Question: Eigen Values and a system of ODEs

Hi All, 

I am working on a numerical simulation of a system of ODEs that blows up when I do a numerical solution. One of my advisors says to multiply  the ODEs in my system by -1 to reverse the role of the eigen values. I only vaguely understand this. Would you be able to point me in the right direction to figure this out? When I do this (-1 in front of two of the ODEs), my results are much more stable. I will attach the ODE system here. The parameter values follow.

 Thanks for any input,

Steve

 > Restart;


> Eq1 := diff(H(t), t) = ((alpha-1)*W(t)+alpha*(Di-R(t))+theta*R(t)+F-E)/(A*S);        

                       
> Eq2 := diff(W(t), t) = -(r*((W(t)+Di-R(t))/k-g/k-Cw)+(diff(R(t), t))/k+(diff(H(t), t))*Cwh-Ch*(alpha-1)/(A*S))/(1/k-Cww);

> Eq3 := diff(R(t), t) = -(r*(-W(t)-Di+R(t)+g-k*Cjr-k*Cr)/k+(diff(W(t), t))/k+Ch*(alpha-theta)/(A*S))/(1/k-Cjrr-Crr);

> sol1 := dsolve([Eq1, Eq2, Eq3, R(0) = 8000, H(0) = 4900, W(0) = 58000], numeric, method = rosenbrock);

> plots[odeplot](sol1, [t, W(t)], t = 0 .. 70, color = red);
 


> Cw := 2*a1*W(t)+2*a2*W(t)*H(t);

> Cww := 2*a1+2*a2*H(t);

> Cwh := 2*a2*W(t);

> Ch := 2*a2*W(t);

> Chh := 0;

> a1 := 500;

> a2 := -0.5e-3;

> Cjr := -b1-2*b2*(Di-R(t));

> Cjrr := 2*b2;

> b1 := 600;

> b2 := -0.4e-2;

> Cr := c1+2*c2*R(t);

> Crr := 2*c2;

> c1 := 500;

> c2 := -0.4e-2;

> A := 45000;

> S := .15;

> r := .10;

> Di := 97000;

> g := 700000*(47036*1/500000);

> F := 85000;

> E := 10000;

> alpha := 0.4e-1;

> theta := 0.6e-1;

> k := (-2059*700000)*1/500000;

> g := 700000*(47036*1/500000);

Please Wait...