Re: In general how to convert a system of 1st order diff equations into higher order diff equation for 1 variable and vice versa?

Thanks a lot Robert.

I digest your pointers. They are really good general methods.

Following your thoughts, I got:

(x1)''' - 6*(x1)'' + 11* (x1)' - 6 x1 = -10 + 36*exp(-t)

x1(t)=c1*exp(t)+c2*exp(2*t)+c3*exp(3*t)+5/3-3/2*exp(-t)

Now in order to find x2(t) and x3(t), enlightened by you, I found that
there is no further integration needed. I just need to find the unknown
coefficient A in
(x1)'' + A*(x1)' , to make x3 terms vanish; then I got

x2(t)=-c1/2*exp(t)-c2/3*exp(2*t)-c3/2*exp(3*t)-4/3+1/4*exp(-t),

Similarly, by finding the unknown coefficient A in
(x1)'' + A*(x1)' , to make x2 terms vanish, I got

x3(t)=c1/2*exp(t)+2/3*c2*exp(2*t)+c3*exp(3*t)+2/3-1/2*exp(-t),

then I found c1, c2, c3 by initial values...

Is this the fastest approach?

This is great! It worked perfectly! And it is general so I can apply
the method to other system equations...

Thanks alot

.