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invlaplace solution
47Categories:
Dear Sir:
Following is the inverse laplace of the given
equation.
restart;
with(inttrans):
X:=-k*(m[1]+m[2])/(s*(k*(m[1]+m[2])+m[1]*s^2*m[2]));
invlaplace(X,s,t);
The result is something like "1-cosh(a t)".
I expect that the result is similar to "1-cos(a t)".
How could I get what I want to get?
Thanks in advance
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Comments
Laplace transform
I am very far from being an expert on Laplace transforms, but when I took the Laplace transform of the result you got when you took the inverse transform of X, I got back an expression which was mathematically the same as X, although it looked different and I had to play around with it a bit. I would suggest you try this. Maybe the answer MAPLE gave is correct.
convert
You could do this:
restart;
with(inttrans):
X:=k*(m[1]+m[2])/(s*(k*(m[1]+m[2])+m[1]*s^2*m[2]));
invlaplace(X,s,t);
convert(%,cos);
Is that what you wanted?
J. Tarr
almost
Dear Prof:
The result is almost what I want to get.
However, I do not know why it has "I" which
is imaginery symbol, I think.
restart;
invlaplace(-1/(s*(s^2+a^2)),s,t);
As a trial, MAPLE gave me "cosine" function
in the result.
All of parameters are constant and positive.
Is it possible to obtaine the result without "I" symbol?
Thank you
Cosh
If you look at the result of the inverse laplace transformation in your first posting above, you will see it is a hyperbolic cosine. You wanted a cosine function and Maple correctly converted that by the well known relationship cosh(x) = cos(i*x). That can be converted to 1/2*(exp(x) + exp(-x)), if you prefer. Please see ?convert.
Maple doesn't know the values of m[1], m[2] or k, so it can only reduce the inverse laplace transformation to a hyperbolic cosine. In your second posting, you used a^2 and so Maple was able to simplify the inverse laplace transformation to a cosine function.
Good luck.
J. Tarr
assuming
I am not sure why this did not work when I tried it about an hour ago, but this should do what you want:
restart;
with(inttrans):
X:=k*(m[1]+m[2])/(s*(k*(m[1]+m[2])+m[1]*s^2*m[2]));
invlaplace(X,s,t) assuming m[1]>0,m[2]>0,k>0;
Does this answer your question?
J. Tarr
Yes
Dear Prof. Mariner:
Yes, that is it.
I have a question about "assume" command.
When should I use the assume command?
Thank you very much.
assume
Please see ?assume and ?assuming. You need to use these commands when the result of the command or action that you are taking depends on the properties of the variables. For example, the solution of a second order ODE depends upon the properties of the parameters - see your text books.
J. Tarr