# A differential Problem

1. Oct 18, 2015

### masterchiefo

1. The problem statement, all variables and given/known data
Chemical reactions being studied in which a body A undergoes transformations
according to the following scheme:
http://prntscr.com/8shuvb

k1, k2, k3 , k4 are the rate constants .
We denote x (t ), y ( t) , z (t) the respective concentrations of the products A, B, C at a given time t
( t expressed in minutes).
The initial conditions x (0) = 1, y (0) = 0 and z ( 0) = 0 .
Is arranged above the vessel where the reaction takes place by a burette which is poured
product A at a constant speed in the tank. Under these experimental conditions,
functions x , y, z defined on the interval [0 ; + infinite [ check the following differential system :

dx/dt (1-2x +y+ z)
dy/dt (x - y)
dz/dt (x - z)

Question 1:
Calculate d/dt( x + y + z) and , using the initial conditions, deduce that :
y(t) + z(t) = 1 + t - x(t)
Question 2:
Demonstrate that x is a solution of the differential equation (E) : dx/dt + 3x = 2 + t, then resolve
Equation ( E) knowing that it validates the initial condition x (0) = 1 .

2. Relevant equations

3. The attempt at a solution

Question 1:
dx/dt+dy/dt+dz/dt= 1-2x+y+z+x-y+x-z
dx/dt+dy/dt+dz/dt= 1

integral dx/dt+ integral dy/dt+ integral dz/dt= integral ( 1 dt)
x(t)+y(t)+z(t)=t*c
y(t)+z(t)=t*c-x(t)

Question 2:
dx/dt+3x=2+t
dx/dt =2+t-3x

dx/dt(x) =1
1= 2+x-3x
1=2+-2x
... does not work. :(
I am not sure what to do here, some tips would help.

2. Oct 18, 2015

### ehild

You need an additive constant when integrating, not a multiplicative one!

3. Oct 18, 2015

### epenguin

Rather a weird question so I wonder if you have reproduced it properly.

First substances are called A, B, C then the same substances are called x , y, z.

Then four rate constants k1, 2, 3, 4, are mentioned, afterwards they have all become 1.

The rate of addition is also 1.

OK if the model has been so simplified, you can simplify it a bit further yourself: since y and z are formed and decompose at exactly the same rate and are initially at the same concentration, y = z at all times it is sufficient to formulate e.g.

Constant supply rate 1 ⇒ x ⇔ w

where the forwards and backward rate constants are both 2; at the end of the calculation get y and z back from y = z = w/2.
____________________

(Corrected)

Last edited: Oct 18, 2015
4. Oct 18, 2015

### masterchiefo

y(t)+z(t)=t+c-x(t)

and for question 2 how do I do it?

5. Oct 18, 2015

### ehild

It is correct. Find the value of c from the initial conditions.
Substitute y(t)+z(t) back into the first equation, you get the desired differential equation for x(t).
It is a first-order linear equation, solve with one of the standard methods you certainly have learnt.

6. Oct 18, 2015

### masterchiefo

this is the first equation ?
dx/dt + 3x = 2 + t

how am I supposed to substitute ?

7. Oct 18, 2015

### ehild

I meant y(t)+z(t) substituted into dx/dt = (1-2x +y+ z).
You should get the equation dx/dt + 3x = 2 + t, but you have to prove it.
The next task is to solve this differential equation and give x(t) using the initial condition x(0)=1.

8. Oct 18, 2015

### masterchiefo

dx/dt=1-2x+(y+z)
dx/dt=1-2x+(1+t-x(t))
dx/dt=2-2x+t-x(t)
dx/dt=2-3x+t
dx/dt+3x=2+t

by doing this I prove that X is a solution ?
and now I just resolve this as a normal linear differential equation ?

9. Oct 18, 2015

### ehild

You derived the de for x(t). Resolve it as a normal linear first order differential equation.

10. Oct 18, 2015

### Staff: Mentor

A differential system is a system of equations, each of which should have = in it somewhere.
Is this the system?
dx/dt = 1 - 2x + y + z
dy/dt = x - y
dz/dt = x - z
This doesn't make any sense to me. You need to have x as some function of t. The differential equation you show above (equation E) is a first-order, nonhomogeneous equation. One technique for solution is to find an integrating factor.

11. Oct 18, 2015

### masterchiefo

okay, 1 and 2 is done thank you very much. This problem confusing me a lot.

Now for Question 3:
Prove that d/dt(y - z) + y - z = 0 and deduce that y = z
d/dt(y - z) = 0 so,.
y-z=0
y = z

Not sure if I am doing this right.

12. Oct 18, 2015

### ehild

How did you get the equation in red?

dy/dt (x - y)
dz/dt (x - z)
Subtract them. What do you get?

13. Oct 18, 2015

### masterchiefo

(x-y)-(x-z)= -y+z

d/dt(y - z) + (y - z) = 0
d/dt(-y+z) + (-y+z) =0

14. Oct 18, 2015

### ehild

Sorry, the original equation missed the equal signs.

So it should have been
dy/dt= (x - y)
dz/dt =(x - z)
When you subtract them, it becomes d/dt(y-z)+(y-z)=0
You can consider y-z as a new function v = y-z, and you have the differential equation dv/dt+v=0. What is the solution for v(t) and what is the initial condition?

15. Oct 18, 2015

### masterchiefo

I dont understand how it becomes d/dt(y-z)+(y-z)=0
when I subtract them I get (x - y)-(x - z) = z-y

16. Oct 18, 2015

### ehild

Subtract both sides. On the left sides, you have the derivatives of y and z : dy/dt and dz/dt. What is their difference?

17. Oct 18, 2015

### masterchiefo

ah Okay,

so its like this: dy/dt - dz/dt = x-y-(x-z)

dy/dt - dz/dt =-y+z
dy/dt - dz/dt +y-z =0
d/dt(y-z) + ( y-z) =0

dv/dt +v =0

I find v(t)= C * e-t

18. Oct 18, 2015

### ehild

Good. What were the initial conditions for y and z?

19. Oct 18, 2015

### masterchiefo

y (0) = 0 and z ( 0) = 0

20. Oct 18, 2015

### ehild

What initial condition does it mean for v?