1. **Stating the problem:** We want to analyze the differential equation $$y' = \sin(y - x)$$ using isoclines and draw approximate integral curves. 2. **Understanding isoclines:** Isoclines are curves in the $xy$-plane where the slope $y'$ is constant. For a given slope $m$, the isocline satisfies: $$m = \sin(y - x)$$ 3. **Finding isoclines:** To find isoclines for various slopes $m$, solve for $y - x$: $$y - x = \arcsin(m) + 2k\pi \quad \text{or} \quad y - x = \pi - \arcsin(m) + 2k\pi, \quad k \in \mathbb{Z}$$ These represent families of straight lines with slope 1 shifted vertically. 4. **Plotting isoclines:** For example, for $m=0$, $y - x = 0 + 2k\pi$, so lines $y = x + 2k\pi$. For $m=1$, $y - x = \frac{\pi}{2} + 2k\pi$, lines $y = x + \frac{\pi}{2} + 2k\pi$. For $m=-1$, $y - x = -\frac{\pi}{2} + 2k\pi$, lines $y = x - \frac{\pi}{2} + 2k\pi$. 5. **Interpreting slopes on isoclines:** On each isocline, the slope $y'$ is constant and equal to $m$. This helps sketch the direction field. 6. **Drawing integral curves:** Integral curves follow the direction field. They cross isoclines at points where the slope matches the isocline's $m$. Since $y' = \sin(y - x)$ is periodic in $y - x$, integral curves will be wavy and roughly parallel to lines $y - x = \text{constant}$. 7. **Summary:** - Isoclines are lines $y - x = c$ where $c$ is chosen so that $\sin(c) = m$. - On each isocline, slope $y' = m$ is constant. - Integral curves can be sketched by following these slopes, showing oscillatory behavior. **Final answer:** The isoclines are lines of the form $$y = x + c$$ where $$c = \arcsin(m) + 2k\pi$$ or $$c = \pi - \arcsin(m) + 2k\pi$$ for integer $k$, and integral curves oscillate around these lines following the slope field defined by $y' = \sin(y - x)$.