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Question Number 82109 by jagoll last updated on 18/Feb/20

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

Commented by mr W last updated on 18/Feb/20

$$u=x+y$$$$y=u-x$$$$\frac{dy}{dx}=\frac{du}{dx}-1$$$$\Rightarrow \frac{du}{dx}-1=\sin u$$$$\Rightarrow \frac{du}{1+\sin u}=dx$$$$\Rightarrow \int \frac{du}{1+\sin u}=\int dx$$$$\Rightarrow -\frac{2}{1+\tan \frac{u}{2}}=x+C$$$$\Rightarrow \tan \frac{u}{2}=-(\frac{2}{x+C}+1)$$$$\Rightarrow \tan \frac{x+y}{2}=-(\frac{2}{x+C}+1)$$$$\Rightarrow y+x+2{\tan }^{-1}(\frac{2}{x+C}+1)=0$$

Commented by jagoll last updated on 18/Feb/20

$$thankyoumister$$

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