- [Teacher] Welcome to several examples on how to write
the form of a particular solution to
a linear second order nonhomogeneous differential equation.
The reason we want to know how to do this is because
in order to apply the method of undetermined coefficients,
we do have to guess the form of a particular solution
which we'll call big Y sub p of x
in order to find the general solution
to the differential equation.
So let's take a look at our examples.
Again, our goal here is to write the form
of the particular solution,
which we'll call big Y sub p of x.
Notice we do have a linear second order
nonhomogeneous differential equation
where the right side of the equation,
often referred to as g of x,
involves both the sine function and the cosine function.
So we're looking for a function big Y sub p
where if we substitute it in the function,
its first derivative and second derivative,
we'd end up with this function here on the right.
So for the form of big Y sub p,
we'll consider one term at a time.
Let's start by considering two sine three x.
Well, it's probably safe to assume it's going to include
an undetermined coefficient times sine three x
so we'll start with that.
So we have A sine three x
but notice how the derivative of this term
would also involve cosine three x
so we also have to include an undetermined coefficient
times cosine three x
so plus B cosine three x.
So just considering the first term,
this would be the correct form for the first part
of the particular solution.
But now we also have to consider the second term
of plus four cosine three x.
So, again, it's going to be safe to assume that we'd have
a constant times the cosine function or cosine three x
So let's call it C times cosine three x.
But now the derivative of cosine also involves sine
so we'll also need plus D sine three x.
So these two terms would be from the second term
involving the cosine function.
But notice how there is some repetition here,
so we can shorten this function.
Notice how we have two cosine functions
and also two sine functions with the same angle.
Notice how all the angles are three x.
If they weren't the same,
we wouldn't be able to shorten this.
So let's write this as big Y sub p of x equals,
let's factor out sine three x
leaving us with A plus D sine three x
and then we'd have plus
the quantity B plus C times cosine three x.
We'll notice that A plus D is just another constant
and so is B plus C.
So let's go ahead and write this as big Y sub p of x equals,
we're left off with D so let's say E sine three x
plus F cosine three x.
This would be the form of the particular solution
for the given differential equation.
Let's take a look at another example.
Looking at the right side of the equation,
notice how we have two x squared
and then we have minus three sine five x plus one.
So if we wanted to, we could rewrite
the right side of this equation
as two x squared plus one
so this would be a quadratic function
minus three sine five x.
So we can think of this as a quadratic function
and then minus a sine function
which means for big Y sub p of x,
our particular solution would take the form of
a quadratic function with undetermined coefficients
which would be Ax squared.
We still have to include an x term so plus Bx plus C
and then for the sine function,
we could have plus D times sine five x
but again, because we're also working with derivatives
and the derivative of sine involves cosine,
we would have to have plus E cosine five x.
There's no repetition here
so this would be the form of our particular solution.
Let's take a look at one more in this video.
Notice in this example,
it says to assume there are no terms in Y sub p,
our particular solution,
that are duplicates in little y sub c
which is the complimentary function
where the first two terms
are the general solution to the nonhomogeneous
differential equation.
This is important because if there was duplication,
it would affect our approach
to forming the particular solution.
We'll take a look at this case in the next video
but for this example, because there's no repetition,
we can go ahead and set up big Y sub p.
So notice how the first term is four
e to the power of two x.
So for the particular solution,
we'd start with A times e to the power of two x.
Remember, A is some unknown coefficient.
Looking at the second term,
we have minus three x times e to the power of negative x.
We need to think of this as a linear function
times an exponential function.
So for a linear function with unknown coefficients,
we would have to have plus the quantity Bx plus C
times the exponential
and then we'd have plus
D, an unknown coefficient, times the exponential e
raised to the power of negative x
but in this case, I think we can simplify this.
Notice if we distribute here,
we actually would have two
e to the power of negative x terms.
Let's go ahead and show that.
Big Y sub p of x is equal to A
e to the power of two x
and then we'd have plus Bx
e to the power of negative x
plus C e to the negative x
and then plus D e to the negative x.
Notice how these are like terms
so we could write this if we wanted to
as A e to the two x
plus Bx e to the negative x
plus quantity C plus D
e to the power of negative x.
Well, C plus D is just a constant
so we don't need both of these.
We'll go ahead and say big Y sub p of x
is equal to A e to the two x
plus Bx e to the negative x.
Let's go ahead and just use C again
because we didn't need this term here.
So plus C e to the negative x.
This would be the form of the particular solution
to the given differential equation.
And again, this is important because if we were going to use
the method of undetermined coefficients,
we would use this as the function that we're guessing
would be a particular solution.
I hope you found this explanation helpful.