In this screencast we are going to show how to solve differential equations using Laplace
Transforms.
For this problem here, we use the equation that is shown with the initial conditions
provided.
In order to solve this equation using Laplace Transforms, we need to individually go term
by term and take the Laplace transform of each of the four terms.
First we're gonna look at the derivatives and the analysis of the derivatives can be
done by taking advantage of the differentiation theorem of which is shown right here.
The real differentiation theorem shows that any derivative can be evaluated using the
equation above.
The first term represents the s raised to the power of the order of the derivative multiplied
by the Laplace transform variable.
All of the remaining terms have evaluations based on the value at time equals zero of
the function and of the appropriate derivatives.
Here, since y(0) and y'(0) both equal zero, these terms can just go away because they
are identically zero.
With this information we can easily take the Laplace transform of the first two terms.
So the Laplace transform of the second derivative, d^2y/dt^2 will be, looking at the first term,
s raised to the order of the derivative, which in this case is to the second power, multiplied
by the Laplace transform variable which is Y for this case here.
For our first derivative, we'll take the Laplace transform of 5dy/dt.
According to the rules of the Laplace transform, 5 can be factored out as if it were a regular
number, which would leave us with 5 times the Laplace transform of the first derivative.
Therefore this will equal 5 multiplied by the s variable raised to the appropriate power
which in this case it's the first order differential so we multiply by the first power multiplied
by the Laplace transform variable, y.
For the third term, the Laplace transform of a variable is the Laplace transform variable.
So therefore, the Laplace transform variable of 6 times y in the time domain, again, we
can factor out the 6 so it's gonna equal 6 times the Laplace transform of y and that
is just going to equal 6 times the Laplace transform variable represented as Y.
Finally x(t) is provided as the step change of magnitude 2.
By looking at a Laplace transform sheet, you'd be able to find the fact that the Laplace
transform of a step of 2 will just be 1/s, which is the Laplace for a step function multiplied
by the constant, 2.
So therefore this will just become 2/s.
Which each term being converted into the Laplace domain, we can now develop our equation to
be solved in the Laplace domain.
So our equation now becomes s^2*Y+5sY+6Y=2/s.
Our goal here is to solve for Y so therefore what we're going to now do is group together
the Y terms leaving us with Y(s^2+5s+6)=2/s and if we were to solve that for Y, that would
leave us with Y=2/(s(s^2+5s+6)).
The next step to solve this equation in the Laplace domain
is we have to break the system down into the roots of the equation because we need to be
able to use the Laplace transform from the table to take the inverse of the Laplace transform.
And there really aren't any Laplace transforms that correspond to s^2+5s+6, however, if we
look at this, we can see the fact that s^2+5s+6 can easily be factored into (s+2)(s+3).
If we look at our Laplace transform, we do have something for a/(s+2) and b/(s+3) where
a and b are constants.
So what we now need to do is find those constants.
So in order to do that, we have to use the method of partial fractions.
So the next step is to use partial fractions and to use partial fractions, we have to break
our function up, which we've already done by separating the roots.
So y=2/(s(s^2+5s+6))=A/s+B(s+2)+C/(s+3).
Our goal now is to find all of the values of A, B, and C that will allow us to use the
Laplace transform table to take the inverse of the Laplace transforms.
In order to do that, we have to find the last common denominator for these functions.
What we're gonna now do is focus solely on the numerator.
So the numerator of Y is gonna equal 2 and that's also gonna equal, when we factor out
A(s+2)(s+3)+B(s)(s+3)+C(s)(s+2).
We now factor out the A, B, and C terms that we've developed here remembering the fact
that it still has to equal 2 so this is going to equal As^2+5As+6A+Bs^2+3Bs+Cs^2+2Cs.
And as a reminder, based on what our numerator was from the beginning, all of this has to
equal 2.
So here we have s^2 terms, s terms, and s^0 terms, in other words, constants.
What we can do to more easily solve this is develop a table separating out the s^2's from
the s's from the constants.
That table can be shown below.
By looking at this table, we can easily use the s^0 terms to solve for A. So we have the
fact that 6A=2 which means the fact that A=1/3.
So what now we're left with is we have the s^2 and s equations that we have to solve
simultaneously.
Our s^2 equation is A+B+C which will be here 1/3+B+C=0 which can be rewritten as B+C=-1/3.
For our s term, we have 5(1/3)+3B+2C=0 and when we rearrange that, that leaves us with
3B+2C=-5/3.
Again remember the fact that here it's negative because the 5/3 term is moving to the other
side of the equation.
Now we can solve for both B and C using any one of a number of different algebraic methods.
Here, we will multiply the first equation, the s^2 term, by -2.
So what we now have is -2B-2C=-1/3*2=-2/3, and then we have just our same equation we
had before for the s term 3B+2C=-5/3.
Now we add these together, the two C terms will cancel out, which will leave us with
B=-1.
To find C, we just substitute it into any of the equations we've developed so we know
the fact that B+C=-1/3 so -1+C=-1/3 so C=2/3.
Now we can substitute A, B, and C into the original equation.
Recall that our equation that we started with was Y=A/s+B/(s+2)+C/(s+3).
We have just found the face that our constants are A=1/3, B=-1, and C=2/3.
Therefore, our equation now becomes Y=1/3/s-1/(s+2)+2/3/(s+3).
We can now take the inverse of the Laplace transform of each of these terms.
The first one is 1/3/s and if we look at a Laplace transform table the inverse Laplace
transform of 1/s is just a constant itself.
We have the 1/3 as a factor so therefore that holds true and what we're left with for this
first term is the inverse Laplace transform is just 1/3.
To look at the Laplace transform of 1/(s+2), this is often represented in a Laplace transform
as 1/(s+A) or 1/(s+B) or some constant you go to the other side of the Laplace transform
to take the inverse Laplace transform of this.
That will just be an exponential raised to the -2t.
The 1/(s+A) Laplace transform is just as valid for the third and final term except for it
being e^-2t, it will be e^-3t.
The 2/3 constant comes out and that leaves us with 2/3*e^-3t.
What we will now do is put back in the signs.
We've got a plus sign in between the second and third term and a negative in between the
first and second term and we've taken the inverse Laplace transform of the righthand
side so we have to do the same thing for the lefthand side.
And the inverse Laplace transform of a variable is itself in the time domain.
So what we're left with now is y(t)=1/3-e^-2t+2/3e^-3t which is the solution to our differential
equation.