Question Number 159428 by mkam last updated on 16/Nov/21
$$\frac{{dy}}{{dx}}={cos}\left({x}+{y}\right)+{sin}\left({x}+{y}\right) \\ $$
Answered by mr W last updated on 16/Nov/21
$${let}\:{u}={x}+{y} \\ $$$$\frac{{du}}{{dx}}=\mathrm{1}+\frac{{dy}}{{dx}} \\ $$$$\frac{{du}}{{dx}}−\mathrm{1}=\mathrm{cos}\:{u}+\mathrm{sin}\:{u}=\sqrt{\mathrm{2}}\mathrm{sin}\:\left({u}+\frac{\pi}{\mathrm{4}}\right) \\ $$$$\frac{{du}}{\:\sqrt{\mathrm{2}}\mathrm{sin}\:\left({u}+\frac{\pi}{\mathrm{4}}\right)+\mathrm{1}}={dx} \\ $$$$\int\frac{{du}}{\:\sqrt{\mathrm{2}}\mathrm{sin}\:\left({u}+\frac{\pi}{\mathrm{4}}\right)+\mathrm{1}}=\int{dx} \\ $$$$\int\frac{{d}\left({u}+\frac{\pi}{\mathrm{4}}\right)}{\:\sqrt{\mathrm{2}}\mathrm{sin}\:\left({u}+\frac{\pi}{\mathrm{4}}\right)+\mathrm{1}}=\int{dx} \\ $$$$\mathrm{ln}\:\frac{\mathrm{tan}\:\left(\frac{{u}}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}\right)+\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{tan}\:\left(\frac{{u}}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}\right)+\sqrt{\mathrm{2}}+\mathrm{1}}={x}+{C} \\ $$$$\frac{\mathrm{tan}\:\left(\frac{{u}}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}\right)+\sqrt{\mathrm{2}}−\mathrm{1}}{\mathrm{tan}\:\left(\frac{{u}}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}\right)+\sqrt{\mathrm{2}}+\mathrm{1}}={ke}^{{x}} \\ $$$$\mathrm{1}−\frac{\mathrm{2}}{\mathrm{tan}\:\left(\frac{{u}}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}\right)+\sqrt{\mathrm{2}}+\mathrm{1}}={ke}^{{x}} \\ $$$$\mathrm{tan}\:\left(\frac{{u}}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}\right)=\frac{\mathrm{2}}{\mathrm{1}−{ke}^{{x}} }−\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\frac{{u}}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{1}−{ke}^{{x}} }−\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$${u}=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{1}−{ke}^{{x}} }−\sqrt{\mathrm{2}}−\mathrm{1}\right)−\frac{\pi}{\mathrm{4}} \\ $$$${y}+{x}=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{1}−{ke}^{{x}} }−\sqrt{\mathrm{2}}−\mathrm{1}\right)−\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow{y}=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{\mathrm{1}−{ke}^{{x}} }−\sqrt{\mathrm{2}}−\mathrm{1}\right)−\frac{\pi}{\mathrm{4}}−{x} \\ $$
Commented by mr W last updated on 17/Nov/21
Commented by mr W last updated on 17/Nov/21
