# Thread: System of differential equations

1. ## System of differential equations

x'(t)=-ax(t)
y'(t)=ax(t)-by(t)

The question is to find x(t) and y(t).
I managed to find x(t):

x(t)=exp(-at)

but I have trouble finding y(t).

I know the solution is y(t)=(a/(b-a)) (exp(-at)-exp(-bt))

I would appreciate very much your help. Thank you!

2. ## Re: System of differential equations

First find the homogeneous solution, then find the specific solution.

3. ## Re: System of differential equations

First, your solution for the first equation is wrong. It should be x(t)= C exp(-at) where C can be any constant. With that x(t), the second equation becomes y'(t)= aCexp(-at)- by. That is the same y'- by= aC exp(-at). That is a linear first order differential equation. There is simple formula for an "integrating factor". Do your textbook have that formula? (Look up "integrating factor" in the index.)

Or you can use the method that SlipEternal refers to. The "associated homogeneous differential equation" is y'- by= 0. That is the same as dy/y= bdx. Integrating both sides, ln(y)= bx+ C. Taking the exponential of both sides gives y= e^(bx+ D)= D'e^(bx) where D'= e^D.

Then, to find the general solution to the entire equation we only have to add any specific equation to the differential equation. Since the "non-homogeneous" part is $\displaystyle aC e^(-at)$, and I know that the derivative of an exponential is just that exponential again, I would try a function of the form $\displaystyle y= He^{-at}$ where H is a constant to be determined. $\displaystyle y'= -aHe^{-at}$. Put those into the equation and you will have a simple equation to solve for H.

4. ## Re: System of differential equations

I used the integrating factor model and obtained this, I am not sure whether I'm correct or not.

y'(t)=aCexp(-at)-by(t)
y'(t)+by(t)=aCexp(-at) (1)
I=exp(integral(b)dt
I=exp(bt) (2)
Multiply I on both sides of (1)

Iy'(t)+Iy(t)= IaCexp(-at)

exp(bt) y'(t)+exp(bt) by(t)=exp(bt) aCexp(-at)

exp(bt) y(t)= integral (exp(bt) aCexp(-at)dt)

exp(bt) y(t)=aC/(b-a) exp(bt-at)
y(t)=aC/(b-a) exp(-at)

5. ## Re: System of differential equations

If you put that x(t) and y(t) into the original equations, are they satisfied?

6. ## Re: System of differential equations

Unfortunately they are not satisfied. I can't really see the mistake I made.

7. ## Re: System of differential equations

Alright. You are attempting to solve the differential equation $\displaystyle y'- by= aCe^{-at}$. An integrating factor is [tex]\phi(t)= e^{-bt}[tex]. $\displaystyle e^{-bt}\left(y'+ by)= (e^{-bt}y)'= aCe^{-(a+b)t}$. Integrating both sides, $\displaystyle e^{-bt}y= -\frac{aC}{a+ b}e^{-(a+b)t}+ D$ so that $\displaystyle y(t)= -\frac{aC}{a+b}e^{-at}+ De^{bt}$.

You keep forgetting the "constant of integration"!

8. ## Re: System of differential equations

I understand but I'm having trouble understanding how y(t)= -aC/(b-a) exp(-at) + Dexp(-bt) is equal to

y(t)=(a/(b-a)) (exp(-at)-exp(-bt)). (This is the solution obtained from the book).

9. ## Re: System of differential equations

Originally Posted by mathfn
I understand but I'm having trouble understanding how y(t)= -aC/(b-a) exp(-at) + Dexp(-bt) is equal to

y(t)=(a/(b-a)) (exp(-at)-exp(-bt)). (This is the solution obtained from the book).
It isn't. However, it is if you take C= 1 and D= -1. Those "undetermined constants", C and D, have to be determined by some additional conditions. What were those additional conditions.

10. ## Re: System of differential equations

I think there are boundary conditions at t=0. One should therefore integrate from 0 to t.

11. ## Re: System of differential equations

We can't help with a problem when you refuse to tell us what the problem is!

12. ## Re: System of differential equations

This is the original problem. It's a Markov process problem and I asked about part (a). I replaced α and β by a and b respectively, p_0(t) and p_1(t) by x(t) and y(t) respectively for simplicity.