Here we're given a homogeneous
differential equation with initial conditions.
And we're first asked to apply the Laplace Transform
and solve for big y of s,
which equals the Laplace Transform of y,
and then we'll solve the initial value problem
by using the inverse Laplace Transform
to determine y of t, which is the particular solution,
where y of t equals
the inverse Laplace Transform of big y of s.
So the first step is to take the Laplace Transform
of both sides of the differential equation.
So we'd have the Laplace Transform
of y double prime
plus six y prime
plus eight y
equals the Laplace Transform of zero.
And now we'll use linearity,
or the properties of Laplace Transforms,
and rewrite the left side.
Let's write the left side as the Laplace Transform
of y double prime
plus six times the Laplace Transform of y prime
plus eight times,
times the Laplace Transform of y
equals the Laplace Transform of zero.
Now determine each Laplace Transform.
The Laplace Transform of y double prime
is s squared times big y of s
minus s times y of zero minus y prime of zero.
And then we have plus six
times the Laplace Transform of y prime,
which equals s times big y of s minus y of zero.
And then we have plus eight
times the Laplace Transform of y, which is big y of s.
We have eight times big y of s
and the Laplace Transform of zero, is zero.
And now let's clear the parenthesis
here by distributing six.
Everything else will remain the same.
So distributing, we have plus six s times big y of s
minus six times y of zero.
And now using the initial conditions,
we'll substitute negative two for y of zero
and zero for y prime of zero.
So we'll substitute negative two here and here,
and substitute zero here.
So we'd have s squared times big y of s.
Here we're going to have minus s times y of zero,
which is minus s times negative two, so we have plus two s.
Y prime of zero is zero, so this term drops out.
And we have plus six s times big y of s,
here we have minus six times y of zero,
which is minus six times negative two,
so we have plus 12
plus eight times big y of s equals zero.
For our next step, notice how these three terms
contain a factor of big y of s,
so we'll factor out big y of s from these three terms,
and then we'll also subtract two s on both sides,
and subtract 12 on both sides,
to move these two terms to the right side.
So we'd have big y of s
times the quantity s squared plus six s
plus eight equals negative two s minus 12.
Let's continue on the next slide.
So our last step to find big y of s,
which is the Laplace Transform of y,
we'll divide both sides of the equation by the quantity
s squared plus six s plus eight.
So simplifying,
we have big y of s equals the quantity
negative two s minus twelve divided by the quantity
s squared plus six s plus eight.
Notice how if we wanted to,
we could also factor the numerator and the denominator.
We could factor the numerator and write this as
negative two times the quantity s plus six,
and the denominator and the product of two binomials
where we'd have s and s,
the factors of eight that add to six
are positive two and positive four.
Let's go ahead and just enter big y of s
in the non-factored form, or this form here.
So going back to the first slide,
big y of s is equal to the quantity
negative two s minus twelve
divided by the quantity s squared plus six s plus eight.
And now we want to solve the initial value problem
by determining the inverse Laplace Transform of big y of s.
Before we do this, though, we do want to write big y of s
as a sum of two fractions by performing
partial fraction decomposition on this fraction here.
So let's go ahead and do that now.
So we have negative two times the quantity s plus six
divided by the quantity s plus two
times the quantity s plus four.
And so now our denominator contains two linear factors,
so we'll have a divided by the quantity s plus two
plus b divided by the quantity s plus four.
So to find the basic equation,
we multiply both sides of this equation
by the least common denominator,
which is the quantity s plus two
times the quantity s plus four.
So we multiply both sides
of the equation by those two factors,
we'd have the basic equation
negative two times the quantity s plus six equals,
when multiplying this first fraction
by the two binomial factors, s plus two would simplify out,
you're left with a times the quantity s plus four.
And then when multiplying by this fraction,
the factor of s plus four would simplify out,
leaving us with plus b times the quantity s plus two.
And now to find the values of a and b,
we'll select convenient values of s,
so let's let s equal negative four.
That would make this first factor equal to zero,
so when s is negative four,
here we'd have negative two times
the quantity negative four plus six, that's positive two.
So we'd have negative four equals this factor would be zero,
and here we'd have b times
quantity negative four plus two,
so we'd have negative two b.
So notice how this tells us that b is equal to positive two.
And now let's select s equals negative two,
which will make this factor zero.
So if s is negative two,
here we'd have negative two times positive four,
which is negative eight equals, this would be
a times positive two or two a and this would be zero.
So we'd have a equals negative four.
Which means you can write big y of s
as a over s plus two,
which would be negative four over the quantity s plus two,
and b is positive two,
so we'd have plus two divided by the quantity s plus four.
And now we'll take this form of big y of s
and 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,
instead of having this negative term first,
let's write this as two over the quantity s plus four
minus four over the quantity s plus two.
Now on the left side,
the inverse Laplace Transform of big y of s is y of t,
so we have y of t equals, let's rewrite the right side
as two times the inverse Laplace Transform of,
the numerator would now be one divided by,
now looking at the table of Laplace Transforms,
notice how we need the denominator
to be in the form of s minus a,
so let's write s plus four as s minus negative four.
And then we'd have minus four times
the inverse Laplace Transform of one
divided by the quantity, again,
let's write s plus two as s minus negative two.
So we have y of t equals two times,
notice how this now fits the form perfectly
of one over the quantity s minus a where a is negative four,
so the inverse Laplace Transform
is e raised to the power of negative four t.
So we have two times e
raised to the power of negative four t.
Then here we have minus four times,
and again notice how again this fits the form perfectly
where a equals negative two,
so the inverse Laplace Transform
is e raised to the power of negative two t.
So here we have minus four times
e raised to the power of negative two t,
which means the particular solution is
y of t equals two times e
raised to the power of negative four t,
minus four times e raised to the power of negative two t.
So going back to the first slide,
using the Laplace Transform
to solve this initial value problem was not the fastest way,
it would have been much easier just to use
a characteristic equation.
But either way, we did find the particular solution is
y of t equals two times e
raised to the power of negative four t
minus four times e raised to the power of negative two t.
I hope you found this helpful.