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

Questions
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1

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.JPG

and with unknown K1, K2, K3 as follows.

Result_K1_K2_K3.JPG

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

Result_analytical.JPG
Result_analytical.JPG951×172 19.9 KB

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

Thanks and best regards!

Jiujiang

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

2

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.

3

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!

4

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 :

FullSimplify_Assumption.JPG

I will check what is the problem.

Thanks!

5

  1. Use double equal signs “==” instead of single equal “=” when expressing equality. Single equal is assignment “=”. See the documentation on “Set” (Equal—Wolfram Documentation) and “Equal” (Equal—Wolfram Documentation) 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’ == f, y, x]

    {{y → Function[{x}, C1 + Integrate[f[K1], {K1, 1, x}]]}}

K[1] here can’t be equal to 0. It wouldn’t make any sense for the integral. http://mathworld.wolfram.com/DummyVariable.html

6

Thanks, Sean Clarke.

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

Thanks again!