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DSolve returns with integration symbol, how to solve it? Thanks!

Hello, everyone! I am using DSolve to solve a differential equation which is somewhat complicated. After several tens of minutes of waiting, I got a result, but with integration symbol.

Result with integration symbol

and with unknown K1, K2, K3 as follows. unknown parameters K1, K2, K3

But when I use DSolve with simpler differential equations, I can get analytical result.

Analytical result for simpler differential equation

Will you please tell me how to get analytical result with my complicated eqation?

Thanks and best regards!


(The software is a trial version V10.0.1.0, 64bit )

POSTED BY: Jiujiang Wang
5 Replies

Thanks, Sean Clarke.

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

Thanks again!

POSTED BY: Jiujiang Wang
  1. Use double equal signs "==" instead of single equal "=" when expressing equality. Single equal is assignment "=". See the documentation on "Set" ( and "Equal" ( for more information.

  2. K[2] is the dummy variable of integration. It can't be equal to 0. Take a look at this example:

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

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

K[1] here can't be equal to 0. It wouldn't make any sense for the integral.

POSTED BY: Sean Clarke

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 : Question of FullSimplify and Assumption

I will check what is the problem.


POSTED BY: Jiujiang Wang

Thanks, Sean Clarke. I really did not see the details of DSolve. And I am not so familiar with the software.

The equation should be evaluatable because I can use every single part of the complicated equations for explicit solution.

I will try the FullSimplify and Assumption you provided.

Thanks again!

POSTED BY: Jiujiang Wang

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.

POSTED BY: Sean Clarke
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