求y"=x+sinx的通解
∵y"=x+sinx
==>y'=∫(x+sinx)dx=x^2/2-cosx+C1 (C1是常数)
==>y=∫(x^2/2-cosx+C1)dx=x^3/6-sinx+C1x+C2 (C2是常数)
∴原方程的通解是y=x^3/6-sinx+C1x+C2.