Question Number 222175 by klipto last updated on 19/Jun/25
$$\boldsymbol{\mathrm{solve}} \\ $$$$\left(\boldsymbol{\mathrm{e}}^{\mathrm{2}\boldsymbol{\mathrm{y}}} −\boldsymbol{\mathrm{y}}\right)\boldsymbol{\mathrm{cosx}}\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}=\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{y}}} \boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{x}} \\ $$$$\boldsymbol{\mathrm{klipto}}−\boldsymbol{\mathrm{quanta}} \\ $$
Answered by som(math1967) last updated on 20/Jun/25
$$\:\int\frac{{e}^{\mathrm{2}{y}} −{y}}{{e}^{{y}} }{dy}=\int\frac{{sin}\mathrm{2}{x}}{{cosx}}{dx} \\ $$$$\int{e}^{{y}} {dy}−\int{ye}^{−{y}} {dy}=\int\frac{\mathrm{2}{sinxcosxdx}}{{cosx}} \\ $$$$\Rightarrow{e}^{{y}} −{y}\int{e}^{−{y}} {dy}+\int\left\{\frac{{dy}}{{dy}}\int{e}^{−{y}} {dy}\right\}{dy} \\ $$$$\:\:\:\:=\mathrm{2}\int{sinxdx} \\ $$$$\Rightarrow{e}^{{y}} +{ye}^{−{y}} +{e}^{−{y}} =−\mathrm{2}{cosx}\:+{C} \\ $$$$\Rightarrow{e}^{{y}} +{e}^{−{y}} \left(\mathrm{1}+{y}\right)+\mathrm{2}{cosx}={C} \\ $$