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Question Number 95068 by Mr.D.N. last updated on 23/May/20

$$\mathrm{Solve}\mathrm{the}\mathrm{differential}\mathrm{equations}-:$$$$1.\frac{\mathrm{dy}}{\mathrm{dx}}=\sin (\mathrm{x}+\mathrm{y})+\cos (\mathrm{x}+\mathrm{y})$$$$$$

Commented by EmericGent last updated on 22/May/20

Would be easy with cos(x-y) instead of cos(x+y)

Commented by EmericGent last updated on 23/May/20

Do we have to use Sturm Liouville in order to solve the second? (I'm sorry I'm learning maths it's hard for me)

Commented by bobhans last updated on 23/May/20

$$\frac{\mathrm{dy}}{\mathrm{dx}}=\sin (\mathrm{x}+\mathrm{y})+\sin (\frac{\pi }{2}-(\mathrm{x}+\mathrm{y}))$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=2\sin \left(\frac{\pi }{4}\right)\cos (\mathrm{x}+\mathrm{y}-\frac{\pi }{4})$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{2}\cos (\mathrm{x}+\mathrm{y}-\frac{\pi }{4})$$$$\mathrm{set}\mathrm{x}+\mathrm{y}-\frac{\pi }{4}=v\Rightarrow 1+\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{dv}{dx}$$$$\frac{dv}{dx}-1=\sqrt{2}\cos v\Rightarrow \frac{dv}{dx}=1+\sqrt{2}\cos v$$$$\frac{dv}{1+\sqrt{2}\cos v}=dx\Rightarrow \int \frac{dv}{1+\sqrt{2}\cos v}=x+c$$$$$$

Answered by mr W last updated on 23/May/20

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

Commented by peter frank last updated on 22/May/20

$$\mathrm{thank}\mathrm{you}$$

Commented by Mr.D.N. last updated on 23/May/20

Thank you mr.w I appreciate your grand hard work but can you present your answer in form of : log{1+ tan (x+y)/2} =x+C.

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