$(dx)/(dy)=cos(x-y)$ ?

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Answer 1 · 2017-05-02T00:47:53

$cot((x-y)/2)=y + C_2$

We have

$(dx)/(dy)=cos(x-y)$ and

$d/(dy)(x-y)=cos(x-y)-1 = cos(x-y)+cospi$

now making $z = x-y$

$dz/dy = cosz+cospi = 2cos((z+pi)/2)cos((z-pi)/2)=-2sin^2(z/2)$

so we have a separable differential equation.

$dz/dy = - 2 sin^2(z/2)$

Solving

$(dz)/sin^2(z/2)=-2dy->-2cot(z/2)=-2y + C_1$

and finally

$cot((x-y)/2)=y + C_2$

This is the differential equation solution in implicit form.

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(dx)/(dy)=cos(x-y)(dx)/(dy)=cos(x-y) ?