But then again, I could be wrong.

[masterchiefo]

Homework Statement

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 .

Homework Equations

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.

 

[ehild]

Homework Statement

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 .

Homework Equations

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

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

 

[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 k_{1, 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)

 

[masterchiefo]

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

oops, made a mistake, but is that a good answer ?
y(t)+z(t)=t+c-x(t)

and for question 2 how do I do it?

 

[ehild]

oops, made a mistake, but is that a good answer ?
y(t)+z(t)=t+c-x(t)

and for question 2 how do I do it?

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.

 

[masterchiefo]

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.

this is the first equation ?
dx/dt + 3x = 2 + thow am I supposed to substitute ?

 

[ehild]

this is the first equation ?
dx/dt + 3x = 2 + thow am I supposed to substitute ?

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.

 

[masterchiefo]

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.

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 ?

 

[ehild]

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

 

[Mark44]

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

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

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

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.

, then resolve
Equation ( E) knowing that it validates the initial condition x (0) = 1 .

 

[masterchiefo]

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

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.

 

[ehild]

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.

How did you get the equation in red?

Start with the original equations in the problem.
dy/dt (x - y)
dz/dt (x - z)
Subtract them. What do you get?

 

[masterchiefo]

How did you get the equation in red?

Start with the original equations in the problem.
dy/dt (x - y)
dz/dt (x - z)
Subtract them. What do you get?

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

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

 

[ehild]

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

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

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?

 

[masterchiefo]

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?

I don't understand how it becomes d/dt(y-z)+(y-z)=0
when I subtract them I get (x - y)-(x - z) = z-y

 

[ehild]

I don't understand how it becomes d/dt(y-z)+(y-z)=0
when I subtract them I get (x - y)-(x - z) = z-y

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

 

[masterchiefo]

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

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

 

[ehild]

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

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

 

[masterchiefo]

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

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

 

[ehild]

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

What initial condition does it mean for v?

 

[masterchiefo]

What initial condition does it mean for v?

v(t)=y(t)-z(t)

v(0)=0

 

[ehild]

v(t)=y(t)-z(t)

v(0)=0

So what is v(t)?

 

[masterchiefo]

So what is v(t)?

v(t)= 0
y(t)-z(t)=0
y(t)=z(t)

You are awesome, thank you so much for the help.
Last thing, I want to push this problem a little bit.
now since we know that y(t) is equal to z(t), if I want to find the equation for them.

y(t)= 1 + t - x(t) - z(t)
z(t)= 1 + t - x(t) - y(t)

is this how I do it?

 

[ehild]

v(t)= 0
y(t)-z(t)=0
y(t)=z(t)

You are awesome, thank you so much for the help.
Last thing, I want to push this problem a little bit.
now since we know that y(t) is equal to z(t), if I want to find the equation for them.

y(t)= 1 + t - x(t) - z(t)
z(t)= 1 + t - x(t) - y(t)

is this how I do it?

Not quite.
You know that y(t)=z(t), and y(t)+z(t)=2y(t) =1+t-x(t). You also know x(t). What is y(t) then?

 

[masterchiefo]

Not quite.
You know that y(t)=z(t), and y(t)+z(t)=2y(t) =1+t-x(t). You also know x(t). What is y(t) then?

x(t)= 1/9 *(4/e^3t+3t+5)
2y(t) = 1+t- x(t)

2y(t) = 1+t-1/9 *(4/e^3t+3t+5)

y(t) = (1+t-1/9 *(4/e^3t+3t+5))/2
and
z(t) = (1+t-1/9 *(4/e^3t+3t+5))/2

both same equation.

 

[epenguin]

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

Now for Question 3:
.

There is some confusion here, so if anything is clear we should see it stated. If the differential equation for x is solved please tell us the solution.

Also because the rest is easy - once we know x the rest is just algebra. We have an algebraic relationship between x, y, z and t and also y=z as pointed out previously.

 

[ehild]

x(t)= 1/9 *(4/e^3t+3t+5)
2y(t) = 1+t- x(t)

2y(t) = 1+t-1/9 *(4/e^3t+3t+5)

y(t) = (1+t-1/9 *(4/e^3t+3t+5))/2
and
z(t) = (1+t-1/9 *(4/e^3t+3t+5))/2

both same equation.

It looks correct, but can be simplified a bit.

 

[masterchiefo]

There is some confusion here, so if anything is clear we should see it stated. If the differential equation for x is solved please tell us the solution.

Also because the rest is easy - once we know x the rest is just algebra. We have an algebraic relationship between x, y, z and t and also y=z as pointed out previously.

x(t)= 1/9 *(4/e3t+3t+5)

 

[epenguin]

Is that same thing as x(t) = (1/9)* (4e^-3t + 3t + 5) and have you checked it works?

 

[ehild]

Is that same thing as x(t) = (1/9)* (4e^-3t + 3t + 5) and have you checked it works?

See post #25. It is the same.