@Preben Alsholm 

Thanks

Now when i run i get this answer

5.275440947 10^6

What means this number

Thanks

@Preben Alsholm 

1. int(1-pf, t=0..infinity) in terms of yrj.

1. int(1-pf, t=0..infinity) in terms of urd.

"t" is the variable of integral. I mean "yrj" and "urd" should appear as constant in final result. the value of them should not be replaced.

Thanks

@Preben Alsholm 

Now I want two extra outputs:

One in terms of yrj and the other in terms of urd. I mean

1. int(1-pf, 0..infinity) in terms of yrj

2. int(1-pf, 0..infinity) in terms of urd

In general output in terms of parameters.

Thanks

@Athar Shahabinejad 

Now I want two extra outputs:

One in terms of yrj and the other in terms of urd. I mean

1. int(1-pf, 0..infinity) in terms of yrj

2. int(1-pf, 0..infinity) in terms of urd

Can you help me?

Thanks

@Preben Alsholm 

Thanks

With that I get the output I said.

but I change the original code.  and I get the same output as you said.

@Preben Alsholm 

Thanks

however when I execute this output is displayed:

-signum(pf-1) infinity

in infinity pf is 1 but 1-pf is zero.  So the output should converges.

Thanks

@Preben Alsholm 

 oOops. Thanks.

this is the correct.

sys_ode := diff(ph(t),t) = (1-yc)*pc(t)+yh*prj(t)+urd*prd(t)+ugd*pgd(t)-yc*ph(t), diff(pc(t),t) = yc*ph(t)-(2-yc)*pc(t), diff(pa(t),t) = ya*pc(t)-pa(t), diff(prj(t),t) = yrj*pa(t)-prj(t), diff(prd(t),t) = -urd*prd(t)+yrd*pa(t), diff(pgd(t),t) = -ugd*pgd(t)+ygd*pa(t), diff(pf(t),t) = (1-ygd-yrj-yrd)*pa(t)+(1-yh)*prj(t)+(1-ya)*pc(t);

ics := ph(0) = 1, pc(0) = 0, pa(0) = 0, prj(0) = 0, prd(0) = 0, pgd(0) = 0, pf(0) = 0;

@Preben Alsholm 

OK.

this is the correct system

sys_ode := diff(ph(t),t) = (1-yc)*pc(t)+yh*prj(t)+urd*prd(t)+ugd*pgd(t)-yc*ph(t), diff(pc(t),t) = yc*ph(t)-(2-yc)*pc(t), diff(pa(t),t) = ya*pc(t)-pa(t), diff(prj(t),t) = yrj*pa(t)+prj(t), diff(prd(t),t) = -urd*prd(t)+yrd*pa(t), diff(pgd(t),t) = -ugd*pgd(t)+ygd*pa(t), diff(pf(t),t) = (1-ygd-yrj-yrd)*pa(t)+(1-yh)*prj(t)+(1-ya)*pc(t);

ics := ph(0) = 1, pc(0) = 0, pa(0) = 0, prj(0) = 0, prd(0) = 0, pgd(0) = 0, pf(0) = 0;

@Rouben Rostamian

those are the diff equations of a marcov chain. I  check several times. I think there's no problem.

Are you familiar with markov chain?

if yes I upload my design.

@Preben Alsholm 

oOops...

ph, pc, pa,prj, pgd, prd, pf

all of them are probability. So they should be >0 and <1...

but here they grow with time with high speed....

Should I change parameters?

but with setting all of them 1 by you, answer doesn't change significantly....

@Markiyan Hirnyk 

thanks.

I need the the answer that converges...

can you help me.

I upload the mw file in som comments.

@Preben Alsholm 

thanks

yes, that is for all values of time, t.

@Preben Alsholm

I forget to mention the initial condition of "ph+pc+pa+pgd+prd+prj+pf=1".

this may effect the convergence of integral.

according to what Markiyan Hirnyk 6431 said one of answers converges. I need it.

Thanks

@Preben Alsholm 

Thanks.

Can you upload your worksheet?

I need these parameters:

params:={yc=1/3600,yh=1,yrd=1/180,ygd=1/180,yrj=1/180,urd=0.8,ugd=0.8,ya=1/180};

Isn't there any solution that converges in infinity?

dsolve.mw

@Rouben Rostamian