By using a suitable substitution, bring the equation…

By using a suitable substitution, bring the equation $y'=\sin(x-y)$ to a form with separated variables, and then integrate it.

Explanation of Method

Substitute $v=x-y$ to remove the dependence on $x$ and $y$ separately, then separate variables using the identity $\dfrac{1}{1-\sin v}=\sec^2 v+\sin v\sec^2 v$.

Step 1: Substitute $v=x-y$

With $v=x-y$, $y'=1-v'$, so $1-v'=\sin v$, i.e. $v'=1-\sin v$.

Step 2: Separate and integrate

\[\frac{dv}{1-\sin v}=dx.\]

Multiplying numerator and denominator by $1+\sin v$ gives $\dfrac{1+\sin v}{\cos^2v}=\sec^2v+\sin v\sec^2v$, whose antiderivative is $\tan v+\sec v$. Hence

\[\tan v+\sec v=x+C.\]

Step 3: Back-substitute

Replacing $v=x-y$ gives the implicit solution

\[\boxed{\,\tan(x-y)+\sec(x-y)=x+C.}\]

Result

Final answer:

\[\boxed{\,\tan(x-y)+\sec(x-y)=x+C,\qquad \cos(x-y)\neq0.}\]