>> Welcome to the Cypress College Math Review
on Laplace Transforms and Differential Equations.
First let's review how to take the Laplace Transform
of derivatives of functions.
Let's take the Laplace Transform of the first derivative.
So n is equal to 1.
So we get s to the first power, times the Laplace Transform
of F, minus -- since n is equal to 1, this term just becomes f
of 0 and there are no other terms.
Now if n is equal to 2, we're taking the Laplace Transform
of the second derivative.
So now we have s to the second power times the Laplace
Transform of F, minus the sum, n is equal to 2.
So this first term becomes s to the first power, times f of 0,
plus -- now we'll have s to the 2 minus 2 power
so now we have no s's, we just have f prime of 0.
And then that's the last of our sum.
If n is equal to 3, we're taking the Laplace Transform
of the third derivative, which should be enough.
So now we have s to the third power,
times the Laplace Transform of F minus the sum.
So now we have s to the second power times f of 0,
plus s to the first, times the derivative of f, plus --
no s's anymore -- and now we have the second derivative
evaluated at 0.
Let's solve this initial value differential equation using
Laplace Transform.
First let's transform the equation
so the Laplace Transform applied to both sides of the equation.
First we have the Laplace Transform
of the second derivative.
So that's s squared times the Laplace Transform
of the function x, minus s times little x
at 0, minus x prime at 0.
Then we have 4 times the quantity --
then we have the Laplace Transform
of the first derivative, which is s times the Laplace Transform
of X minus little x at 0.
Then we have plus 5 times the Laplace Transform
of X. That's equal to the Laplace Transform of 25t.
Looking at a table of transforms,
we notice that the Laplace Transform of t
to f power is the quotient of n factorial
over s to the n plus 1.
Well, n in this case would be 1, so we would have the constant,
which is 25 times one factorial,
over s to the 1 plus 1 or s squared.
There we go.
Now let's substitute in the values that were given.
x of 0 is negative 5.
And x prime at 0 is 7.
Multiplying both sides of the equation by s squared
and factoring out cap X of s.
Over on the left hand side we have s
to the fourth plus 4s cubed, plus 5s squared.
Moving all terms that do not involve cap X of s
to the other side -- and once again we're calling
that we're multiplying by s squared.
We have 25 minus 13s squared minus 5s cubed.
Solving for cap X of s we have 25 minus 13s squared minus 5s
cubed, factoring out an s squared in the denominator,
we have s squared plus 4s plus 5.
Next we set ourselves up for a partial fraction decomposition.
So we have A over s, plus B over s squared, plus C s plus D
over s squared plus 4s plus 5.
OK, multiply by the original denominator
and we have 25 minus 13 s squared minus 5 s cubed equals A
times s, times the quantity s squared plus 4s plus 5.
Plus B times the quantity s squared plus 4s plus 5.
Plus the quantity Cs plus D times s squared.
We can let s be 0 and we immediately get
that 25 is equal to 5B, so B is 5.
But that's the only easy value we're going to get that way.
So next I think we'll equate coefficients.
So negative 5 s cubed, minus 13s squared plus 0s plus 25 is equal
to As cubed plus 4As squared, plus 5As, plus B s squared,
plus 4Bs, plus 5B, plus Cs cubed, plus Ds squared.
Now let's group like terms.
For s cubed, we have A plus C.
For s squared we have 4A plus B plus D. For s
to the first power we have 5A plus 4B.
And our constant is simply 5B.
We already know that B is 5 so we don't need that.
The next one that's easiest to look at would be one
that involves B and only one other unknown,
which would be this equation.
So 5A plus 4B is equal to 0, by equating coefficients,
so 5A plus 4 times 5 is equal to 0, which gives us
that A is equal to negative 4.
Since A is negative 4, the next one that we should probably look
at would be the s cubes.
A plus C is equal to negative 5.
Since A is negative 4, we get that C is negative 1.
The final equation will give us D.
This is equal to negative 13 by equating coefficients.
So D is equal to negative 13 minus B minus 4A,
so that's equal to negative 13 minus 5 plus 16,
so we get a value of negative 2 for D. So cap X of s is equal
to A, which is negative 4 over s. Plus B,
which is 5 over s squared.
Plus C, which is negative 1 times s. Plus D,
which is negative 2 and all of that is
over s squared plus 4s plus 5.
Let's rewrite this.
Put the coefficients out front.
So negative 4 times 1 over s, plus 5 times 1
over s squared minus 1 times quantity s plus 2
and then we'll complete the square and write this
as s plus 2 quantities squared plus 1.
Now we want to take the inverse Laplace Transform
of each of these three terms.
Here's a table of transforms that will help us
with the first two terms.
So we want the Laplace Transform of 1 over s.
So in this case, n would be 0.
So the inverse Laplace Transform of 1 over s, would simply be t
to the 0 power and the Laplace Transform of 1 over s squared --
in this case n would be 1.
The Laplace Transform would be t to the first.
So, so far we have for our answer --
x of t is equal to negative 4 plus 5 times t. Now we look
at our next term.
Our final term is a composition of two functions.
So since we have the composition,
we're going to have e to the ct times f of t. So we're going
to have e to the -- what is c?
c is negative 2 because we have s minus negative 2, yes?
So that's going to be negative 2t
and that's being multiplied times f of t. f
of is the inverse Laplace Transform of s
over the quantity s squared plus a squared
and the inverse Laplace Transform of that is cosine
of a times t. a is this value here, 1.
So I have cosine of just 1t.
And there's the answer to our problem.