#(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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