Here, we're given a second-order,
homogeneous differential equation,
with initial conditions.
We have to apply the Laplace transform
and solve for big Y of S, which equals
the Laplace transform of Y.
And then we want to solve the initial value problem
by using the inverse Laplace transform,
which they give us Y of T, where Y of T
equals the inverse Laplace transform of big Y of S.
So our first step is to take the Laplace transform
of both sides of the differential equation,
which would give us the Laplace transform
of Y-double-prime, minus 10 Y-prime plus 25Y
equals the Laplace transform of zero,
and now we use linearity, or the properties
of Laplace transforms, and rewrite the left side.
We'll write the left side as the Laplace transform
of Y-double-prime minus 10 times the Laplace transform
of Y-prime plus 25 times the Laplace transform of Y
equals the Laplace transform of zero.
And now we know the Laplace transform of Y-double-prime
equals S squared times big Y of S,
minus S times Y of zero, minus Y-prime of zero.
And then we have minus 10 times the Laplace transform
of Y-prime, so we'd have minus 10 times the quantity
of the Laplace transform of Y-prime equals S times big Y
of S minus Y of zero.
Then we have plus 25 times the Laplace transform of Y,
which will give us plus 25 times big Y of S
equals the Laplace transform of zero is zero.
Now let's clear the parenthesis here by distributing.
The other terms will remain the same.
So we'll have minus 10S, big Y of S, plus 10 Y of zero,
and then we have plus 25 big Y of S, equals zero.
For our next step we'll use the initial conditions
and substitute two for Y of zero,
and a negative one for Y-prime of zero.
So we'll substitute two here and here.
And we'll substitute negative one here.
So that would give us S squared big Y of S,
and then we'll have minus S times two,
so minus two S.
Here, we'll have minus negative one,
which would be plus one,
and then minus 10S big Y of S,
and then plus 10 times two, so I have plus 20.
Plus 25 big Y of S,
equals zero.
Notice how we have three terms that have a factor
of big Y of S, here,
here,
and here.
So for our next step, we'll factor our big Y of S
from these three terms and move the remaining terms
to the right side of the equation.
So that would give us big Y of S times the quantity
S squared minus 10S, plus 25, equals,
we'd add two S to both sides, so we'd have two S,
and then notice that one plus 20 is 21,
so we subtract 21 on both sides,
so the right side of the equation is two S minus 21.
Now, let's continue on the next slide.
To solve for big Y of S, we'll divide both sides
by the quantity S squared minus 10S plus 25.
Simplifying, we now know the Laplace transform of Y,
or big Y of S, equals the quantity two S minus 21
divided by the quantity S squared minus 10S plus 25.
Notice how, if we wanted to,
we could factor the denominator.
This is a perfect square trinomial,
so we'd have the quantity two S minus 21
divided by the quantity, we'd have two binomial factors,
or the factors of S squared are S and S,
and the factor of 25 that add to negative 10,
are negative five and negative five,
so notice how we can write the denominator
as the quantity S minus five squared.
Our question does ask for big Y of S,
which we could enter in this form, or this form.
I'm going to go ahead and enter this form, here.
Again, big Y of S equals the quantity two S minus 21
divided by the quantity S squared minus 10S plus 25.
Now, we actually want to solve the initial value problem
by determining little Y of T,
by finding the inverse Laplace transform of big Y of S.
However, before we do this, we do want to perform
partial fraction decomposition on this fraction, here.
So we can use a table of Laplace transforms,
in order to find the inverse Laplace transform.
Let's go ahead and perform partial fraction decomposition
on this fraction, here.
So we have the quantity two S minus 21 divided by
the quantity S minus five squared.
Notice how we have two linear factors, here,
so we'll have a sum of two fractions
where we'll first have A divided by one factor
of S minus five, and then because this linear factor
is repeated, we'll have plus B divided by
the quantity S minus five squared.
And now, we'll form the basic equation by multiplying
both sides by the quantity S minus five squared.
Well, that would give us the equation two S minus 21
equals, when multiplying the square of S minus five
times this fraction, one factor of S minus five
would simplify out, leaving us A times the quantity
S minus five, and we're multiplying
the quantity S minus five squared.
In this fraction, both factors of X minus five
would simplify out, giving us just plus B.
And now I'll select convenient values of S
in order to find the values of A and B,
so we can write big Y of S as a sum of these two fractions.
So let's let S equal five,
because that would make this factor zero.
So, if S is five here, we'd have two times five
minus 21, that's negative 11.
Equals, this would be zero, and therefore we'd just have B,
so we know B must equal negative 11.
And now we'll select another value of S
to determine the value of A.
Let's select S equals zero.
So if S equals zero, we'd have the equation
negative 21 equals, if S is zero,
this would be negative five A, and then plus B,
but we know B equals negative 11,
so we'd have negative 21 equals
negative five A minus 11.
Adding 11 to both sides, we'd have negative 10
equals negative five A, so A equals positive two.
So using the values of A and B, we can now write big Y of S
as the sum of two fractions.
Big Y of S is equal to, well, A is positive two,
so we'd have two divided by the quantity S minus five,
and then because B is negative, we'd have minus 11,
divided by the quantity S minus five squared.
So now, to find the inverse Laplace transform of big Y of S,
which will give us Y of T, we'll take the inverse
Laplace transform of both sides of this equation.
Let's do this on the next slide.
We'd have the inverse Laplace transform
of big Y of S equals the inverse Laplace transform
of two divided by the quantity S minus five,
minus 11, divided by the quantity S minus five squared.
Well, we know the inverse Laplace transform of big Y of S
is Y of T, which is what we're looking for,
so we have Y of T equals,
let's rewrite the right side using properties
of inverse Laplace transforms,
as two times the inverse Laplace transform
of, because we factored out the two,
this would now be one over the quantity S minus five,
and then minus 11 times the inverse Laplace transform
of one over the quantity S minus five squared.
Now, let's look at the table of Laplace transforms below.
Notice how the second column has big F of S,
in our case, big Y of S,
and the first column is F of T,
which in our case would be Y of T.
So because we're determining inverse Laplace transforms,
we'll be using this table in this direction here,
looking at this column, let's see if we can find a form
that fits this inverse Laplace transform.
Notice how here, we have one over the quantity S minus A,
which fits this form perfectly.
Where A is equal to positive five.
Which means this inverse Laplace transform
is E raised to the power of five T.
So we know that Y of T equals two times E
raised to the power of five T.
And now, looking at this inverse Laplace transform,
because the denominator is
the quantity S minus five squared,
notice how it fits this last form, here,
where the denominator is in the form of the quantity
S minus A raised to the power of N plus one.
So because this is squared, that means N would
have to be one, since one plus one is two,
and therefore the numerator must be one factorial,
which equals one, which we have.
And therefore this fits the form perfectly.
Again, where N equals one.
And therefore, the inverse Laplace transform
is two to the first, times E raised to the power of five T.
So our second term here is minus 11T, E to the five T.
So this is our solution to the initial value problem.
If we wanted to, we could factor out the common factor
of E raised to the power of five T,
and write this as Y of T equals E to the five T
times the quantity two minus 11T.
Either of these two forms are acceptable.
So, going back to our first slide,
again we have Y of T equals two E to the five T
minus 11T, E to the five T.
I'm not sure if you noticed, but we could have found
this particular solution much faster
by using a characteristic equation
where we'd have the repeated root, or repeated solution.
I hope you found this helpful.