{the}\:{math}+{physics}\:{in} \\ $$$${coming}\:{exam}… \\ $$