The solution of the equation dydx=sec⁡(x+y) is given by

Put x+y=u⇒1+dydx=dudx. The given equation reduces to dudx−1=sec⁡u⇒du1+sec⁡u=dx⇒1−11+cos⁡udu=dx⇒u−tan⁡u/2=x+C⇒x+y−tan⁡x+y2=x+C⇒y−tan⁡x+y2=C.

The solution of the equation dydx=sec⁡(x+y) is given by

Put x+y=u⇒1+dydx=dudx. The given equation reduces to dudx−1=sec⁡u⇒du1+sec⁡u=dx⇒1−11+cos⁡udu=dx⇒u−tan⁡u/2=x+C⇒x+y−tan⁡x+y2=x+C⇒y−tan⁡x+y2=C.

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The solution of the equation $\frac{d y}{d x} = \sec (x + y)$ is
given by

see full answer

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$y - \tan \frac{1}{2} (x + y) = C$

$y - \tan (x + y) = C$

$y - \frac{1}{2} \tan (x + y) = C$

$y + \frac{1}{2} \tan (x + y) = C$

answer is C.

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Detailed Solution

Put $x + y = u \Rightarrow 1 + \frac{d y}{d x} = \frac{d u}{d x}$ . The given equation

reduces to

$\begin{array}{l} \frac{d u}{d x} - 1 = \sec u \\ \Rightarrow \frac{d u}{1 + \sec u} = d x \\ \Rightarrow (1 - \frac{1}{1 + \cos u}) d u = d x \\ \Rightarrow u - \tan u / 2 = x + C \\ \Rightarrow x + y - \tan \frac{x + y}{2} = x + C \\ \Rightarrow y - \tan \frac{x + y}{2} = C . \end{array}$