John2020

220 Reputation

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2 years, 234 days

MaplePrimes Activity


These are replies submitted by John2020

@Carl Love 

Dear friend,

Thanks a lot for your help.

These codes also work very well.

Thanks again.

@Kitonum

 

Dear friend,

Many thanks for your help.

 

@Christian Wolinski 

Dear friend,

Thanks a lot. These codes work very well.

Many thanks for your help.

@Carl Love 

Dear friend,

Thanks a lot for your explanations.

The answer of @acer 26216 was exactly what I want.

@Ronan 

Dear friend,
Thanks a lot for your efforts.

@Rouben Rostamian  

Dear friend,
Thanks a lot for your help and excellent solutions.

@tomleslie 

Dear friend,
Thanks a lot for your help and excellent solutions.

@acer 

Dear friend,
Thanks a lot for your help and excellent solutions.

@dharr 

Dear friend,

Thank you very much.

@Rouben Rostamian  

Dear friend,

Thanks a lot for your comments and helps.

I'll consider the problem again.

@dharr 

Dear friend,

Thanks a lot for your help.

Two conditions correspond to the first and third time intervals.

I want that maple to distinguish this and solve the ode at the first interval with the first bc and solve the ode at the third interval with the second bc and give me the true result, but ....

@tomleslie 

Yes, you and dear carol are right.

I realized my mistake a couple of hours ago.

sorry!

@Carl Love 

Dear friend,

Thanks a lot for consideration.

But, x[0](tau) and y[0](tau) are independent from each other. For this reason, I have this expectation that maple understand this and use separation method and then splits the eq to 2 equations as:

eq[1]:=diff(x[0](tau), tau) + x[0](tau)=0;

eq[2]:=- diff(y[0](tau), tau) - y[0](tau)=0;

then the solutions with that boundary conditions are exactly exp(-tau) as I mentioned.

Is there any way to get this?

@acer 

Dear friend,

Thank you very very much.

It is exactly what I wanted.

@acer 

Hi,

one (or 2 or 3) real non-zero (if possible) root(s) for each variable (a[1],a[2],...) is sufficent. 

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