need help with differential equation please

If you integrate repeatedly you get

y''' = cx + c1
y'' = c/2 x^2 + c1 x + c2 but you know that c2 = 0 because y''(0) = 0
y' = c/6 x^3 + c1/2 x^2 + c3
y = c/24 x^4 + c1/6 x^3 + c3 x + c4 but you know that c4 = 0 because y(0)=0

so y = c/24 x^4 + c1/6 x^3 + c3 x

Since y''(a)=0 , c/2 a^2 + c1 a = 0
That means that either a=0 and y(a/2)=y(0)=0 or a=-2 c1/c so you could write c1 = -ac/2

Substituting: y = c/24 x^4 -ac/12 x^3 + c3 x
but y(a) = 0. Plug in and solve for c3 in terms of a and c

Now plug back in for c3 and you have an equation in terms of a and c. Solve for a/2.