1. **State the problem:** We need to solve the integral equation $$y'(x) = x^2 + \int_0^x y(t) \cos(x - t) \, dt$$ with the initial condition $$y(0) = 4$$. 2. **Recognize the type of equation:** This is a Volterra integral equation of the second kind with a convolution kernel $$\cos(x-t)$$. 3. **Use differentiation to simplify:** Differentiate both sides with respect to $$x$$: $$y''(x) = 2x + \frac{d}{dx} \int_0^x y(t) \cos(x - t) \, dt$$ 4. **Apply Leibniz rule to the integral:** $$\frac{d}{dx} \int_0^x y(t) \cos(x - t) \, dt = y(x) \cos(0) + \int_0^x y(t) \frac{\partial}{\partial x} \cos(x - t) \, dt = y(x) - \int_0^x y(t) \sin(x - t) \, dt$$ 5. **Substitute back:** $$y''(x) = 2x + y(x) - \int_0^x y(t) \sin(x - t) \, dt$$ 6. **Differentiate again to eliminate integral:** Differentiate $$y''(x)$$: $$y'''(x) = 2 + y'(x) - \frac{d}{dx} \int_0^x y(t) \sin(x - t) \, dt$$ 7. **Differentiate the last integral:** $$\frac{d}{dx} \int_0^x y(t) \sin(x - t) \, dt = y(x) \sin(0) + \int_0^x y(t) \frac{\partial}{\partial x} \sin(x - t) \, dt = 0 + \int_0^x y(t) \cos(x - t) \, dt$$ 8. **Recall original integral:** $$\int_0^x y(t) \cos(x - t) \, dt = y'(x) - x^2$$ from the original equation. 9. **Substitute:** $$y'''(x) = 2 + y'(x) - (y'(x) - x^2) = 2 + y'(x) - y'(x) + x^2 = 2 + x^2$$ 10. **Simplify:** $$y'''(x) = x^2 + 2$$ 11. **Integrate to find $$y(x)$$:** Integrate thrice: $$y''(x) = \int (x^2 + 2) \, dx = \frac{x^3}{3} + 2x + C_1$$ $$y'(x) = \int y''(x) \, dx = \frac{x^4}{12} + x^2 + C_1 x + C_2$$ $$y(x) = \int y'(x) \, dx = \frac{x^5}{60} + \frac{x^3}{3} + \frac{C_1 x^2}{2} + C_2 x + C_3$$ 12. **Apply initial condition $$y(0) = 4$$:** $$y(0) = C_3 = 4$$ 13. **Use original equation at $$x=0$$ to find $$C_2$$:** At $$x=0$$, $$y'(0) = 0^2 + \int_0^0 y(t) \cos(0 - t) dt = 0$$ From $$y'(x)$$ formula: $$y'(0) = C_2 = 0$$ 14. **Use $$y''(0)$$ to find $$C_1$$:** From step 11: $$y''(0) = C_1$$ From step 3 original equation differentiated: $$y''(x) = 2x + y(x) - \int_0^x y(t) \sin(x - t) dt$$ At $$x=0$$: $$y''(0) = 0 + y(0) - 0 = 4$$ So, $$C_1 = 4$$ 15. **Final solution:** $$y(x) = \frac{x^5}{60} + \frac{x^3}{3} + 2 x^2 + 4$$