## differential equation...

i want to solve this equation,

y''(x)=5*exp(-10/y'(x)) on ]0,15[ with y(0)=0,y(15)=2

can any one help me ? thank you

## how do I remove the error "initial Newton iteratio...

Respected member!

 >
 >
 >
 >
 >
 >
 >
 (1)
 >
 (2)
 >
 >

## Computational physics with DE equations ...

I need help with these two questions. anything is helpful

## How to take the integral from a function solved nu...

3.mw

There are 2 questions actually. The first as the title says is about taking the integral. I have 2 functions that were found numerically from the system of differential equations (see the file), and I need to take the integral of the expression that includes both of them. Maple gives me something like Int () = Int (), so it doesn't solve anything. Why can it be?

The second question is about varying the boundary conditions. If, for example, I have the system with the condition like R(x_0)=R_0, can I get the plot  of R(x,R_0)? In my case I need to vary conditions on R and mu (R(0)=R_0 and mu(0)=mu_0) and then get the plot of the integral in relation to R_0 and mu_0. Is it even possible?

## Fourth order problem showing use midpoint method...

Dear sir,

I tried to solve a fourth order problem. But I got the error message as better to use midpoint method. Can I know what is midpoint method and here I uploading the problem please verify it if I did anything mistake?program.mw

## how i can remove error ["unable to store %1 when d...

mar.mw
hi..how i can dsolve these differential equations? omega is unknown and fir solving i add an extra boundary condition, but the error "unable to store %1 when datatype=%2" is appear!!!

how i can remove this error?

thanks

## differential equations with user defined function...

I would like to solve the following differential equation with a relatively complicated function that is best declared.

myfun := proc (x::float)

local output;

output := 4*x^2;

end proc;

de := diff(y(x), x)+myfun(x)*y(x) = 0.;

However, this gives the following message:

Error, invalid input: myfun expects its 1st argument, x, to be of type float, but received x

Any suggestions?

## coupled non linear odes solution in the different ...

i am interested to numerically solve the 3 non-linear coupled ODE's in the 3 different intervals of rho(define in attached file) for the different corresponding parameters alpha and beta at i=0..2.

## numeric dsolve error...

Hello,

when trying to solve my equation I always get this error-message:

Error, (in DEtools/convertsys) invalid input: degree expects 1 or 2 arguments, but received 3

To solve, I use this one:

Anfangsbed := phi(0) = 0,  D(phi)(0) = 10;

Einst :=             range = 0..10,            #Integrationsgrenzen
relerr = 1e-3,            #Toleranz zur Bestimmung der Integrationsschrittweite
optimize = true,          #Optimierung der DGL
maxfun = 0,               #Anzahl der Integrationsschritte -- 0 = keine Obergrenze
output = operator,        #LÃ¶sungsfunktionausgabe: damit Definition als Funktion mÃ¶glich
stiff = true:

Lösung_Bewgl := dsolve({Bewegungsgl_mit_Para, Anfangsbed}, numeric, Einst);

The equation to solve is this one: http://imgur.com/a/SFxI8

What am I doing wrong?

Edit:

This is the worksheet, problem is the last point.

WS.mw

## find value for t where ca(t)=0.2 (very simple requ...

Hey there!

this is the formula:

diff(ca(t), t) = -3.600000000*10^20*exp(-15098.13790/(340-20*ca(t)))*ca(t), ca(0) = 2

I can easily plot it by solving DE numerically, however I can't seem to find a way to determine what 't' is at ca(t)=0.2

thank you so much in advance!

Kind regards,

Gerard

## How do I get rid of the error?...

`Download 2222222222222222222.mw`

```> restart;
> A[0] := 10^(-3); a := 10^5;
> sys := diff(R(theta), theta) = A[0]*exp(2*mu(theta))*sin(theta)/(2*a), R(theta) = 2*exp(-2*mu(theta))*(1-(diff(mu(theta), `\$`(theta, 2)))-cot(theta)*(diff(mu(theta), theta)));
> cond := R(0) = 10^(-5), mu(0) = 118.92, (D(mu))(0) = 0;
> F := dsolve({cond, sys}, [R(theta), mu(theta)], numeric);
> with(plots);
> odeplot(F, [theta, R(theta)], 0 .. 3.14, color = black, thickness = 3, linestyle = 4)
> odeplot(F, [theta, mu(theta)], 0 .. 3.14, color = blue, thickness = 3, linestyle = 1)```

After last two lines maple writes:

Warning, cannot evaluate the solution past the initial point, problem may be complex, initially singular or improperly set up

And gives me empty plots. I can't figure out where an error can be. Some things I noticed:

Maple doesn't calculate the system before and after zero. If I change the range from 0..3.14 to -10..10 or to 0.00001..0.00001, it gives me 2 errors for 1 plot.

Also if I change the condition mu(0) = 118.92 to mu(0) = 1 or mu(0) = 50 or mu(0) = 80, it works. After ~80 it gives an error. I can't imagine where could appear a division by 0 or some other mistake.

## Polarplot warning for a semistable limit cycle...

Hey, i'm trying do demonstrate that a nonlinear system has a semistable limit cycle but i get a warning at the plot command saying "Warning, unable to evaluate the function to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct" and i dont understand it. So i wonder if someone here could help me?

restart; with(PDEtools); with(plots);
eq1 := diff(x(t), t) = x(t)*(x(t)^2+y(t)^2-1)^2-y(t);
2
d              /    2       2    \
--- x(t) = x(t) \x(t)  + y(t)  - 1/  - y(t)
dt
eq2 := diff(y(t), t) = y(t)*(x(t)^2+y(t)^2-1)^2+x(t);
2
d              /    2       2    \
--- y(t) = y(t) \x(t)  + y(t)  - 1/  + x(t)
dt
tr := {x(t) = r(t)*cos(theta(t)), y(t) = r(t)*sin(theta(t))};
{x(t) = r(t) cos(theta(t)), y(t) = r(t) sin(theta(t))}
eq1b := dchange(tr, x(t)*eq1+y(t)*eq2, [r(t), theta(t)], simplify);
/ d      \       2 /        4         2\
r(t) |--- r(t)| = r(t)  \1 + r(t)  - 2 r(t) /
\ dt     /
eq1b := expand(eq1b/r(t));
d                    5         3
--- r(t) = r(t) + r(t)  - 2 r(t)
dt
eq2b := dchange(tr, y(t)*eq1-x(t)*eq2, [r(t), theta(t)], simplify);
2 / d          \        2
-r(t)  |--- theta(t)| = -r(t)
\ dt         /
eq2b := simplify(eq2b/(-r(t)^2));
d
--- theta(t) = 1
dt
sol1 := dsolve({eq1b, r(0) = r[0]}, r(t));
/      /  /     2  \
|      |  | r[0]   |          2     2
r(t) = exp|RootOf|ln|--------| (exp(_Z))  r[0]
\      \  \r[0] - 1/

2     2
- ln(r[0] + 1) (exp(_Z))  r[0]

/             2\
|(exp(_Z) - 1) |          2     2            2        2
- ln|--------------| (exp(_Z))  r[0]  + (exp(_Z))  _Z r[0]
\ exp(_Z) - 2  /

/     2  \
2     2         | r[0]   |             2
+ 2 (exp(_Z))  r[0]  t - 2 ln|--------| exp(_Z) r[0]
\r[0] - 1/

2
+ 2 ln(r[0] + 1) exp(_Z) r[0]

/             2\
|(exp(_Z) - 1) |             2                    2
+ 2 ln|--------------| exp(_Z) r[0]  - 2 exp(_Z) _Z r[0]
\ exp(_Z) - 2  /

/     2  \
2       | r[0]   |          2
- 4 exp(_Z) r[0]  t - ln|--------| (exp(_Z))
\r[0] - 1/

/             2\
2     |(exp(_Z) - 1) |          2
+ ln(r[0] + 1) (exp(_Z))  + ln|--------------| (exp(_Z))
\ exp(_Z) - 2  /

/     2  \
2                   2       | r[0]   |
- (exp(_Z))  _Z - 2 t (exp(_Z))  + 2 ln|--------| exp(_Z)
\r[0] - 1/

/             2\
|(exp(_Z) - 1) |
- 2 ln(r[0] + 1) exp(_Z) - 2 ln|--------------| exp(_Z)
\ exp(_Z) - 2  /

2                                    2
- (exp(_Z))  + 2 _Z exp(_Z) + 4 t exp(_Z) + r[0]  + 2 exp(_Z)

\\
||
- 1|| - 1
//
sol1 := simplify(sol1);
/      /   /     2  \
|      |   | r[0]   |               2
r(t) = exp|RootOf|-ln|--------| exp(2 _Z) r[0]
\      \   \r[0] - 1/

2
+ ln(r[0] + 1) exp(2 _Z) r[0]

/             2\
|(exp(_Z) - 1) |               2                    2
+ ln|--------------| exp(2 _Z) r[0]  - exp(2 _Z) _Z r[0]
\ exp(_Z) - 2  /

/     2  \
2         | r[0]   |             2
- 2 exp(2 _Z) r[0]  t + 2 ln|--------| exp(_Z) r[0]
\r[0] - 1/

2
- 2 ln(r[0] + 1) exp(_Z) r[0]

/             2\
|(exp(_Z) - 1) |             2                    2
- 2 ln|--------------| exp(_Z) r[0]  + 2 exp(_Z) _Z r[0]
\ exp(_Z) - 2  /

/     2  \
2       | r[0]   |
+ 4 exp(_Z) r[0]  t + ln|--------| exp(2 _Z)
\r[0] - 1/

/             2\
|(exp(_Z) - 1) |
- ln(r[0] + 1) exp(2 _Z) - ln|--------------| exp(2 _Z)
\ exp(_Z) - 2  /

/     2  \
| r[0]   |
+ exp(2 _Z) _Z + 2 t exp(2 _Z) - 2 ln|--------| exp(_Z)
\r[0] - 1/

/             2\
|(exp(_Z) - 1) |
+ 2 ln(r[0] + 1) exp(_Z) + 2 ln|--------------| exp(_Z)
\ exp(_Z) - 2  /

2
+ exp(2 _Z) - 2 _Z exp(_Z) - 4 t exp(_Z) - r[0]  - 2 exp(_Z)

\\
||
+ 1|| - 1
//
sol2 := dsolve({eq2b, theta(0) = theta[0]}, theta(t));
theta(t) = t + theta[0]
theta[0] := (1/4)*Pi;
1
- Pi
4
plot1 := polarplot([subs(r[0] = .1, rhs(sol1)), rhs(sol2), t = 0 .. 10], color = red);
Warning, unable to evaluate the function to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct
plot2 := polarplot([subs(r[0] = 2, rhs(sol1)), rhs(sol2), t = 0 .. 10], color = blue);
Warning, unable to evaluate the function to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct
display({plot1, plot2}, scaling = constrained, tickmarks = [4, 3], view = [-2 .. 2, -2 .. 2]);

## can Maple find solution to eigenvalue boundary val...

newbie here. When I give Maple 2016.2  a boundary value ODE with an eigenvalue in it, it returns the trivial solution. I was wondering if Maple supports finding non-trivial solution and also give the eigenvalue values associated with the non-trivial solution?

```restart;
ode:=diff(y(x),x\$2)+lambda*y(x)=0;
bc:=y(0)=0,y(L)=0:
assume(L>0,L,'real'):
sol:=dsolve({ode,bc},y(x));
```

The outtput is `y(x)=0`

In Mathematica, it gives both trivial and non-trivial solution:

```Clear[L0, lam, x, y, r]
ode = y''[x] + lam y[x] == 0;
bc = {y[0] == 0, y[L0] == 0};
sol = Assuming[Element[L0, Reals] && L0 > 0,
DSolveValue[{ode, bc}, y[x], x]]```

If Maple does not currently support this, any one knows if this will added to Maple 2017?

## How to solve delay differential equation by method...

How to solve delay differential equation by method of steps in MAPLE software.

## help appreciated...

Hi I am getting this message while soliving my first order non-linear initial value problem.