In this segment, we'll talk about finding an exact solution of ordinary differential equation which
has polynomial forcing function. So, I'm going to take this example this is a first-order ordinary
differential equation it has only first derivative terms in there, and if you look at the right-hand
side which is 16x squared the forcing function it is a polynomial. Of course, we are given
the initial condition y sub 0 equal to 5, and we need to figure out what y is as a function
of x. That's the goal. So, how do we go about doing this is? We're going to use the classical
solution technique in this particular example, which means that we have to find the homogenous
part and we have to find the particular part. And once we have found these two, we will be able to
find out what the value of what is the expression for dependent variable of Y as a function of the
independent variable X. So, let's go and find out what the homogenous part. The homogenous
part of the solution will be, let's look at the characteristic, should be 10m to the power 1 that
is this is the first derivative so it's m to the power 1 plus 4 m to the power 0 because that's the
zeroth derivative right here. Will be equal to 0. So, what we get is 10m 10 times M plus 4 is equal
to 0 and we get M is equal to minus 4 divided by 10 which is minus 0.4, and hence the homogeneous
part is k e to the power to the power minus 0.4x . Let’s go and see how we can find the particular
part. For the particular part what you have to do is you have to think about hey what is the form
of the right-hand side or the forcing function, and what are all its possible derivatives. That
becomes your form of the particular part of the solutions so it will be ax squared and what is
the derivative of X square it is 2x so it will be of the form B times X what is the derivative of X
it's 1, so it's a constant, what is the derivative of a constant? Is 0. so there's no need to add
that so that is the form of the particular part of solution always that's going to be the case except
for an exception where if this any part of this is any part of the homogenous part, then you have to
look at next independent solutions, but that's not the case here so let's not muddle anything here,
so that is your particular part of the solution. The question now is to how to find A, B, and C so
now to find A, B, and C what we basically do is we plug it back into our differential equation
to be able to find. So, what we had was 10 dy by dx plus 4y is equal to 16x squared and we
have our particular solution to be Ax squared plus Bx plus C form so we're going to plug it
back in here I'm gonna say 10 d by dx is equal to Ax squared plus Bx plus C plus 4 Ax squared
plus Bx plus C will be equal to 16x squared. So, you might now think that hey yes surely I'm able
to set up an equation now to find A, B and C but there's only one unknown a one equation and three
unknowns. How do I go about finding A, B, and C? Let's first find the derivative so we get 10 times
2Ax plus B plus 4 times Ax squared plus Bx plus C is equal to 16x squared. So, somehow what we do
is we have to find out what A, B, and C are, but we only have one equation at hand. So, the way to
do it is to let's see what the coefficients of the different terms of the polynomial are what is the
coefficient of x squared what is the coefficient of x and what is the coefficient of the constant
terms so that we can equate those and maybe we'll get three equations three unknowns. So, what I'm
going to do is I'm going to right x squared here and what are the terms of x squared right here 4A
that's the only x squared term. Let's look at X terms I get an X term from here which is 20A an
X term from here which is 4B and let's collect all the constant terms right here… here which is 10 B
and here I have 4C is equal to 16x squared. Now, what I can do is I can write this as 0 times X
plus 0 now how does that help me it helps me that hey if I equate the coefficients of x squared term
if I equate the coefficient of the X term and if I equate the coefficients of the constant term,
there are now three equations three unknowns which basically implies that 4A is equal to 16,
20A plus 4B is equal to 0 and 10B plus 4C is equal to 0. So, I basically have my three equations and
three unknowns. And I should be able to find A, B, and C. If you look at equation number one you
directly have 4A is equal to 16, which gives you A is equal to 16 divided by 4 and that's 4. Let's
look at a second equation, we have 20A plus 4B equal to 0 so 20, A is 4, plus 4B is equal to 0,
and then that gives me 80 plus 4B is equal to 0 and what does that give me? So, I have 80 plus 4B
is equal to 0 so for B is equal to minus 80 and B is equal to minus 80 divided by 4 which is minus
20 let's go and find/use equation 3. Now, equation 3 is 10B plus 4C is equal to 0. I just found B to
be minus 20, and I will be able to find what C is from here, so I get minus 200 plus 4C is equal to
0, 4C is 200, C is 200 over 4 and that's 50. So, by solving these three equations three unknowns
I have been able to find out what the what the values of A, B, and C are so my particular part
of solution is Ax squared plus Bx plus C. A turned out to be how much? 4. B turned out to be minus 20
and C turn out to be 50 so this is my particular part of the solution. Now, in order to find the
full solution y is equal to yh plus yp, and what is yh? We already found out the homogenous part
of the solution is something like this, the particular part of solution we just found out,
and that gives me the complete solution. However, I still have to find out what the value of K is.
And the value of the K can be found by using our initial condition. So, our initial condition was
given as Y sub 0 is equal to 5. So, Y we know is Ke to the power minus 0.4x plus 4x squared minus
20x plus 50 I am rewriting it because of my new screen here so Y sub 0 is 5 will be equal to Ke
to the power minus 0.4 times 0 plus 4 times 0^2 minus 20 times 0 plus 50. So, if that's the case
I get 5 is equal to K plus 0 minus 0 plus 50, and that gives me K is equal to 5 minus 50, and that's
minus 45. And, now, what that means is that I have found out all the things, which I need to find.
So, Y will be equal to minus 45 because that's the value of K, e to the power of minus 0.4x
plus 4x squared, minus 20x, plus 50. And that is my complete solution to this ordinary differential
equation. If you have to find the value of y at a particular value of x, all you have to do is to
substitute that value of x and you will get your value of y. And that is the end of this segment.