differential equation

http://en.wikipedia.org/wiki/Linear_differential_equation#Homogeneous_equations_with_constant_coefficients

First, solve the differential equation...

You guess that the solution has form y(x) = e^(zx) for some complex number z. Plug that into the differential equation, divide by e^(zx), and solve the resulting quadratic equation for z.

There should be two answers for z: "a" and "b".

Overall your answer will be y(x) = Ae^(ax) + Be^(bx) for some A and B.

Second, solve the initial value problem...

Use the fact that y(0) = y'(0) = 0 to figure out what A and B have to be.

can you please solve

I tried, after you posted the second time.

The method I told you to use doesn't work well for this particular problem. I will hopefully be able to give you a better answer in a day or two. Maybe. I have an idea what to do.

Don't quote me on this but it may be as simple as y(x) = 0.

My reasoning is that your answer should be a unique function, and y(x)=0 works.