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Question Number 159428 by mkam last updated on 16/Nov/21

$$\frac{dy}{dx}=cos(x+y)+sin(x+y)$$

Answered by mr W last updated on 16/Nov/21

$$letu=x+y$$$$\frac{du}{dx}=1+\frac{dy}{dx}$$$$\frac{du}{dx}-1=\cos u+\sin u=\sqrt{2}\sin (u+\frac{\pi }{4})$$$$\frac{du}{\sqrt{2}\sin (u+\frac{\pi }{4})+1}=dx$$$$\int \frac{du}{\sqrt{2}\sin (u+\frac{\pi }{4})+1}=\int dx$$$$\int \frac{d(u+\frac{\pi }{4})}{\sqrt{2}\sin (u+\frac{\pi }{4})+1}=\int dx$$$$\ln \frac{\tan (\frac{u}{2}+\frac{\pi }{8})+\sqrt{2}-1}{\tan (\frac{u}{2}+\frac{\pi }{8})+\sqrt{2}+1}=x+C$$$$\frac{\tan (\frac{u}{2}+\frac{\pi }{8})+\sqrt{2}-1}{\tan (\frac{u}{2}+\frac{\pi }{8})+\sqrt{2}+1}={ke}^x$$$$1-\frac{2}{\tan (\frac{u}{2}+\frac{\pi }{8})+\sqrt{2}+1}={ke}^x$$$$\tan (\frac{u}{2}+\frac{\pi }{8})=\frac{2}{1-{ke}^x}-\sqrt{2}-1$$$$\frac{u}{2}+\frac{\pi }{8}={\tan }^{-1}(\frac{2}{1-{ke}^x}-\sqrt{2}-1)$$$$u=2{\tan }^{-1}(\frac{2}{1-{ke}^x}-\sqrt{2}-1)-\frac{\pi }{4}$$$$y+x=2{\tan }^{-1}(\frac{2}{1-{ke}^x}-\sqrt{2}-1)-\frac{\pi }{4}$$$$\Rightarrow y=2{\tan }^{-1}(\frac{2}{1-{ke}^x}-\sqrt{2}-1)-\frac{\pi }{4}-x$$

Commented by mr W last updated on 17/Nov/21

Commented by mr W last updated on 17/Nov/21

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