Good afternoon every one. So, we are in this
class, we will be looking into the specific
examples of Eigenvalue problems and how this
Eigenvalue problems will be used for solving
the chemical engineering application oriented
process equations.
So, we have already seen one example in the
last class - how to solve a non-homogeneous
set of non-homogenous algebraic equations.
We have taken up two examples: in the first
example the Eigenvalues were real and the
solutions an eigen and all the elements of
the Eigenvectors are real and you are getting
a real solution of the system. In the next
example, we have considered a problem where
the Eigenvalues were imaginary; so, therefore,
the elements of the Eigenvectors are also
imaginary, but ultimately you will be getting
a real valued solution, as a, as a final answer,
because the chemical engineering problem that
was - that we have considered, they are basically
real of a real valued.
Next, we have taken up the theory for solution
of a set of homogeneous or in differential
equations. And we have taken up one example,
we have developed the theory for it, and we
have taken up an example how to solve them,
and we are almost at the last step of the
solution.
So, if you remember that we have the initial
value set of the conditions where 2 and 3;
these are the initial value problem, at time
t is equal to 0; these are there values of
the, of variable. And we are looking into
the solution and that for the solution you
are going to get the coefficients C i naught
and these coefficients are defined as the
inner product of X naught and Y i and this
is divided by inner product of Z i and Y i.
If you remember that Y is are the Eigenvectors
of coefficient matrix 
transpose of coefficient matrix A transpose
and Z i where the Eigenvectors of A.
So, therefore, this will be nothing but X
0 transpose Y i Z i transpose Y i. And C 1
0 becomes X 0 inner product of X 0 Y 1divided
by Z 1inner product of Z 1 and Y 1; so, this
will be X 0 transpose Y 1 divided by Z 1 transpose
Y 1. So, you have the coefficient C 1 0, if
you remember X 0 transpose, where it was 2
3, it is the transpose of the vector 2 3 and
Y 1 is nothing but eigenvector 1 1 divided
by Z 1was the vector 2 1 and this will be
Y 1 will be 1 1. And this will be 2 plus,
2 into 1, 2 plus, 3 into 1, 3 divided by,
2 into 1, plus 1 into 1; it will be 5 by 3.
Similarly, we have, we had seen, we have already
seen that C 2 naught will be the inner product
of X 0 T Y 2 divided by Z 2 T Y 2. And this
will be 2 3 and it will be Y 2 will be 1 minus
2 divided by 1 minus 1 and 1minus 2; and ultimately
it will be 2 minus 6 divided by 1 plus 2;
so, it will be minus 4 by 3. So we are able
to estimate the coefficient C 1 and C 2 and
then we have solved almost everything and
the, all the quantities are known to you.
Now, we will write down the final solution,
the final solution will be, if you remember
the final solution is in this form, final
solution becomes x 1 x 2 is equal to C 1 0
e to the power lambda 1 t Z 1 plus C 2 0 e
to the power lambda 2 t Z 2. And if we remember
the thing becomes the lambda 1 trans out to
be, we just put the values of lambda 1 and
lambda 2, so this becomes x 1 x 2 is equal
to C 1 0 e to the power lambda 1 t and this
becomes Z 1 was 2 and 1 the vector of 2 and
1 and this becomes C 2 0 e to the power minus
3 t, and Z 2 was 1 minus 1.
So, therefore, the x 1 and x 2 becomes put
the value of C .1 C 1 is 5 by 3, lambda 1
was 0, so it will be one and this will be
2 1 plus C 2, C 2 will be minus 4 by 3 e to
the power minus3 t 1 and minus 1. So, therefore
we can construct the final solution as x 1
is equal to 5 by 3 into 2 minus 4 by 3 e to
the power minus 3 t times 1; so, this becomes
10 by 3 minus 4 by 3 e to the power minus
3 t. And x 2 becomes 5 by 3 into 1 minus 4
by 3 e to the power minus 3 t multiplied by
minus 1; so, it will be 5 by 3minus plus 5
by 3 e to the power minus 3 t; so that is
the final solution of, what are the problem
we are looking into. So, this gives a direct
application of Eigenvalue, eigenvector problem,
to solve the set of ordinary differential
equations which are homogenous.
Now, let us talk about the advantage of this
method, compare to any numerical scheme. If
you look into the numerical scheme 
for the numerical scheme, then you could have
preferred any integration subroutine to solve
this type of problem; these subroutines include
Runge Kutta 4, forth order or Euler's equation,
Euler's method or Adam- Bashforth method or
predictor corrected method.
So, therefore, but all these numerical schemes
will be having the disadvantage of instability,
there may be numeric, if the, where the step
size of numerical integration is not properly
selected that instability may be a problem.
If the equations are stiff equations, for
stiff ordinary differential equations, none
of this method will be, can be utilized; then
you have to use the special algorithm like
Gear's, Gear's method may be used. The stiff
ODE, means, if the variation of dependent
variable, a small variation of independent
variable gives a large defect variation of
the dependent variable, so that gives a, that
is called whether stiff ordinary differential
equation.
So, you require special algorithm for them;
so, you could have of all these short coming
can be overcome, by using this Eigenvalue
method. Eigenvalue - eigenvector method 
that gives us more accurate and elegant method
of solution trouble free; so, what are the
various subroutines, one has to write, one
has to have a subroutine of evaluation of
the Eigenvalues that is not a problem; one
can use the Honor's method, they, one can
use the eigen the subroutines for evaluation
of Eigenvectors. So, Eigenvalues - Eigenvector
then matrix transpose of, you have to write
a couple of line routines for matrix transpose
and then one has to go for the simple calculations
like n matrix multiplication.
So, using four such basics subroutines, one
can utilize the Eigenvalue - Eigenvector method,
in order to solve the complicated problems
of linear ordinary differential equation,
in chemical engineering. And can do away with
the a numerical, the complicacies or you know
short comings of numerical integration.
Next, we talk about the non-homogenous solution
of, non-homogenous first or first order ordinary
differential equation, but it is, they are
non-homogenous; in earlier case we have solve
the homogenous. And in this case we will be
taking about the solution of non-homogenous
first order ODEs; of course they are IVPs.
In fact, even boundary value problem can be
broken down into series of ordinary differential
equations, by assuming the unknown, the initial
condition and then matching with the corresponding
boundary condition anyway.
So, let us write down the set of non-homogenous
ODEs in a compact form, in a compact form
and that will be d x d t is equal to A x plus
b of t. In general this term corresponds to
non-homogenous term. So, this can be the non-homogenous
term, can be constant term or it can be function
of time, but this cannot be function of X;
then, this term, the extra term in the additional
term on the right hand side is called a non-homogenous
term, then this set of equation has to be
solved by subject to the boundary condition,
the initial conditions that, at X vector at
time t is equal to 0 is given by this represented
by this vector x naught.
Now, A is a real non-singular matrix 
and we know how to evaluate the Eigenvalues
and Eigenvectors of this matrix; and for this
matrix, you can evaluate the Eigenvalues.
And let us say Z I, forms a set of vectors
which are nothing but the Eigenvectors. And
since Eigenvectors are, they form the independent
set of vectors, we have already proved earlier
these are independent set of vectors and they
constitute a basis set. Therefore any other
vector in the domain can be expressed as a
linear combination of the basis set vectors
or the Eigenvectors of the matrix A.
Now, we assume the vector b t, b which is
a function of t is expressed as a summation
of beta I, as a summation of, as a linear
combination of this vector the i Eigenvectors
Z i. And the solution X t is expressed as
combination of these Eigenvectors Z I. So,
we have to assume the form of the solution
as X t is equal to Z e to the power lambda
t would have proved that, if this is the form
of the equation, a form the solution, we have
proved earlier, that lambda i corresponds
to Eigenvalues of A and Z i correspond to
the Eigenvectors of A.
So, therefore, we have the Eigenvalue problem
A Z i is equal to lambda i Z I; so, lambda
i is i th Eigenvalue and Z i is the corresponding
eigenvector and this Eigenvalue problem is
defined.
So, next we write down the equation d x d
t is equal to A x plus b. And write down the
solution, write down the expression of X,
X is summation of C i Z I; so, therefore this
will be summation d C i d t, if you remember
C i was a sole function of time, so it becomes
d c e i d t Z i is equal to A summation of
C i which is a function of t Z i plus summation
of beta i times Z i. So, therefore one can
bring this up on the left hand side and put
everything under the summation, so these becomes
d C i d t Z i minus summation of C i t A Z
I, so matrix is an operator, so operate on
Z plus minus summation of beta i Z i will
be equal to 0. So, we have summation of d
c d t Z i and we identify that this A Z i
is nothing but lambda i Z I, so therefore
this becomes C i t lambda i Z i minus summation
beta i Z i is equal to 0 and all these summations
are over the index i runs from 1 to n for
n-dimensional problem.
Next, we bring everything under the summation,
so it becomes so d C i d t minus lambda i
c i minus beta i times Z i equal to 0 and
if we remember that what this Z i's are this
Z i's are Eigenvectors and they are independent
vectors and they form a basis set. So, therefore,
if these vectors are independent vectors and
they are in this form the coefficient of all
these vectors has to be independently equal
to 0, in order to satisfy this equation.
So, therefore, we have the condition d C i
d t minus lambda i C i minus beta i should
be equal to 0; in order to have, in order
to satisfy this equation because the Eigenvectors
are set of independent vectors, to satisfy
this equation all the corresponding individual
coefficient must be equal to 0.
So, we get an ordinary non-homogeneous first
order equation, non-homogeneous first order
ODE as d C i d t as a condition d C i d t
minus lambda i C i minus beta i, which is
in general function of time.
Now, this is a non-homogeneous PDE, this method
this ODE can be non-homogenous ODE ordinary
differential equation, this ODE can be solved
by the method of integrating factor, integrating
factor. So, we multiply both the sides by
integral e to the power e to the power minus
lambda i d t, this is the integrating factor,
so this becomes e to the power minus lambda
i t; so, you multiply both side by e to the
power minus lambda i t, so this becomes e
to the power minus lambda i t d C i d t minus
lambda i e to the power minus lambda i t C
i minus beta i e to the power minus lambda
i t is equal to 0.
So, we can combine these two parts, so it
will be, d d t of e to the power minus lambda
i t times c is equal to beta i e to the power
minus lambda i t. So, we can integrate this
out, so it will be 0 to t beta e is in general
function of time, so 0 to t d t, so we can
we should write it in the next step. So, it
will be in the next step, we will be getting,
0 to t d of e to the power minus lambda i
t times C is equal to integral beta i t e
to the power minus lambda i t d t 0 to t.
So, we can get this integral evaluated, so
this will be giving you that, after evaluation
this integral will be getting C I, we will
be having the sub script i in all the cases
of C i of the c's; so, therefore, after integration
we will be getting C i e to the power minus
lambda i t minus C i at time t is equal to
0, at time t is equal to 0 e to the power
0 becomes one is equal to 0 to t beta i t
e to the power minus lambda i t d t. And you
will be getting C i as a function of time
as C i naught e to the power lambda i t plus
e to the power lambda i t 0 to t beta i e
to the power minus lambda i t d t.
So, we will be getting these expressions as
the final solution. And let us see how the
C i naughts are obtained C i naught, the vector
C i naught is obtained from the initial condition,
obtained from initial condition; and what
is initial condition, we know the vector X
naught, that means, all the solutions were
known as the initial condition at initial
time t is equal to 0. So, these vector can
be expressed as a linear combination of independent
vectors Z I; so, therefore, one can obtain
this by combining this with the Eigenvectors
of A transpose.
So, now, if Y i is a set of Eigenvectors for
A transpose, then inner product of X naught
and Y 1 should be given as you take the inner
product of this equation with respect to Y
1, so this becomes C i naught inner product
of Z i and Y 1; in fact if you open up this
series, then we will be getting C 1 0 inner
product of Z 1 Y 1 plus C 2 0 inner product
of Z 2 Y 1 plus like that and C n 0 inner
product of Z n Y 1.
Now, Z 1 and Y 1, Z we have already proved
that Eigenvectors of A and A transpose will
form an a bi orthogonal set, so inner product
of Z 2 and Y 1 will be 0, inner product of
Z and Y 1 will be Y 1 equal to 0; so, only
one term will survive in this summation that
will be C 10 inner product of Z 1 and Y 1.
So, in that case what we will be getting as
C 1 0 is that, so you will be getting C 1
0 is equal to inner product of X naught Y1
divided by inner product of Z 1 and Y 1. And
we can get as X 0 transpose Y 1 divided by
Z 1 transpose Y 1. And ultimately you can
get the generic form of C i naught as inner
product of X naught and Y i divided by inner
product of Z i and Y i. So, you can put into
the matrix in vector multiplication form,
matrix multiplication X 0 t Y i Z i transpose
Y I; so that will be the form of coefficient,
since C is in the above solution.
And next we will be talking about the other
parts of the solution. So, the C i naught
vector will be determined by this, because
we can evaluate the Eigenvectors of A transpose,
we can evaluate the Eigenvectors of A, we
know the initial condition vector, so all
of these coefficients C i naught will be calculated.
Next is the calculation of beta I, next step
is calculation of beta i, so if you remember
the vector b as a function of time can be
expressed as linear combination of the Eigenvectors
of a as beta i Z i.
Then we can take the inner product of this
equation with respect to Y i of above equation
with respect to Y I; so, you will be getting
Y i inner product of Y and b is summation
beta i Z i and Y i beta i being a constant,
so it will be summation beta i inner product
of Z i and Y i. Again, one can open up this
summation series and see what you get so it
will be, let us considered for one, so beta
t Y i inner product of Y and b, you will be
getting beta 1 inner product of Z 1 Y 1 plus
beta 2 Z 1 Y i, so beta 2 inner product of
Z 2 Y i plus beta 3 inner product of Z 2 and
Y i, like that you will be getting beta i
inner product of Z i and y i. So, since Z
i's are the Eigenvectors of A and Y i are
the Eigenvectors of A transpose, they form
the biorthogonal set and we have already proved
that except i is equal to j, they will be
for the inner product of the individual Eigenvectors
will be equal to 0; this will vanish, this
will vanish, this will vanish, only this term
will survive out of this summation.
So, therefore, in general one can evaluate
beta i is summation of Y i b divided by inner
product of y i Z I, inner product of Z i Y
i is identical to inner product of Y i and
Z I, so these will be nothing but inner product
the multiplication of Y i transpose and the
vector b and this will be multiplication of
Y i transpose and vector Z i.
So, that is how the coefficients beta i can
be evaluated. So, once you obtain the expression
of beta i, once you obtain the expression
of C i, then you can construct the complete
solution as. So, the complete solution, if
you look into the complete solution, we wrote
C i t is equal to C i naught e to the power
lambda i t plus e to the power lambda i t
0 to t beta i e to the power minus lambda
i t d t.
So, therefore, where the coefficient C i naught
can be expressed as inner product of X naught
and Y i divided by inner product of Z i and
Y I; so, this will be X naught t Y i we have
already written this, so this will be Z i
transpose Y i. And beta i we have already
obtained as, Y i transpose b divided by Y
i transpose Z i. So, once we get this, we
can get the complete the solution of C i t
and will be getting the complete solution.
Next, we will be with this theoretical formulation
of solution of non-homogenous ordinary differential
equation by Eigenvalue - Eigenvector method,
we will be taking up one particular example
a chemical engineering application example
and see how this problem will be solved by
using this Eigenvalue - Eigenvector method
So, the example that we will be talking about
is highly relevant for chemical engineers
and this is the problem of reaction engineering.
And we will be taking up a series reaction,
considered a series reaction in a CSTR and
we have already, I have already discussed
what a continuous start tank reactor is in
earlier classes.
So, considered a series reaction 
in a CSTR, so you will be having a reaction
like this A going to B going to C and they
are elemental reaction let us say and the
rate expressions are given as first order
rate expression. The feed concentrations was
given C A F as 1gram mol per c c and initial
conditions where at time t is equal to 0 the
concentration of A, B, C where given by this
vector 1 gram mol per liter 0 gram mol per
liter and 0 gram mol per liter.
So, at time t is equal to 0, there was nothing
in the system except C, A feed that was one
gram mol per cc and the K 1 and K 2 values
are given. K 1 is given by 2 second inverse
K 2 is given as 3 second inverse. And tau
that is residence time and it is defined as
volume divided by the volumetric flow rate.
So, volume will be in meter, volumetric flow
rate will be in meter cube per second, so
it will be second and this tau will be 4 second.
So, we have now in a position to write down
the species balance equation, we write it
as tau residence, the time. So, write down
the species balance equation, this will be
d C A d t is equal to one by tau C A F minus
C A minus K 1 times C A; that is the A balance.
B balance will be one by tau C B F feed minus
C B plus k 1 C A minus K 2 C B; so, this is
the accumulation term, this is net material
in minus, net material out, minus, plus rate
of generation here. And in the case of A,
it will be consumption; in fact this is a
combination of rate of generation and rate
of consumption, because B is produced at the
rate K 1 C A and it will be consumed at the
rate K 2 times C B.
So, d C C d t can be written as one by tau
C C final, if C C in the feed minus C C plus
rate of generation K 2 C B. So, we can write
it more amenable form d C A d t will be minus
1 by tau res minus K 1 times C A plus C A
f divided by tau; d C B d t can be written
as K 1 C A plus minus 1 by tau minus K 2 times
C B plus C B feed divided by tau resistance.
And d C C d t can be written as K 2 C B minus
C C by tau plus C C f by tau; so, you will
be having the form of set of non-homogeneous
ordinary differential equation.
So, this will be 3 the dx dt part or the component
were C A C B and C C. These are the coefficient
matrix and these terms correspond to the non-homogeneous
part.
So, therefore one can get these equations,
so we can write it as d C d t in a more compact
form matrix notation minus 1 by tau minus
K 1 0 0 K 1 minus 1 by tau minus K 2 0 0 K
2 minus 1 by tau 
times C plus the vector C A f divided by tau
C Bf divided by tau and C C C f divided by
tau; where the matrix C is that the vector
C is basically given by C A C B C C; so, it
has three elements.
Now, we can write down this in a compact form
d C d tau is equal to A C plus b; where the
b vector is given by this, so this is a b
vector. And the coefficient matrix A can be
given as just put all the values, so if you
put all the values, this becomes minus 2.25
0 0 2 minus 3.25 0 0 3 minus 0.25, put all
the values there; so, you will 
be landing up with the coefficient matrix
like this. Now this is a 3 by 3 matrix, so
we can use, we can evaluate the Eigenvalues
of this matrix and we can, we can obtain the
Eigenvectors.
So, for the Eigenvalues what we do is we put
determinant of A minus lambda i is equal to
0; so, you will be getting a polynomial of
lambda and this polynomial will be a third
order polynomial and you will be getting 3
roots, those will be corresponding to the
Eigenvalues. So, lambda 1 is equal to minus
2.25; lambda 2 is equal to 2 minus 3.25 and
lambda 3 is equal to minus 0.25 and the corresponding,
so these are the Eigenvalues. Corresponding
Eigenvectors will be Z 1 and it will be 1
2 minus 3 transpose and at again deriving
it. We have already seen in the earlier examples,
how to evaluate the Eigenvalues and corresponding
Eigenvectors of a square matrix Z 2 is Z 1
minus 1 transpose and Z 3 is 0 0 1transpose.
We can get the Eigenvalues of the transpose
matrix will be identical that we have already
proved earlier. And we can get the Eigenvectors
of A transpose and this will be Y 1 is 1 0
0 transpose; Y 2 is minus 2 1 0 transpose;
and Y 3 will be 1 1 1 transpose; so, this
will be the Eigenvectors of A transpose.
So, we can write the compact notation that
d C d t is equal to minus 2.25 0 0 2 minus
three 3.25 0 0 3 minus 0.25 times the vector
C plus the vector 0.25 0 0. So, therefore,
this is the set, this is the non-homogeneous
term, this is the A, this is b; so, d C d
t equal d x d t equal to in the form of A
X plus b, this is the non-homogeneous term.
Now, we are going to solve this equation,
so our non-homogeneous vector becomes 1 by
4 0.25 0 0. And we write beta b is summation
of beta i Z i. We write X as summation of
C i Z I, where X is the solution is, X is
the solution matrix solution vector. So, therefore
you can evaluate since all these Eigenvalues
and Eigenvectors of A and A transpose have
already been develop and derived, we need
not to, you just go ahead with the solution
procedure.
So, if you write down the expression of beta
i this becomes Y i transpose times b Y i transpose
times Z i. And the expression of C i as a
function of t, if you remember this becomes
C i 0 e to the power lambda i t plus e to
the power lambda i t 0 to t beta i e to the
power minus lambda i t d t.
So, therefore, we can, if you remember the
expression of the vector C i naught the element
of the vector C i, it will be Y i transpose
X naught divided by Y i transpose Z I; where
X naught where nothing but the vector which
C A f C B f and C C f the initial values.
So, therefore C 1 0 will be nothing but Y
1 transpose X naught divided by Y 1 transpose
Z 1, so this becomes 1 0 0, this becomes 1
0 0 that is the initial value vector divided
by 1 0 0 that is Y 1 transpose and Z 1 will
be 1 2 minus 3 that is the first eigenvector
of the matrix A. And this becomes C 1 0 becomes
1 plus 1 divided by 1, so it will become 1;
so, all the other terms will be 0.
Similarly, we can evaluate that C 2 0 and
C 3 0, if you do that C 2 0 trans out to be
Y 2 transpose X naught divided by Y 2 transpose
Z 2; and this becomes minus 21010 0 and this
will be minus 2 1 0 0 1 minus 1. So, this
becomes minus 2, this becomes 1 divided by
1, so it becomes minus 2. And C 3 0 becomes
Y 3 transpose X 0 divided by Y 3 transpose
Z 3. And this becomes 1 1 1 1 0 0 divided
by 1 1 1 0 0 1 that is the third Eigenvector;
so this becomes 1 and this also 1, so it becomes
also 1.
So, now you are in a position to evaluate
the coefficient beta 1, so beta 1 becomes
Y 1 transpose times b divided by Y 1 transpose
times Z 1; so, this will be 1 0 0 that is
Y 1 transpose and b will be 1 by 4 0 0 divided
by 1 0 0 1 2 and minus 3; so, this becomes
1 by 4 divided by 1; so, this becomes 1 by
4. Similarly, one can get beta 2 as Y 2 transpose
b divided by Y 2 transpose Z 2; and this becomes
minus 2 1 0 will be 1 by 4 0 0 divided by
Y 2 transpose and Z 2, so minus 2 1 0 and
Z 2 will be 0 1 minus 1that is the second
Eigenvector; this will be equal to minus half
divided by 1, so it will be minus 1 by 2.
Similarly, we can have beta 3, beta 3 is Y
3 transpose b Y 3 transpose Z 3, so this will
be 1 1 1 1 by 4 0 0 divided by 1 1 1 that
is Y 3 transpose it will be 0 0 1; and this
will be 1 by 4 divided by 1so it will be 1
by 4. Now, you are in a position to get the
coefficient C 1 t, so C 1 t will be e to the
power minus 2.25 t plus e to the power minus
2.25 t, if we look into the solution this
becomes 0 to t 1 by 4 e to the power 2.25
t d t.
So, ultimately after getting into the integration
carried out, the solution becomes e to the
power minus 2.25 t plus 1 by 9 1 minus e to
the power minus 2.25 t. And you will be getting
C 2 t as minus 2 e to the power minus 3.25
t minus 2 by 13 1 minus e to the power minus
3.25 t. And C 3 t will be e to the power minus
0.25 t plus 1 minus e to the power minus 0.25
t.
So, we are in a position to get the coefficients
C 1, C 2, C 3, next set we are going to get
the final solution. If you remember the final
solution 
was given by the solution that vector X is
summation C i Z I; so you will be getting
C A C B C C is equal to C 1 Z 1 plus C 2 Z
2 plus C 3 Z 3. So, you have the coefficient
that we have already evaluated C 1 as a function
of time, you have just evaluated just now,
just put the value of Z 1 1 2 and 3 plus,
C 2 you have already evaluated that, 0 1 minus
1 plus C 3, we have already evaluated the
expression of C 3 and Z 3 will be 0 0 1.
So, therefore, you are in a position to get
the value of final solution. Now, we will
be putting C 1 is e to the powe,r actually
both it will be, C 1 times one plus C 2 times
0 plus C 3 times 0, so they will contribute
nothing, so it will minus 2.25 t plus 1 by
9 1 minus e to the power minus 2.25 t, that
is the expression of C 1.
And if you look into the expression of C B,
it will be 2 times C 1 plus 1 times C 2 plus
C 3 into 0, so we have already evaluated to
C 1 and C 1 and C 2 put them here, 2 e to
the power minus 2.25 t minus e to the power
2 minus e 2 e power minus 3.25 t plus 2 by
9 1 minus e to the power minus 2.25 t minus
2 by 13 1 minus e to the power minus 3.25
t; so, we can just simplify this equation
and get the final expression of C C.
The C C will be simply, 3 times C 1 minus
C 2 1 times C 2 plus 1 times C 3; so ultimately
you will be getting minus three e to the power
minus 2.25 t plus 2 e to the power minus 3.25
t plus e to the power minus 0.25 t minus 3
by 9 1 minus e to the power minus 2.25 t plus
2 by 13 1 minus e to the power minus 3.25
t plus 1 multiplied by 1 minus e to the power
minus 0.25 t; so that gives the final solution
of C C. So, we get the expression of concentration
of A B and C as a function of time.
So, we can get the steady state solution also,
steady state solution is obtained when t tends
to infinity; so, once we just put t tends
to infinity in the solution, we get the steady
state solution. And the steady state values
of C A at the steady steadies 1 by 9 molar
gram mole per liter. And C B steady state
is 8 by 117, and C C steady steadies 32 by
39.
So, one can get the transient or non-homogenous
ordinary differential equation and the Eigenvalue
- Eigenvector method can be fundamentally
utilized, to solve elegantly the set of ordinary
differential equation homogenous and non-homogenous
both and also there should be the eigen, Initial
Value Problem; so, IVP homogeneous and non-homogeneous,
ordinary differential equation can be solved
by this using this a Eigenvalue - Eigenvector
methods.
Next, in the, we will be, we can see that
since this the solution contains the form
e to the power minus lambda t; now depending
on the sign of lambda one can expect the solution,
how you get the solution grows in time or
decays with time, because the solution will
be in the form e to the power lambda t; so
sign of lambda plays a crucial role and so
therefore this Eigenvalue or Eigenvector method
plays a very crucial role in determining the
stability of ordinary differential equations.
And which are basically representing the chemical
engineering system; so, chemical engineering
systems can also be directly related to the
signs of the Eigenvalues.
So, therefore, in the, so we stop in this
class and the next class we will start looking
into the stability of the chemical engineering
system depending on the sign of the Eigenvalues.
And how these will be quantified and put in
a more formulized mathematical way to evaluate
the stability of the system. Thank you very
much.
