To define the matrices A[k]:

assign(seq(A[k] = <k,k^2;2*k,2*k^2>, k=1..10));

To make a block-diagonal matrix out of these:

M := LinearAlgebra:-DiagonalMatrix([seq(A[k], k=1..10)]);

To view M:

evalm(M);

Aside: If you have no need to refer to the individual matrices A[k], then you may bypass the middleman and make the block-diagonal matrix directly:

M := LinearAlgebra:-DiagonalMatrix([seq(<k,k^2;2*k,2*k^2>, k=1..10)]);

The differential equation itself expresses y''(t) in terms of y(t) and y'(t).  Therefore once you have obtained y(t) and y'(t), you can evaluate y''(t) directly, like this.

Download mw.mw

Regarding your statement:
I know function f(x) for y>1.4266 is monotically decreasing and there is no turning points,(actually i can compute infection points of above function: y=1.4266 and x=4/5 ).

That's correct but the calculation in your worksheet may be simplified to one line:
solve({diff(F(x,y),x) = 0, diff(F(x,y),x,x) = 0});

As to your statement:
I know function for 1.4266>y>1.0144 has two turning points a minimum for x<4/5 and a maximum for x>4/5.

I don't see the significance of the 1.0144.  The function has a minimum and a maximum for all values of y in 0 < y < 1.4266.  You can see that by letting
F := (x,y) -> x^2+y*(1-x)^(3/4);
and then plotting F(x,y) for 0 < x < 1 and various values of y.

I can't tell what you mean by "at least one of the terms is derivative free".  Perhaps you mean "when the derivative is zero"?  If so, then the following sample may help.

>  restart; >  de := diff(U(z),z,z) + diff(U(z),z) + U(z) = 0; >  ic := U(0)=0, D(U)(0)=1; >  E := events=[[diff(U(z),z)=0, halt]]; >  dsol := dsolve({de, ic}, numeric, E); >  plots:-odeplot(dsol, [z,U(z)], z=0..2); Warning, cannot evaluate the solution further right of 1.2091995, event #1 triggered a halt

mw.mw

Add an extra point at the origin, as in:

plot([ piecewise( x<5, 100, x>5,200), [[0,0]] ], x=0..10);

You asked:  Second example gives complex answer. Is it possible to get trigonometric answer with real constants c1 and c2?

The answer is no.  Here is why.

Download mw.mw

restart;
with(plottools): with(plots):
s := display(sphere([0,0,0],1), style=surface, transparency=0.8);
c := display(cuboid([-1,-1,-1], [1,1,1]), transparency=0.3, style=surface);
display([c,s], axes=none, orientation=[-65,72,0]);

 

Your calculations assume that the functions { C(n,x), S(m,x) } form an orthogonal family with respect to the weighted integral (weight = 1/x) on the interval (1,2), but that's not true.  Let's check:

We see that the result is not zero in general.

You may already know the difference between the colon (":") and a semicolon (";")  command terminators.  In both cases the command is executed, but in the case of the colon the result is not shown.

Important:  Always use the semicolon terminator when developing new code, especially when you are having trouble with it.  Suppressing the result through a colon deprives you of the assurance that your command works as intended.

With this in mind, go back to your code and replace the colon terminators with semicolons, and then come back here if you still have problems.

Just by looking at your equations it is clear that y(x)≡0 is a solution.  I suppose you are looking for nonzero solutions.

Your equations arise in the analysis of buckling of an Euler beam.  A combination of the parameters P, K(x), and L that yield nonzero solutions correspond to the beam's buckling modes.

You must be familiar with the analysis corresponding to a uniform beam, that is, when K(x) is a constant.  There one shows that buckling modes exist only when L*sqrt(P/K) = nπ, where n is a positive integer.  When K is not a constant, there is no general formula for buckling modes.  The best you can hope for is some numerical experiments for specific choices of K(x).

restart;
m := f__1(x,y)*A + f__2(x,y)*B + f__3(x,y)*C;
f := <coeffs(m, [A,B,C])>;

I don't know how to do this in Maple but that sort of thing is easily done using specialized image manipulation tools.  I would do it with Imagemagick's command-line tools but I am sure there are many other alternatives, perhaps gimp or photoshop.

It takes only two imagemagick commands to produce the desired composite.  Let's call your two images fig.png and fig-zoomed.png.

• Step 1:  Scale fig-zoomed.png  to half its original size and call the result fig-zoomed-scaled.png:
      convert -scale 50% fig-zoom.png fig-zoom-scaled.png
• Step 2: Produce the composite which we call out.png:
      composite -geometry +150+50 fig-zoom-scaled.png fig.png out.png

The "-geometry +150+50" option says that the inset image will be offset relative to the upper-left corner by 150 pixels (horizontally) and 50 pixels (vertically).  Here is the result:

Name the plots, as in
p1 := plot(<a,c,e,g,i,a>, <b,d,f,h,j,b>);
...
p2 := plot(<a,c,e,g,i,a>, <b,d,f,h,j,b>);
and then do:
plots:-display([p1,p2], scaling=constrained);

Download mw.mw