restart:

Eq1:=diff(f(x),x$4)+(f(x)*diff(f(x),x$3)-diff(f(x),x$1)*diff(f(x),x$2))=0;

Eq2:=diff(f2(x),x$2)+f(x)*diff(f2(x),x$1)=0;

bcs:=f(0)=0,(D@@2)(f)(0)=0,f(1)=0,D(f)(1)=1,f2(-1)=0.5,f2(1)=1;

dsolve({Eq1,Eq2,bcs},numeric,output = listprocedure);

Error, (in dsolve/numeric/process_input) boundary conditions specified at too many points: {-1, 0, 1},

can only solve two-point boundary value problems

How to handle such problem using dsolve?

Cheers!

restart:

lambda1:=(1/(K+2))*(S+sqrt(2*alpha*(K+2)+S^2));

lambda2:=(1/(K+2))*(S-sqrt(2*alpha*(K+2)+S^2));

where K>=0, S (-15, 15) and alpha (-15, 15). While plotting for small values of S and alpha, I get complex roots.

• How we can avoid the complex roots?
• Is it possible to impose a condition in plotting? 

  solve(lambda1=lambda2,S);

  solve(lambda1=lambda2,alpha);

  solve(lambda1=lambda2,K);

• How to single out the range of S and alpha for which we have complex roots?

Thanks

Hello everyone,

I need help to type in the following type of initial condition.

diff(1/x*diff(F(x),x),x)=0 at x =0.

Thanks

Hi,

restart:with(plots):

L1 := (1/2)*(S+sqrt(S^2+4*a*b))/b;

L2 := (1/2)*(S-sqrt(S^2+4*a*b))/b;

solve(L1=L2,S);

         2*sqrt(-a*b), -2*sqrt(-a*b)

solve(L1=L2,a);

             -(1/4)*S^2/b

solve(L1=L2,b);

                -(1/4)*S^2/a

Here is my question. 

Now plotting L1, L2 

a:=1:b:=1: # say

plot({L1,L2},S=0..1);

Is there a way to utilize the conditions which we can get through solve and use it while plotting (not 3d plotting) for any given choice of the values of a, b, S? 

Hi,

I have an ode like this

ODE:=(diff(T(x), x, x))+P*(S+a*(1-exp(-L*x))/L)*(diff(T(x), x))=0;

bcs:=T(0)=1,T(infinit)=0;

where P, S, L a, are all constants.

let assume that 

z=P/L^2*exp(-L*x);

subing z into the ode, we can have

ode1:=diff(T(z), z$2)+(1+z*a-P)*(diff(T(z), z)) = 0;

bcs1:=T(P/L^2)=1,T(0)=0;

Is it possible to find a closed form solution (T(x) in a compact form)?