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Electric Circuit ODE solving

ttokoro
21-Topaz I

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.

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ACCEPTED SOLUTION

Accepted Solutions
Werner_E
25-Diamond I
(To:ttokoro)


@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:

undefined

View solution in original post

8 REPLIES 8
Werner_E
25-Diamond I
(To:ttokoro)

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

undefined

ttokoro
21-Topaz I
(To:Werner_E)

Problem 1 can solve only one line.

undefined

 

Is problem 2 also can solve one or two steps?

 

Werner_E
25-Diamond I
(To:ttokoro)


@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:

undefined

LucMeekes
23-Emerald III
(To:ttokoro)

You did NOT do that in one step:

LM_20180801_ODE1.png

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

LM_20180801_ODE2.png

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

 

Success!
Luc

Werner_E
25-Diamond I
(To:LucMeekes)

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.

LucMeekes
23-Emerald III
(To:Werner_E)

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

Werner_E
25-Diamond I
(To:LucMeekes)


@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:

undefined

-MFra-
21-Topaz II
(To:ttokoro)

 
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