Question #2e6ef

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Question #2e6ef

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Answer 1 · 2017-02-18T11:59:57

$y = x - 2 (arccot (\frac{x + 2 + C}{x + C}) + k π)$ $y = x-2("arccot"((x+2+C)/(x+C))+k pi)$ , $k = 0, 1, 2, \dots$ $k = 0,1,2,cdots$

Explanation:

Making $z = x - y$ $z = x-y$ we have $\frac{d z}{d x} = 1 - \frac{d y}{d x}$ $(dz)/(dx)=1-(dy)/(dx)$ so

$\frac{d y}{d x} = sin (x - y) \to \frac{d z}{d x} = 1 - sin (z)$ $(dy)/(dx)=sin(x-y)-> (dz)/(dx)=1-sin(z)$ This differential equation is separable so

$\frac{d z}{1 - sin (z)} = d x$ $(dz)/(1-sin(z))=dx$ integrating

$\frac{2 sin (\frac{z}{2})}{cos (\frac{z}{2}) - sin (\frac{z}{2})} = x + C$ $(2sin(z/2))/(cos(z/2)-sin(z/2))=x + C$ or

$\frac{2}{cot (\frac{z}{2}) - 1} = x + C$ $2/(cot(z/2)-1)=x+C$ or

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

Finally

$y = x - 2 (arccot (\frac{x + 2 + C}{x + C}) + k π)$ $y = x-2("arccot"((x+2+C)/(x+C))+k pi)$

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Question #2e6ef

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