Thanks, Sean Clarke.

I will try other methods/functions to solve my equation.

Thanks again!

I used "texp" as the full result with the integration part, and used this clause: tfexp = FullSimplify[texp, Assumptions -> K[2] = 0],

But , it returns with :

I will check what is the problem.

Thanks!

From the documentation on DSolve:

Solutions given by DSolve sometimes include integrals that cannot be carried out explicitly by Integrate. Variables K[1], K[2], ... are used in such integrals.

Here's a simple example which produces output with an integral:

DSolve[y'[x] == f[x], y, x]

{{y -> Function[{x}, C[1] + Integrate[f[K[1]], {K[1], 1, x}]]}}

There's no way to just get rid of the integral. It has to exist in this output.

Do you believe that the integral it gives should be evaluatable? If so, what is the integral? Maybe the integral is evaluatable under some kind of assumption like "C[1] is a positive real number". If that's the case, you might be able to use FullSimplify and Assumptions to get your answer. If not, then the Integral is an inherent part of the symbolic solution.