Hello,

I am trying to solve the equation (non-linear) with one variable t2 (please see bellow):

(35595.29412*(52040.0-2400.0*t2))*(11-t2)*exp(ln(-360.0*t2+29600.0)-(0.9803921570e-5*(-360.0*t2+59200.0))*t2)*exp(ln(52040.0-2400.0*t2)-8.804313725+.5176470590*t2)*(360000*(exp(ln(-360.0*t2+29600.0)-(0.9803921570e-5*(-360.0*t2+59200.0))*t2))^2+(620*(-360.0*t2+29600.0))*(1200*exp(ln(-360.0*t2+29600.0)-(0.9803921570e-5*(-360.0*t2+59200.0))*t2)-2.232000*10^5*t2+1.83520000*10^7))*exp(ln(-360.0*t2+29600.0)-(0.9803921570e-5*(-2760.0*t2+59200.0))*t2)/(360000*exp(ln(52040.0-2400.0*t2)-8.804313725+.5176470590*t2)*(exp(ln(-360.0*t2+29600.0)-(0.9803921570e-5*(-360.0*t2+59200.0))*t2))^2+(372000*((52040.0-2400.0*t2)*exp(ln(-360.0*t2+29600.0)-(0.9803921570e-5*(-2760.0*t2+59200.0))*t2)+(-360.0*t2+29600.0)*exp(ln(52040.0-2400.0*t2)-8.804313725+.5176470590*t2)))*exp(ln(-360.0*t2+29600.0)-(0.9803921570e-5*(-360.0*t2+59200.0))*t2)+(384400*(52040.0-2400.0*t2))*(-360.0*t2+29600.0)*exp(ln(-360.0*t2+29600.0)-(0.9803921570e-5*(-2760.0*t2+59200.0))*t2))^2 = 1/150

During evaluation of the solve command I received a warning that solutions may have been lost. How can I overcome this problem? Also, I need that t2>0.

Thanks in advance,

Dmitry 

Hi,

I have a system of differential equations with boundary conditions:

diff(S(t), t) = -K(t)*S(t)/N, diff(K(t), t) = K(t)*S(t)/N, where S(T)=10, K(T)=20; If I would like to solve this system backward in time, is it right to re-write the system of original diff. equations in the following way:

diff(S(t), t) = K(t)*S(t)/N, diff(K(t), t) = -K(t)*S(t)/N and S(0)=10, K(0)=20 ( I simply changed the sign of the right hand of the equations and changed the boundary conditions to the inotoal ones).

Thanks,

Dmitry

Hello all, 

During my last attempt to solve ODE system (autonomous system which includes 3 first order diff. equations) with initial conditions, Maple had performed the solution which includes d_z1 parameter as follows below (I present the solution of one of the equations):

S(t)=S(0)∫(QN_z1+A)d_z1, where integral ∫ is defined integral from 0 to t, S(0) is the initial value of S, Q, N and A are the parameters. I would like to ask, what does it mean _z1 and d_z1? Why if the ODE system is only time dependent, I received the integral with other differential, that is d_z1? Does it mean that the integral can't be evaluated or maybe something else?

Thanks in advance,

Dmitry

Hello all, 

I have the system which consists of three linear first-order differential equations (see below). First, I solved this system numerically backward in time by implying the final condition for the variables at time t=T (i.e, x(T=4)=10 and so on) and depicted the graph. Then, from the obtained solution I found the values of the variables at time t=0 and solve once again the system of ODE's but forward in time with...