janhardo

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11 years, 65 days

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These are replies submitted by janhardo

@Alfred_F 
This is userfriendlier by using y(x) explicit 

NULL

with(Student:-ODEs)

ode1 := y(x)*(diff(y(x), x)) = a

y(x)*(diff(y(x), x)) = a

(1)

ODESteps(ode1)

"[[,,"Let's solve"],[,,y(x) ((ⅆ)/(ⅆx) y(x))=a],["•",,"Highest derivative means the order of the ODE is" 1],[,,(ⅆ)/(ⅆx) y(x)],["•",,"Integrate both sides with respect to" x],[,,∫y(x) ((ⅆ)/(ⅆx) y(x)) ⅆx=∫a ⅆx+`c__1`],["•",,"Evaluate integral"],[,,((y(x))^2)/2=a x+`c__1`],["•",,"Solve for" y(x)],[,,{y(x)=sqrt(2 a x+2 `c__1`),y(x)=-sqrt(2 a x+2 `c__1`)}],["•",,"Redefine the integration constant(s)"],[,,{y(x)=sqrt(2 a x+`c__1`),y(x)=-sqrt(2 a x+`c__1`)}]]"

(2)
 

restart

"maple.ini in user"

(3)

with(Student:-ODEs)

ODE2 := {y(x)*(diff(y(x), x)) = a, y(0) = b}

{y(x)*(diff(y(x), x)) = a, y(0) = b}

(4)

ODESteps(ODE2)

"[[,,"Let's solve"],[,,{y(x) ((ⅆ)/(ⅆx) y(x))=a,y(0)=b}],["•",,"Highest derivative means the order of the ODE is" 1],[,,(ⅆ)/(ⅆx) y(x)],["•",,"Integrate both sides with respect to" x],[,,∫y(x) ((ⅆ)/(ⅆx) y(x)) ⅆx=∫a ⅆx+`c__1`],["•",,"Evaluate integral"],[,,((y(x))^2)/2=a x+`c__1`],["•",,"Solve for" y(x)],[,,{y(x)=sqrt(2 a x+2 `c__1`),y(x)=-sqrt(2 a x+2 `c__1`)}],["•",,"Redefine the integration constant(s)"],[,,{y(x)=sqrt(2 a x+`c__1`),y(x)=-sqrt(2 a x+`c__1`)}],["•",,"Use initial condition" y(0)=b],[,,b=sqrt(`c__1`)],["•",,"Solve for" `c__1`],[,,`c__1`=b^2],["•",,"Substitute" `c__1`=b^2 "into general solution and simplify"],[,,y(x)=sqrt(2 a x+b^2)],["•",,"Use initial condition" y(0)=b],[,,b=-sqrt(`c__1`)],["•",,"Solve for" `c__1`],[,,`c__1`=b^2],["•",,"Substitute" `c__1`=b^2 "into general solution and simplify"],[,,y(x)=-sqrt(2 a x+b^2)],["•",,"Solutions to the IVP"],[,,[y(x)=sqrt(2 a x+b^2),y(x)=-sqrt(2 a x+b^2)]]]"

(5)
 

with(plots); p1 := plot(sqrt(2*x), x = 0 .. 5, color = red, thickness = 2, legend = "y = √(2x), a=1, b=0"); p2 := plot(sqrt(x), x = 0 .. 5, color = green, thickness = 2, legend = "y = √(x), a=0.5, b=0"); p3 := plot(sqrt(-2*x), x = -5 .. 0, color = blue, thickness = 2, legend = "y = √(-2x), a=-1, b=0"); display([p1, p2, p3], title = "Solutions of y(x) = sqrt(2*a*x + b²) with b = 0 (positive branch)", labels = ["x", "y(x)"], axes = boxed)

 
 

``

Download ode_via_odesteps_mprimes24-8-2025.mw

The dependent variable cannot be easily isolated. ?

If b= 0 in ic y(0)=0

y1 := sqrt(2*x);
y2 := sqrt(2*0.5*x);
y3 := sqrt(2*(-1)*x);
p1 := plot(y1, x = 0 .. 5, color = red, thickness = 2, legend = "a = 1");
p2 := plot(y2, x = 0 .. 5, color = green, thickness = 2, legend = "a = 0.5");
p3 := plot(y3, x = -5 .. 0, color = blue, thickness = 2, legend = "a = -1");
display([p1, p2, p3], title = "Solutions of y(x) = sqrt(2*a*x) with Initial Condition y(0) = 0", labels = ["x", "y(x)"], axes = boxed);


@salim-barzani 
I don't think I fully understand you.

@salim-barzani 

Do you want to calculate all this manually?

Then you need to extract the procedures from the wkptest module.

The procedures must be reduced from general to a single calculation.

Can Maple give a another answer than this ?
sin^2(a)+cos^2(a) =1 
sin^2(a)+cos^2(a) +1 = 2 
sin^2(a)+1 = - cos^2(a) +2  
Pythagoras in the unit circle.

@acer 
Thank you, don't make the calculation bigger, but more efficient.

That's what I understand from it.

@acer 
I can't manage to create such a nice short procedure like you do. 

splitsen_in_lineair_en_niet_linear_werkend_def_CORR2procedure-12-8-2025.mw

@acer 
Would it be correct to say that t and K are degree 1 and a number is degree 0  and both lineair ?
linpred (graad 0 óf 1 ⇒ “lineair”)  

splitsen_in_lineair_en_niet_linear_werkend_def_CORR1_12-8-2025.mw

@salim-barzani 

For a pde you can select lineair en not llinear terms by hand 

L := convert(expand(pde1), list, `+`)
lineair := L[1] + L[2]

A procedure , numeric, symbolic sphere equation for given sphere centre and 3 circles 

A phytagoras circles example

res := SphereFromSections([8, 4, 0], 4, 3, 5, symbolic = true, plot = true);
res[equation];
res[center], res[radius];

"Sphere equation: ", (x - 3.000000000)^2 + (y - 4.000000000)^2 + z^2 = 25.00000000

opgave_rouben_bol_mprimes_procedure_nmeriek_en_symbolisch_11--8-2025DEF.mw

@Rouben Rostamian  
Ai says  , its ill posed?
At first, the Ai said that the task was impossible, but later AI said that it was possible.( again question) 



.

Download opgac_rouben_bol_mprimes_11--8-2025.mw

 

@sand15 
The ai  is improved lately, and seems to be much better , making this is not a achievement
bolvergelijking_opstellen_mprimes_10-8-2025.mw

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