1. **State the problem:** We need to solve the differential equation $$y' = (1 + y^2) + \sin x$$ where $y'$ denotes the derivative of $y$ with respect to $x$. 2. **Rewrite the equation:** The equation is $$\frac{dy}{dx} = 1 + y^2 + \sin x$$. 3. **Check if separable:** The right side is $1 + y^2 + \sin x$, which is not separable into a product of a function of $x$ and a function of $y$. 4. **Try substitution or rearrangement:** This is a nonlinear first-order ODE. Notice that $y' - y^2 = 1 + \sin x$. 5. **Recognize the Riccati form:** The equation is of the form $$y' = y^2 + f(x)$$ with $f(x) = 1 + \sin x$. 6. **Solve Riccati equation:** Riccati equations can be transformed if a particular solution is known or by substitution $y = -\frac{u'}{u}$. 7. **Use substitution:** Let $$y = -\frac{u'}{u}$$ then $$y' = -\frac{u''u - u'^2}{u^2}$$. 8. **Substitute into original equation:** $$-\frac{u''u - u'^2}{u^2} = 1 + \sin x + y^2 = 1 + \sin x + \left(-\frac{u'}{u}\right)^2 = 1 + \sin x + \frac{u'^2}{u^2}$$ 9. **Multiply both sides by $u^2$:** $$- (u''u - u'^2) = (1 + \sin x) u^2 + u'^2$$ 10. **Simplify:** $$-u''u + u'^2 = (1 + \sin x) u^2 + u'^2$$ 11. **Cancel $u'^2$ on both sides:** $$-u''u = (1 + \sin x) u^2$$ 12. **Divide both sides by $u$ (assuming $u \neq 0$):** $$-u'' = (1 + \sin x) u$$ 13. **Rewrite:** $$u'' + (1 + \sin x) u = 0$$ 14. **Interpretation:** This is a second-order linear differential equation for $u(x)$. 15. **Summary:** To solve the original nonlinear ODE, solve the linear ODE $$u'' + (1 + \sin x) u = 0$$ and then find $$y = -\frac{u'}{u}$$. 16. **Final answer:** The general solution to the original equation is $$y = -\frac{u'}{u}$$ where $u$ satisfies $$u'' + (1 + \sin x) u = 0$$. This second-order ODE may require numerical or special function methods to solve explicitly.