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Jul 28, 2018
08:20 PM

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Jul 28, 2018
08:20 PM

Electric Circuit ODE solving

Is their more smart way to solve ODE question No. 2 by Mathcad Prime 4.0?

My answer is attached.

T.Tokoro.

Solved! Go to Solution.

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Aug 01, 2018
05:27 AM

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Aug 01, 2018
05:27 AM

@ttokoro wrote:

Problem 1 can solve only one line.

Is problem 2 also can solve one or two steps?

How about this, using time shifting:

8 REPLIES 8

Jul 29, 2018
06:57 AM

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Jul 29, 2018
06:57 AM

For the ODE of first order maybe a utility function like this:

Aug 01, 2018
04:40 AM

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Aug 01, 2018
04:40 AM

Problem 1 can solve only one line.

Is problem 2 also can solve one or two steps?

Aug 01, 2018
05:27 AM

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Aug 01, 2018
05:27 AM

@ttokoro wrote:

Problem 1 can solve only one line.

Is problem 2 also can solve one or two steps?

How about this, using time shifting:

Aug 01, 2018
05:54 AM

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Aug 01, 2018
05:54 AM

You did NOT do that in one step:

This proves that y1(t) is a solution to ODE1.

This poves that y2(x) is a solution to ODE2.

Success!

Luc

Aug 01, 2018
06:54 AM

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Aug 01, 2018
06:54 AM

Luc, you probably know that your MC11 solution will not work in MC15 or Prime.

Using the time shift as in my response above I guess that with MC11 it should be possible to get the solution for ODE2 in one go, too.

Aug 01, 2018
07:46 AM

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Aug 01, 2018
07:46 AM

Sure, I know!. For Mathcad 15 I'm certain.

It was just that I was surprised how far the OP got with solving the 2nd order ODE. Maybe my example could give some idea how it could be done in Prime.

Can you define in Prime:

laplace(f,t):=f(t)->laplace,t

(that is...symbolic ) and have it work like the laplace operator?

Luc

Aug 01, 2018
10:13 AM

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Aug 01, 2018
10:13 AM

@LucMeekes wrote:

Sure, I know!. For Mathcad 15 I'm certain.

It was just that I was surprised how far the OP got with solving the 2nd order ODE. Maybe my example could give some idea how it could be done in Prime.

Can you define in Prime:

laplace(f,t):=f(t)->laplace,t

(that is...symbolic ) and have it work like the laplace operator?

Luc

No, thats not possible.

When the symbolic laplace is applied to a function y(t) the symbolics return "laplace(y(t),s,t)" as in the good old maple days, but this function is of no value and obviously treated as something special (probably simply a remainder from the Maple time).

Its not possible to use "substitute" to replace "laplace(y(t),s,t)" for something else nor does it help if we define our own function "laplace" with three argument before the symbolic laplace is applied. The two are treated as something completely different. Furthermore "laplace" is even treated by the symbolics like an unknown or undefined function:

Jul 29, 2018
12:38 PM

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Jul 29, 2018
12:38 PM