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

Option: 1

-\ln \left|1+\tan \frac{x+y}{2}\right|=x+c


Option: 2

\\ {\ln \left|1+\tan \frac{x+y}{2}\right|=x+c}


Option: 3

\\ {\ln \left|1-\tan \frac{x+y}{2}\right|=x+c}


Option: 4

\\ {-\ln \left|1-\tan \frac{x+y}{2}\right|=x+c}


Answers (1)

best_answer

Putting x+y=t
\text { we get }1+ \frac{d y}{d x}=\frac{d t}{d x} \Rightarrow \frac{d y}{d x}=\frac{d t}{d x}-1
\text { Therefore, } \frac{d t}{d x}-1=\cos t-\sin t

\\\Rightarrow \frac{d t}{1+\cos t-\sin t}=d x \\\Rightarrow \frac{d t}{1+ \frac{1-tan^2(\frac{t}{2})}{1+tan^2(\frac{t}{2})} -\frac{2tan(\frac{t}{2})}{1+tan^2(\frac{t}{2})}}=d x \\\Rightarrow \frac{\sec ^{2} \frac{t}{2} d t}{2\left(1-\tan \frac{t}{2}\right)}=d x

\\ {\Rightarrow-\ln \left|1-\tan \frac{t}{2}\right|=x+c}

\\ {\Rightarrow-\ln \left|1-\tan \frac{x+y}{2}\right|=x+c}

Posted by

seema garhwal

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