So, now let's solve a differential equation using
the Laplace transform technique. The differential equation is
going to be x double dot plus x equals sine 2t with
two initial conditions, x of zero equals two and x dot
of zero equals one. We know how to solve this in homogeneous differential equation
already by a method of ansatz. But here I want to solve it using this Laplace transform
technique and show you that you can get the same solution. So, we take the Laplace transform
of this differential equation. So, we get the Laplace transform of x double dot plus
the Laplace transform of x, equals the Laplace
transform of sine 2t. We don't have to do any work. We don't have to do any integrals. What we need to do is
to look at the table and read these Laplace
transforms off of the table. So, let's have a look
at the table now. So, to Laplace transform
the second derivative of x, we use line 16. So, that will be equal to
S squared times capital X minus S over the initial value
of x, which is two. So, minus 2S minus x dot
of zero, which is one. Then to Laplace transform sine 2t, we use line six, which will be b is equal to two. So, b two S squared plus
two squared, which is four. So, making use of the table then, we can go ahead and do the transform. So, we have S squared capital X, minus 2S minus one, plus, there is no first
derivative in this equation. In a general equation, you would have a first derivative, but here there is none. So, plus X is equal to
the Laplace transform of sine 2t, which from the table is two
over S squared plus four. Okay. We need to solve for x. So, we have S squared plus one
times X equals the right-hand side, which is going to be 2S plus one, plus two over S squared plus four. Finally, we get our capital X, which will be 2S plus one, over S squared plus one plus two over S squared plus
one times S squared plus four. Okay. So, we've transformed the equation and then solve
the algebraic equation. So now we have the solution
for X in S space, making use of the initial conditions. The remaining step in
this solution procedure then is to take the inverse
Laplace transform of this solution to get x of t.
This is the most difficult step. Let's look at the table
now and see what exactly we need to
transform these two terms. The first term is a 2S plus
one over S squared plus one. The second term is a 2 divided by S squared plus one times S
squared plus four. So, let's look at the table. Okay. To transform the 2S plus
one over S squared plus one term, we're going to make use of
line seven and line six. So, here b is equal to one and we're going to have a two times
S over S square plus one, which will be using line seven. We're going to have one
over S squared plus one, which will be using line six. Then to transform the
two over S squared plus one times S
squared plus four term, we're going to have to use
a partial fraction decomposition to write that in terms of something
divided by S squared plus one, plus something divided
by S squared plus four. Let me show you that partial
fraction decomposition now. So, to do the partial
fraction decomposition, we need to write two
over S squared plus one, times S squared plus four as
something over S squared plus one, that something will be an unknown
coefficient a times s plus b, plus something over
S squared plus four, which will be an unknown
coefficient c times s plus d. This is the standard form for
a partial fraction decomposition. To determine the coefficients a, b, c, and d, you can either stare at
it really hard or you can do some algebra is a little bit easier to just do some algebra, it doesn't hurt your brain so much. You multiply this side by top and
bottom by S squared plus four. So, putting under
this denominator on the left, you multiply the second term, top and bottom by S squared plus one. Then you set the S cubed, S squared and S term to zero, and set the constant term to two. When you do that, you can get a solution that you
find a equals c equals zero, b is equal to two-thirds, and d is equal to minus two-thirds. Okay. Putting everything together. Now, we have our X, is going to be, we're going to have to write that as two times S over S squared plus one, plus one over S squared plus one. Then we have this extra term. So, we have a is zero,
b equals two-thirds. So, we have a two-thirds plus two-thirds over one over
S squared plus one, and d is minus two-thirds. So, if we have an S squared
plus four in the denominator, we will see from the table that
we need a two in the numerator. So, I will write that as minus one-third times two over
S squared plus four. Okay? This one over S squared
plus one term will combine. So, one of them plus two-thirds of them will give us
five-thirds of them. So, we'll end up with
three pieces here, and then we'll be able to
write down our solution for X of t. After we consult a table, you still need the table to do
the inverse Laplace transform. So, let's have a second look
at the table now. Okay. So, we're trying to transform the term S over S squared plus one. There will come from line
seven, b equals one. So that will be a cosine
t. Then transform the term one over S squared plus one. That will be from line six, that will be a sine t. Finally, transform the term two
over S squared plus four. That will come also from line six, but b will be equal to two. Okay. So now, we're using
the table to do this transform, we get the first term is going to be two times cosine t.
The second term, that's from here. The second term will come
from plus five-thirds sine t. Then this last term
here will be minus one-third times sine 2t. Okay. So, all of this from
the table of the taking the inverse Laplace transform using the table of the Laplace transform. Okay. Let me review then
what I've done here. We have then the
initial value problem, one that we already know how to
solve using the method of ansatz. But here, I'm showing
you how to solve this by Laplace transform method. We take the Laplace transform
of the differential equation, and then using the table
and after doing that, then we can solve for x in S space. The complicated procedure is that we have to then take the inverse Laplace transform of this to get
back X of t. The first term, we can figure out from the table. The second term is not in the table. So, what we need to do is a partial fraction decomposition
that requires a bit of algebra. But you can if you remember
your partial fraction decomposition, you know how to write
this expression, and then the algebra will tell
you what these coefficients are. Then we have our X of S
with these four terms, two of the terms combined. So, we actually have three terms, and they're all written
in a way that we can read the inverse transform
off of the table. When we look at the table and
read the inverse transform, then we get our solution here. X of t is two cosine t plus five-thirds sine t minus
one-third sine 2t. From our previous knowledge of
solving differential equations, you'll recognize
the first two terms are the homogeneous solution and the last term is
the particular solution. The coefficients of
the homogeneous solution are chosen so that you satisfy
the two initial conditions. I'm Jeff Chasnov. Thanks for watching, and I'll see you in the next video.