Welcome to an example on how to find
a particular solution to a linear second order
nonhomogeneous differential equation
using the method of variation of parameters.
In this problem, we're given y sub one equals t squared,
and y sub two equals t to the negative one.
Satisfy the corresponding homogeneous equation
of the given nonhomogeneous differential equation,
which means the general solution
to the nonhomogeneous differential equation can be written
as y of t equals c sub one times y sub one,
plus c sub two times y sub two, plus big y of t.
Well, the first two terms make up what's called
the complementary function,
and big y of t is a particular solution.
So in this problem we're asked
to use the variation of parameters to find big y of t.
Normally, when solving a linear second order
nonhomogeneous differential equation
using the method of variation of parameters,
we follow these three steps.
We first solve the corresponding
homogeneous differential equation,
which gives us y sub c, the complementary function,
which again is the sum of these two terms here.
Then we find a particular solution
using the formula as shown here,
which we derived in a previous lesson.
Once we find the complementary function,
y sub c, and a particular solution given by big y sub p,
we then form the general solution as shown here.
Now, for our example, we're already given y sub one
and y sub two, and we're only asked
to find big y sub p of t, but before we find
big Y sub p of t here, it is important to recognize
that the given differential equation
is not in the correct form.
We need it to be in this form here,
the first term needs to be y double prime,
so let's first put the equation in the correct form.
Let's divide both sides by t squared.
Simplifying, we have y double prime,
minus two, divided by t squared, times y equals,
on the right, we have negative three t,
minus t, raised to the power of negative two.
It's important that we recognize it's not in correct form,
because notice how the formula for big y sub p of t
does use g of t which is the right side of the equation
once it's in the correct form.
Now let's work on finding big y sub p of t.
Notice how the formula for big y sub p of t
does use the Wronskian of y sub one comma y sub two,
so let's find this first, before setting up this formula,
so again, we're given y sub one is equal to t squared,
and y sub two is equal to t
raised to the power of negative one.
We also need to find the first derivatives,
so y sub one prime is equal to two t,
and y sub two prime is equal to negative t
to the negative two.
Now, let's find the Wronskian of y sub one comma y sub two,
which is the Wronskian of t squared
comma t to the negative one,
which equals this two by two determinant
where the first row is y sub one, y sub two,
which is t squared, t to the negative one.
Second row are the first derivatives,
two t, and negative t to the negative two.
This determinant is equal to this product
minus this product.
Well, t squared, times negative t to the negative two,
would be negative two to the zero, or negative one.
When we have minus t to the negative one,
times two t would be two t to the zero, which is two,
so we have one minus two, which equals negative three,
so now we can find big y sub p of t,
which equals the opposite of y sub one,
which would be negative t squared, times the integral
where the numerator is y sub two, times g of t,
which would be t to the negative one,
times g of t is negative three t,
minus t to the negative two.
The denominator is the Wronskian on y sub one
comma y sub two, which we know is negative three.
Then we have plus y sub two, which is t to the negative one,
times the integral where the numerator is y sub one,
times g of t, which would be t squared,
times the quantity negative three t,
minus t to the negative two.
The denominator is a Wronskian where we have d t.
Let's continue on the next slide.
For the next step,
let's go ahead and factor out negative 1/3
to eliminate the fraction from the integrated function.
So if we factor out negative 1/3 here,
we'd have positive 1/3 t squared.
Let's also distribute here.
T to the negative one, times negative three t
would just be negative three,
and then we have minus t to the negative one,
times t to the negative two, is t to the negative three.
Factor out negative 1/3 here, we'd have minus 1/3
t to the negative one, times the integral.
Again, let's distribute, so we have
negative three t to the third,
and then we'd have minus t to the zero, or minus one.
Now we integrate.
So the integral of negative three with respect to t
would be negative three t,
and there we'd have minus the integral
of t to the negative three,
which would be minus t to the negative two,
divided by negative two,
which becomes plus two to the negative two, divided by two.
We have minus 1/3 t to the negative one,
times the integral of negative three t cubed,
with respect to t, would be negative three,
times t to the fourth, divided by four.
Negative 3/4 t to the fourth.
The integral of one, with respect to t, would be t.
Now, we'll distribute and combine like terms.
So we'll distribute here,
and we'll distribute here.
So negative 1/3 t squared, times negative three t,
would just be negative t cubed,
and then 1/3 t squared, times t to the negative two,
divided by two, that'd be plus 1/6 t to the zero,
or just plus 1/6.
Now, because of the subtraction,
we can think of distributing
negative 1/3 t to the negative one,
so negative 1/3 t to the negative one,
times negative 3/4 t to the fourth,
notice how the threes would simplify out,
so we have 1/4 t to the third.
Then finally, we have negative 1/3 t to the negative one,
times negative t, that's positive 1/3 to the zero,
or just plus 1/3.
We're combining like terms, so negative one t cubed,
plus 1/4 t cubed, would be negative 3/4 t cubed,
and 1/6, plus 1/3, is equal to 1/6, plus 2/6,
so 3/6, which equals 1/2.
So this is the particular solution
our question is asking for.
So going back to the first slide,
again, we're not asked to enter the general solution,
only the particular solution given by big y of t,
which, once again, is negative 3/4 t to the third, plus 1/2.
We do have all the information we need
to form the general solution, but again,
this question only asked for big y of t.
I hope you found this helpful.