What is the solution of the differential equation \(\dfrac{dy}{dx} = 1 + x\cot(y-x)\)?
\(\sec(y-x) = ce^{x^{2}/2}\)
Let \(v=y-x\), so \(\frac{dv}{dx}=\frac{dy}{dx}-1 = x\cot v\). This separates as \(\tan v\,dv = x\,dx\). Integrating, \(-\ln|\cos v| = \frac{x^2}{2}+C\), which gives \(\sec v = ce^{x^2/2}\). Substituting back \(v=y-x\) gives \(\sec(y-x)=ce^{x^2/2}\).
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