For the unique solution y(x) to the initial-value problem {y' + y = |x|, x (-3, 3) y(0) = 0 Verify by direct calculation that your solution y(x) indeed satisfies the differential equation. According to your expression, what's y(1), y(-2), y'(1), y'(0)? Is y' continuous at 0? Prove your answer. Is y twice differentiable at 0? Prove your answer. Can you express y(x) in closed (i.e. integral-free) form?