Suppose there is a continous function on which is twice countinously differenciable on such that for any we have that passing through , i.e. . Then by the fundamental theorem for it means (differenciate both sides) . Thus, . Thus, . Thus, . This means the twice continously differencial functions which solve the integral equation above are: and . Using the initial condition (and continuity) that we get that in the first class and in the second class. Thus, the two solutions are: for .