Question Number 82109 by jagoll last updated on 18/Feb/20
$${what}\:{is}\:{solution} \\ $$$$\frac{{dy}}{{dx}}\:=\:\mathrm{sin}\:\left({x}+{y}\right) \\ $$
Commented by mr W last updated on 18/Feb/20
$${u}={x}+{y} \\ $$$${y}={u}−{x} \\ $$$$\frac{{dy}}{{dx}}=\frac{{du}}{{dx}}−\mathrm{1} \\ $$$$\Rightarrow\frac{{du}}{{dx}}−\mathrm{1}=\mathrm{sin}\:{u} \\ $$$$\Rightarrow\frac{{du}}{\mathrm{1}+\mathrm{sin}\:{u}}={dx} \\ $$$$\Rightarrow\int\frac{{du}}{\mathrm{1}+\mathrm{sin}\:{u}}=\int{dx} \\ $$$$\Rightarrow−\frac{\mathrm{2}}{\mathrm{1}+\mathrm{tan}\:\frac{{u}}{\mathrm{2}}}={x}+{C} \\ $$$$\Rightarrow\mathrm{tan}\:\frac{{u}}{\mathrm{2}}=−\left(\frac{\mathrm{2}}{{x}+{C}}+\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{tan}\:\frac{{x}+{y}}{\mathrm{2}}=−\left(\frac{\mathrm{2}}{{x}+{C}}+\mathrm{1}\right) \\ $$$$\Rightarrow{y}+{x}+\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{{x}+{C}}+\mathrm{1}\right)=\mathrm{0} \\ $$
Commented by jagoll last updated on 18/Feb/20
$${thank}\:{you}\:{mister} \\ $$