Solving a differential equation
using the Laplace transform, you obtain
big Y of s equals the quantity of s squared plus 21
divided by the quantity s cubed plus s squared.
Left to right, y of s in the form shown here,
which means you want to perform
partial fraction decomposition on this fraction here.
And then we're asked to find little y of t
by finding the inverse Laplace transform of big Y of s.
Let's first write this fraction in factored form.
So we'd have the quantity s squared plus 21 divided by,
notice how the greatest common factor
or denominator is s squared.
We can make the denominator as s squared
times the quantity s plus one.
So notice how we have three linear factors,
the linear factor of s occurs twice.
So we can write this as a sum of three fractions,
where the first fraction would be
a divided by the linear factor of s,
and because the linear factor of s is repeated,
we then have plus b divided by s squared,
and then plus c divided by the linear factor of s plus one.
So the way I've written this,
notice, A is equal to f of s,
B is equal to g of s,
and C is equal to h of s.
The next step is to form the basic equation
by multiplying both sides of the equation
by the least common denominator,
which is s squared times the quantity s plus one.
When we do this, we'd have the equation
s squared plus 21 equals,
when multiplying the right side of the equation
by s squared times the quantity s plus one,
For this first fraction, one factor of s would simplify out,
so we'd have A s times the quantity s plus one,
when multiplying by this fraction,
s squared would simply out,
so we'd have plus B times the quantity s plus one.
Plus, when multiplying by this fraction,
the factor of s plus one simplifies out,
so we'd have plus C s squared.
And now we'll select convenient values of s
to determine the values of A, B and C,
so we can write y of s as the sum of these three fractions.
So let's first select s equals zero.
Notice when s is zero, the left side would be 21,
and the right side, this would be zero,
and this should be zero.
So we'll just have B times one or B,
so we know B or g of s must equal 21.
Now we select s equals negative one.
If s equals negative one,
the square of negative one is one
plus 21, that's 22.
One the right side, if s is negative one,
this should be zero, and this should be zero,
and the square of negative one is one.
So we just have C times one or C.
So know we know that C equals 22.
And now to find the value of A.
Left-foot s equals let's say one.
So if s is one, the left side is 22
equals the right side,
we'd have A times one times two, that's two A
plus B times two, that would be two B
plus C times one squared, that would be C.
And now we'll substitute the values of B and C
into the equation and find the value of A.
So we'd have 22 equals two A plus
two times B would be 42
plus C, which is 22.
Notice 42 plus 22 is 64.
We subtract 64 on both sides.
We'd have negative 42 equals two A.
Dividing both sides by two,
we have A equals negative 21.
So now we can write Y of s
as a sum of these three fractions,
since we now know the value of A, B and C.
Before we do this, though,
f of s is equal to A,
which is negative 21,
g of s equals B, which is 21,
and h of s equals C, which is 22.
And now, let's write big Y of s
as a sum of these three fractions.
We would have big Y of s
is equal to A divided by s,
which is negative 21, divided by s,
plus B divided by s squared,
that would be 21 divided by s squared,
plus C divided by the quantity s plus one,
which would be plus 22 divided by
the quantity s plus one.
Now, to find y of t,
we'll take the inverse Laplace transform
of both sides of the equation.
So we'd have the inverse Laplace transform of big Y of s
equals the inverse Laplace transform of
negative 21 divided by s
plus the inverse Laplace transform of
21 divided by s squared
plus the inverse Laplace transform of
22 divided by the quantity s plus one.
So on the left side, the inverse Laplace transform
of Y of s gives us y of t,
which is what we're looking for.
And now to identify the formulas we'll use
to find these inverse Laplace transforms.
Looking at the table below,
we have Y of s in the second column
and y of t in the first column.
So we'll be using the table in this direction.
And because every denominator here is just s,
we'll use this first row to determine
the first inverse Laplace transform.
Notice how we have the denominator of s here.
Here we have the denominator of s squared,
so we'll use this row here,
where the denominator is s raised to the power of n plus 1,
and finally here we have a denominator of s plus 1,
which doesn't look like it's in the form of
s minus A, where we can write
this sum as a subtraction,
we can write this as s minus negative one.
So we'll use these first three rows
to determine these inverse Laplace transforms,
which will give us y of t.
Let's write this first inverse Laplace transform
as negative 21 times the inverse Laplace transform of
one divided by s.
Notice, in this form, it fits the formula perfectly.
First the next inverse Laplace transform,
let's write this as plus 21
times the inverse Laplace transform of
one divided by, let's try s squared as
as s raised to the power of one plus one.
So you can easily see that n is equal to one.
And for this last inverse Laplace transform,
let's write this as plus 22
times the inverse Laplace transform of
one divided by, let's try it,
s plus one as s minus negative one.
Again, in this form, it fits our formula perfectly.
So now we know that y of t is equal to
negative 21 times the inverse Laplace transform of
one over s is equal to one.
So we just have negative 21 times one.
So you can leave this as negative 21,
and we have plus 21 times the inverse Laplace transform of
one over s raised to the power of one plus one.
So notice how our numerator here is n factorial.
But because n is one, one factorial is one,
so this does fit the formula perfectly
and, therefore, the inverse Laplace transform
is just t to the first, or t.
So I have plus 21 t, and we have plus 22
times the inverse Laplace transform of
one divided by the quantity s minus negative one.
So it's this form where A equals negative one,
so we have e raised to the power of negative one times t,
or just e raised to the power of negative t.
So this is plus 22 e to the negative t.
So this is the inverse Laplace transform
of big Y of s.
I hope you found this helpful.