I have an assignment for Q that after subsequent other assignments and substitutions  results in 

-XC1*R1^2/((R1^2 + XC1^2)*(R1*XC1^2/(R1^2 + XC1^2) + 12960.54302))

when I type Q.

I would like to solve this for XC1, for values of Q that make  XC1 is real. 

How do I do this?  Can I rearrange this assignment?   

I guess I could do something like this:

eq1:= -XC1*R1^2/((R1^2 + XC1^2)*(R1*XC1^2/(R1^2 + XC1^2) + 12960.54302))

solve(eq1=Q,XC1)

but Q as a function of XC1 is a derived from other relationships.

The worksheet probably makes what I'm asking more clear.    I was able to get the result, but I'm sure there is a better, more elegant  way to do what I needed to do...

Thanks
 

Download pi-filter_anal_copy.mw

EDIT TIME: 14:30 CET

where P(1) and P(2) are NxN matrix functions.

My trial code for STEP 0

 alpha times integral w.r.t. x

 Int_x__alpha:=proc(term,alpha): 
 return
select(has,term,x).P(alpha)^T.remove(has,term,x)
end proc: 

 alpha times integral w.r.t. t

Int_t__alpha:=proc(term,alpha):   #alpha times integral w.r.t. t
 return
remove(has,term,t).P(alpha).select(has,term,t)
end proc:

when I run the last procedure for the testing

 Int_t__alpha(Psi(x)^T.C. Psi(t),2);

I get

But it must be 

Because the multiplication is not commutative in Matrices. So, the last procedure must be corrected. 

DETAILS for the procedures:

-------------------------------------------------------------------------------------------------------------------------------------

MAIN QUESTION:

Suppose that we have a PDE as follows 

                                             ...(3)

subject to appropriate Initial and Boundary conditions.

-------------------------------------------------------------------------------------------------------------------------------------

STEP 1

• Find the highest derivative w.r.t. x and w.r.t. t. Then, Let the trial function be the summation derivative of these highest derivatives. I mean

Trial Function:                                           ...(4)

where Psi(x), Psi(t) are Nx1 vectors and C is a NxN matrix.

I can't write a maple code for selecting the trial function.  May be you can.

trial_function:=diff(u(x,t),x,t)=Psi(x)^T.C. Psi(t); 

# I deliberately used ^T instead of ^+ for Transpose.
# If I use ^+, the transpose sign doesn't appear in 2d output. May be you have an other idea.

-------------------------------------------------------------------------------------------------------------------------------------

STEP 2

• Integrate the Eq.4  w.r.t. t from 0 to t, we have

STEP2:=int(  lhs(trial_function) ,t=0..t)=Int_t__alpha(rhs(trial_function),1);


The code must be improved. Firstly, substitute t=s in lhs(trial_function) and then integrate s=0..t 

-------------------------------------------------------------------------------------------------------------------------------------

STEP 3

int(lhs(STEP2),x=0..x)= Int_x__alpha(rhs(STEP2),1);

The code must be improved.

-------------------------------------------------------------------------------------------------------------------------------------
STEP 4

• Integrate Eq.4 w.r.t x

-------------------------------------------------------------------------------------------------------------------------------------
STEP 5

• Substitute Eq. (5), Eq. (6), Eq. (7) to Eq. (3),
• I mean substituting  u_x(x,t), u_t(x,t), u(x,t) to PDE.

-------------------------------------------------------------------------------------------------------------------------------------

STEP 6

DOWNLOAD ALL MAPLE CODE: all_code.mw

Any ideas why the attached package won't install?

We've been using Maple Cloud previously, but that doesn't work anymore apparently (same problem as stated here, https://www.mapleprimes.com/questions/226154-Installing-Packages-From-The-Cloud-Hangs, support case filed previously).

But I can't even get it installed manually either.

NODELibrary.zip

how I can sort differential equation [q] in terms of u(x,y,z,t) and its derivatives.

In other words, we should find three relations including L11(u) ,L12 (v), and L13(w). 

For instance in L13{w) only w(x,y,z,t) and its derivatives would appear...

For example;

sort.mw

Download sort.mw

This is the context. I am doing reduction of order on an ODE. This sometimes converts the ode to form  where common factor containing only  x can be moved outside, making the ode looks like  f(x)*(y''(x)+....etc...) = 0 then now I can eliminate f(x)  and only solve (y''(x)+....etc...) = 0 which is simpler.

I found if I can call factor on the ode, it works. It does remove any common terms. The problem I am having is how to cleanly determine the factors obtained. In the above example, it will be f(x) and (y''(x)+....etc...) 

For an example, the ODE   x^2 y'' + x y' = 0 can be written (using factor) as  x ( x y'' + y') =0 and now canceling x<>0, the ode becomes simpler x y''+ y' = 0.

I am now trying to find the two factors, using using op() on the result of factor.

But it does not work for all cases. There should be a way more robust way to obtain the factors.  I give 2 examples to better explain.

Example 1

ode:=x^2*diff(y(x),x$2)-(x-3/16)*y(x)=0:
ode:=expand(algsubs( y(x)=v(x)*x^(1/4)*exp(2*sqrt(x)),ode));
ode:=factor(ode);

After factoring the original ODE, there is common term found, which is the one shown UP. I need to find this to cancel it and keep the rest.

Currently I do this, which does not work for call cases

#check it is factored OK. If the type is `*` then Maple
#found common factor.
if type(lhs(ode),`*`) then 
   #add code to extract the two parts
   LHS:=op([1..-2],lhs(ode));
   RHS:=op(-1,lhs(ode));
fi;

Even though this worked here. I can now check it is the LHS which needs to be canceled. But all what I have are the operands. I do not know how to reconstruct the LHS from the operands. They could be + or *.  If it was `*` between the oprands for LHS, then I can do LHS:=mul(LHS) and this gives 

But I got lucky here. I do not know if it will be `*` all the time for LHS operands.

example 2

ode:=x^2*diff(y(x),x$2)-x*(x+2)*diff(y(x),x)+(x+2)*y(x)=0;
ode:=algsubs( y(x)=v(x)*x,ode):
ode:=factor(ode);

For this, the code I have works, but this is only because the LHS was simple.

if type(lhs(ode),`*`) then #factored OK
   LHS:=op([1..-2],lhs(ode));
   RHS:=op(-1,lhs(ode));
fi;

My question is: I expect factor, if there is common factor, to generate 2 expressions with `*` between them. I am looking for a good way to find what these two factors are. Once I do, it is easy for me to find which is the ODE and which is not and cancel the one which is not out.

I tried collect, but this does not work in general. Since I do not know beforehand, what is the common term, if any, present.

If this needs more clarification please feel free to ask. I have many more examples. This is all done by coding. Non-interactive. So solution based on looking at the output then do something, will not work for me.

I have been working on some code today and I keep getting errors in my code trying to solve it. I am trying to solve this problem:

y' = (-2xy) + 1 , y(0) = 0

Solve this problem numerically using step size of h=0.1 (101 values) and h = (0.05) (201 values). Create a plot of the two approximate solutions and the analytic solution. Show all work!

I think I got the plot code correct, I must be entering the beginning code incorect as It says:

"Error, (in DEtools/DEplot/CheckDE) extra unknowns found: xy"

Here is my code. Somebody please help me or lead me in the right direction! Thank you!

TEST_CODE_1.mw

Dear all

I compute the solution of a system of ordinary differential equaiton using Picard iteration. I have the following error:

Error, invalid input: f uses a 2nd argument, y, which is missing

My code 

Picard_iteration.mw

Thank you for any help 

Mac OS 11 Big Sur ist not listed as a compatible system with Maple 2020. Are there known issues? Could someone who already upgraded give feedback?

How can I plot the following function?

delta := t -> piecewise(t = 0, 1, 0)
Suspect that the point  at Y=1 automatically scales when zooming in.

Thanks

Currently to parse initial conditions of ODE entered bu user in the form y(x0)=y0 in order to determine x0 and y0 I use patmatch on lhs and rhs separatly which is not a big deal. I wanted to see if I can use an equation inside patmatch.

I could not figure how to use patmatch on both sides of equation on one call.  Help says expression to patmatch is algebraic expression. Does this means an equation can't be used as whole? Here is an example to make it more clear

restart;
ic:=y(0) = 1;
patmatch(ic,y(x0::anything)=y0::anything,'la')

Which gives false. i.e. no match found.

Is there a trick to make the above work if possible? This is just a basic example. I might want to apply this on more complicated equation where more things are on lhs and rhs of =  but patmatch does not like `=` in the middle there. 

Why won't Maple solve any of these inequalities for Q?

At first I tried solving the system of equations, but then I tried solving the inequalities individually for Q, and those too could not be solved by Maple.  What am I doing wrong?
 

Download pi-filter_anal_copy.mw

howca i determine the range of gains if g(x)=1/(x+x^2+x^3+2)

the answer is like closed loop

I need to use the fact that the expression under the radical in a larger expression is >0.

How do I programmaticaly "grab" the radical term to include it in an equation such as for example:

eq1:= b^2-4 a c>0

where lhs term would have come from a previous derivation or solution ?

Thank you...

How can I get Maple to simplify expressions into more meaningful forms?

For example, 

xc1 := -(2*Q*R1 + sqrt(4*Q^2*R1*RL - R1^2 + 2*R1*RL - RL^2))*R1/(4*Q^2*R1 + R1 - RL)

The numerator, under the radical, is more meaningful as sqrt(4 Q^2 R1 RL-(R1-RL)^2).

Similarly, the denominator can be simplified to Rs(4 Q^2+1)-RL.  

How do I get Maple to get me there?

Hi,

i have plotted a histogram and then fitted a uniform distribution curve. I need to compare how well the distribution fits the histogram. We are allowed to use the statistics package but i am nit sure how to do that. Not my image but just showing what I get approximately. I just need to quantify how well the red line fits the histogram. 

Many thanks 

Thanks