Exiu

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These are questions asked by Exiu

If I have an equation:

-2*a[0]*((x^3*a[1] + 42)*diff(y(x), x, x) - 6*y(x)*x*a[1])*sqrt(-a[0]*(a[1]^2 + 3)) - 3*x^2*((1/3*a[1]^3 + a[0] + a[1])*x + (8*a[1]^2 + 24)*a[0])*diff(y(x), x, x)

I want to obtain a differential algebraic equation for y(x)without the parameters a[1] and a[0]. My idea is to first compute the derivative of U to get:

V:=-2*a[0]*(3*x^2*a[1]*diff(y(x), x, x) + (x^3*a[1] + 42)*diff(y(x), x, x, x) - 6*diff(y(x), x)*x*a[1] - 6*a[1]*y(x))*sqrt(-a[0]*(a[1]^2 + 3)) - 6*x*((1/3*a[1]^3 + a[0] + a[1])*x + (8*a[1]^2 + 24)*a[0])*diff(y(x), x, x) - 3*x^2*(1/3*a[1]^3 + a[0] + a[1])*diff(y(x), x, x) - 3*x^2*((1/3*a[1]^3 + a[0] + a[1])*x + (8*a[1]^2 + 24)*a[0])*diff(y(x), x, x, x)

Then, compute the resultant: resultant(U,V,a[0]). However, since a[0] is algebraic, we cannot compute the resultant directly. I want to substitute:

algsubs((−a[0](a[1]2+3))2=c2,U)

Then, eliminate c, a[0], and a[1] using the resultant. However, I find that this substitution command does not work. Is there a method to help eliminate the parameters a[0]and a[1]?

Suppose I have a list of differential equations:

L:=[x*diff(f(x),x)+2*x+1=0, 4*(diff(f(x),x$2))^2+7*diff(f(x),x)-2=0, x*diff(f(x),x)+2*x+1=0, 3*(diff(f(x),x))^2-f(x)-1=0];

how can I rewrite this list such that 
L:=[[x*diff(f(x),x)+2*x+1=0,2], [4*(diff(f(x),x$2))^2+7*diff(f(x),x)-2=0,1], [3*(diff(f(x),x))^2-f(x)-1=0,1]];
i.e., rewrite this list as the [equation, Number of occurrences in the list]

if I have arational function f(x) in C(x), where C is the complex field, then the partial fraction of f can be written as 

f(x)=p(x)+\sum_{i=0}^n a[i]/(1-b[i]*x)^c[i]
i.e., every denominator is linear in x.

How can I get this factorization in maple, and I think when the degree of the denominator is too high, we can not get the explicit a[i] and b[i], in this situation, can we get their minimal polynomials?

I have a polynomial where one of its coefficients contains an undetermined parameter c. I want to identify which coefficient contains c and solve for c by setting that coefficient equal to zero. Since c might appear in any coefficient corresponding to any power of x, I cannot use a direct solve command when I creat a procedure. 

I have some code, but I think here should have some simple command to do this.


n := 3;
L := [3*a0 + 2*a1, -a1/3 + a0/2 + 12*a3];
vol := proc(L)

local i, res; i := 1;

while i < nops(L) + 1 do

if coeff(L[i], a3, 1) = 0

then i := i + 1;

else res := solve(L[i], a3);

return res;

end if;

end do;

end proc;
vol(L);

 

the output is   a1/36 - a0/24

          

How can I subtitute a truncated series in nonlinear differential equation? like what I do in mathematica.

In maple I can only get:

How can I get a reduced result like what mathematica did? I think it has the relation with the arithmetic operations of truncated series.

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