In this lecture, we will talk about finding exact solution of a second-order ordinary differential
equation and the special thing about the ordinary differential equation which will take is example;
its characteristic equation are going to have identical roots.
So, let's take the example of y double prime plus 4y prime plus 4y is equal to 10t.
Initial conditions are given as y of 0 is equal to 5, y prime of 0 is equal to 2. Now we know
that if we're going to solve it by the classical solution techniques that the full solution will be
equal to the homogeneous part plus the particular part. Let's go and look at the homogeneous part;
how we're going to calculate that. Homogeneous part meaning that hey it is the solution to the
differential equation with no forcing function. So, you would have no 10t in there. So,
let's see what the solution will be. For that what we do is we calculate the characteristic equation
which will be hey this is second derivative so it's m squared, this is first derivative here
so will be four m; four being the constant there and this is zero derivative so it'll be just four
and put that equal to zero. Now, we can solve this particular quadratic equation by
either doing the factoring which is m plus 2 whole square equal to 0, hence implying m is equal to
minus 2 and minus 2 or what you can do is you can solve it by using the quadratic equation which
will be minus b which is in this case is minus 4 plus minus b squared which is 4 squared minus 4ac
divided by 2 times a and if you look at this one this one gives you minus 2 and minus 2 as well.
So, we have 2 roots of the characteristics minus 2 and minus 2. So, that gives us the homogeneous
part of solution will be of the form K1 e to the power minus 2t plus K2 e to the power minus 2t,
but, since this is the same as this one we have to choose the next independent solution corresponding
to the second root of the characteristic solution will be K2t e to the power -2t.
So, we have found the homogeneous part let's concentrate on the particular part.
The particular part of the solution will look like the form of the right-hand side
and all its possible derivatives. So, in this case it is 10t is the forcing function. So, the
form will be A times t and then the derivatives of A times t is a constant; let's suppose A so we
will write as B. So, that is the form of the particular part of the solution. If
that is the case now how do we find out A and B? We find out A and B by just saying that hey
let's put that back into the differential equation we have y double prime plus four y prime plus four
y equal to 10t and what we're going to do is we're going to plug it back in here.
We'll get second derivative of this quantity, first derivative of
the quantity the assumed solution plus 4 times the sum solution is equal to 10 times t.
So, if we solve that we get 0 from the first term, we get 4A from the second term
and we get 4At plus 4B from the last term is equal to 10t. Now, here we have
two unknowns and only one equation, but we know that if we equate the coefficients of
the polynomial of the left-hand side and the right-hand side, we will be able to
set up two equations and two unknowns. So, what I mean by that is that let's
isolate the terms of t, which is 4A times t coming from here, then the constant terms
are 4A plus 4B and on the right side we have 10 times t so this 10 will be equal to 4A because the
terms have to match and then let's put a 0 here so that we can say hey this will be equal to this.
So, we get 4A times t is equal to 10t, I shouldn't write it like that,
I should simply say the coefficients of the t to the power one term are equal. So,
4A is equal to 10 and 4A plus 4B is equal to 0. So, we have two-equations-two-unknowns and
those can be solved easily. This one will directly give you A is equal to plus 2.5 and then this one
will give you 4 times 2.5 plus 4 times B is equal to 0, that is the second equation and that will
give you B is equal to minus 2.5 from there. So, those are the two unknowns which we have from
there which tells us that the particular part of the solution will be 2.5 times t minus 2.5.
So, we've done that part we've found the particular part exactly and we have found the form
of the homogeneous part and now what we're going to do is to complete solution will be equal to
the homogeneous part plus the particular part as we said. The homogeneous part is as follows
and the particular part which we just found out is as follows. So, the only thing which is
remaining is that how do we find K1 and K2 and we find those by using the initial conditions.
So, the initial condition here is y of 0 is equal to 5. So, let's plug it in. We get 5 equal to K1
e to the power minus 2 raised to the power 0 plus K2 times 0 e to the power minus 2
times 0 plus 2.5 times 0 minus 2.5 and that gives me K1 here, gives me 0 here, 0 here and minus 2.5
and that eventually gives me K1 is equal 7.5. So, we have at least found out what K1 is. So,
let's go and find out what K2 will be. So, for that case we know that y prime of 0 is given as 2.
So, y prime of 0 is given as 2, what does that mean? That means that our y
prime will be the derivative of our current form of our solution
and this one will give us minus 2K1 e to the power minus 2t, this part has
to be done by differentiation by parts, not differentiation by parts, differentiation of
product of two functions. So, that will give you K2 e to the power minus 2t
and minus 2K2t e to the power -2t. So, that is coming from differentiating this part
and then differentiating this part gives you 2.5 and differentiating this part of course,
gives you 0. So, having said that then let's go and see what this gives us.
So, what we have is that y prime is now given as this quantity right here.
So, y prime of 0 is 2. So, that means 2 is equal to;
by putting in the value of t equal to 0, let's see what happens.
So, we get 2 is equal to minus 2K1 plus K2 plus 0 plus 2.5 and now we have 2 is equal to -2K1.
And let's see, yeah, we already know the value of K1. So, K1 is how much? 7.5 and
we want to find out what K2 is and from here we'll find K2 is equal to 14.5. So,
we have found K1 and K2 and we have a complete solution available to us. So, y will be equal to
yH plus yP and we have found that out to be,
I don't need to write that down because I already wrote that down. So, let me just write down y
we had as K1 e to the power minus 2t plus K2t e to the power minus 2t plus
this quantity here. So, we found out that K1 is equal to 7.5, we found out K2 is 14.5
and so that becomes our complete solution and that is the end of this segment, thank you.