- [Instructor] Welcome to an intro lesson
to initial value problems.
The objectives of this video are to
define an initial value problem,
and also to solve initial value problems
given the general solution to a differential equation.
Because this is an intro video,
we will be given the general solution.
In other videos, we will be finding
the general solution ourselves,
and then solve the initial value problem.
A differential equation that has given conditions
allows us to find the specific function that
satisfies a given DE rather than a family of functions.
These types of problems are called
initial value problems, abbreviated using IVP.
The solution to an IVP is called a particular solution.
Here's an example of a first order IVP.
We're given the differential equation,
DY DX equals F of X Y,
but we are also given that Y of
X sub zero equals Y sub zero.
It's this information here that
allows us to find the particular solution.
Let's take a look at an example.
We want to find the actual or particular solution
to the initial value problem, DY DX minus Y equals zero,
where Y of three is equal to one,
if we know that Y of X equals C times E to the power of X
is a general solution to the differential equation.
Again, we're given this is the general solution
to the differential equation,
but since we are also given that Y of three equals one,
we can find the particular solution to this IVP.
Remember, if Y of three equals one,
we can substitute three for X into our general solution,
and this function must equal positive one.
Using this, we know that C times E
to the power of three must equal positive one.
This allows us to solve for the constant, C,
and therefore find the particular solution.
We'll divide both sides by E to the third.
This tells us that C is equal to
one divided by E to the third,
which we can also write using properties of exponents,
as E to the power of negative three.
That means our particular solution,
Y of X is going to be equal to, instead of C,
we will have E to the negative three,
times E to the X,
and now we can apply the property of exponents again,
because our bases are the same,
so our particular solution is going to be
Y of X equals E raised to the power of
negative three plus X, or just the power of X minus three.
This would be our particular solution to this IVP.
If we wanted to check this, we could find DY DX,
and sub it into our differential equation.
Well, DY DX is going to be equal to just
E to the power of X minus three.
Remember, think of this as E to the U,
and apply the chain rule, U prime is just going to be one.
So now, subbing this into our differential equation,
we know that DY DX is E to the power of X minus three,
but so is Y, so we would have E
to the power of X minus three,
minus E to the power of X minus three equals zero,
which is true.
Again, this is DY DX, and this is our function, Y.
Their difference is zero, which verifies
our particular solution is correct.
Let's take a look at a second example.
Again, we want to find the actual or particular solution
to the IVP problem, Y double prime plus four Y equals zero,
where Y of zero equals zero,
and Y prime of zero equals one
if we know that Y of X equals two times C sub one
times cosine two X minus two times C sub two
times sine two X is a general solution
to the differential equation.
So, given this general solution,
and these two initial conditions,
we can find the particular solution.
Let's start by using Y sub zero equals zero.
This means if we sub zero in
for X into our general solution,
this function must equal zero.
So, two times C sub one, times the cosine
of two times zero, or zero,
minus two times C sub two times the sign of two times zero,
which is zero, must equal zero.
Well, the sine of zero is equal to zero,
so this product here is zero.
The cosine of zero is equal to one,
so this tells us that two times C sub one must equal zero.
Well, if we divide both sides by two, we know
that C sub one must equal zero.
Now we know that C sub one is equal to zero.
Even though we don't know C sub two,
we can simplify our function, Y of X.
We know Y of X has to be equal to
negative two C sub two sine two X.
Now, if we can find the derivative of this,
and then use the fact that Y prime of zero
is equal to one, to find C sub two.
Y prime of X is going to be equal to
the derivative of this function here,
so we will have negative two times C sub two,
times the derivative of sine two X,
which will be cosine two X times two,
using the chain rule.
Y prime of letter X is equal to negative four C sub two
cosine two X.
Since Y prime of zero is equal to one,
we can replace X with zero,
and know this function value must be equal to one.
So, negative four C sub two cosine of two times zero,
which is zero, must equal the function value of one,
and cosine zero is equal to one,
so we have negative four C sub two equals positive one.
Divide both sides by negative four,
and we know that C sub two must equal negative one fourth.
Now we can find our particular solution,
since we know that C sub one is equal to zero,
and C sub two is equal to negative one fourth.
Let's do that on the next slide.
Again, we know that C sub one is equal to zero,
C sub two is equal to negative one fourth,
so Y of X must equal two times zero,
times cosine two X, minus two times
negative one fourth sine two X.
So, Y of X is equal to --
Again, this is zero,
and this is going to be positive one half sine two X.
Now, we will check this particular
solution into the given differential equation.
I'll have to find Y double prime, and add four times Y,
and see if this sum is equal to zero.
Y prime is going to be equal to
the derivative of our particular solution here,
so we'll have one half times the derivative of sine two X,
which will be cosine two X, times two,
again, here's the chain rule.
Y prime is equal to, this will just be cosine two X,
so Y double prime will be the
derivative of our derivative,
which is going to be negative sine two X times two,
which is equal to negative two sine two X.
Let's perform the substitution into
our differential equation on the next slide.
Again, we know that Y equals one half sine two X,
and we know Y double prime is equal
to negative two sine two X.
To check this, Y double prime, or negative two sine two X,
plus four times Y, which is one half sine two X,
must equal zero.
So, we have negative two sine two X, plus two sine two X
equals zero, which is true, zero equals zero.
This verifies that our particular solution here is correct.
Okay, I hope you found this intro video helpful.