- [Teacher] Welcome to a lesson on standard form
and differential form
of a first-order differential equation.
The objectives of this video are to define
standard and differential form of a first-order DE
and also to write differential equations
in both standard and differential form.
The standard form of a first-order differential equation
is dy/dx equals f of x comma y
where dy/dx appears on the left side of the equation.
So for most first-order DEs,
we can write the equation in standard form
by solving for dy/dx.
So once we have an equation in standard form,
if we let dy/dx equal M of x comma y
divided by negative N of x comma y,
the above standard form can be written
in an equivalent differential form
M of x comma y dx
plus N of comma y dy equals zero.
Now, it's important to recognize that when writing a DE
in differential form, the differential form is not unique.
However, if we ever get stuck, we can always let
M of x comma y equal f of x comma y
and then let N of x comma y equal negative one.
So using this, we can always write a first-order DE
from standard form to differential form.
Let's take a look at our examples.
We want to write two x y prime
minus y to the third equals zero in standard form.
So, again, the main idea here is if we can solve this
for y prime or dy/dx, we'll have this in standard form.
So if we start with two x y prime
minus y the third equals zero,
let's go ahead and add y to the third
to both sides of the equation.
That would give us two x y prime equals positive y the third
and now we can solve for y prime
by dividing both sides by two x.
So on the left side, we have y prime equals y the third
divided by two x.
If we want to, we can replace y prime
with Leibinz notation or dy/dx,
so we'd have dy/dx
equals y the third divided by two x.
So now the first-order differential equation
is in standard form.
Let's take a look at another example.
Again, to write this in standard form,
we need to solve this equation for y prime,
which is here.
So if we start with the given equation,
if we want to solve for y prime,
we need to start by adding four e to the three x y
to both sides of the equation.
So this would be zero,
so we have e to the two x y prime
equals four e to the three x y
plus cosine x,
and now we'll divide everything by e to the power of two x.
So on the left side, this simplifies to one times y prime
or just y prime.
Here we have a common base,
so we can subtract the exponents.
Three x minus two x is x,
so this simplifies to four e to the x y plus,
and then for this term here,
if we don't want this in fraction form,
remember, if we move this up to the numerator,
it's going to change the sign of the exponent,
so we can write this as e to the power
of negative two x cosine x.
And now we have the differential equation in standard form.
Now let's write some first-order differential equations
in differential form.
So notice for our first example this differential equation
is given in standard form.
We're going to write this in differential form.
A reference is given down here below in red,
and to do this we're going to set dy/dx
equal to this quotient here,
which means we'll set M of x comma y
divided by negative N of x comma y
equal to dy/dx, which is equal to y-squared
divided by x plus y.
Now, remember, there are several ways to do this.
One way is just to let M of x comma y
equal the entire quotient here
and then let N of x comma y equal negative one.
So let's start by doing that,
and then we'll look at this another way.
Again, we'll let M of x comma y
equal this entire quotient.
If we do that, then we can just let N of x comma y
equal negative one.
Notice, if this was negative one,
we'd have the opposite of negative one,
which would just be one.
So in differential form, we would have the quotient
y-squared divided by the quantity x plus y times dx,
and then since N of x comma y is equal to negative one,
we could put plus negative one times dy
or just minus one dy.
This one, of course, is optional as well.
Equals zero.
Let's take a look at another way
of writing this in differential form.
Again, we'll start by setting the quotient M of x comma y
divided by negative N of x comma y
equal to y-squared
divided by the quantity x plus y.
We could just let M of x comma y
equal the numerator y-squared
and then let N of x comma y
equal the opposite of the denominator
or the opposite of the quantity x plus y.
Notice in this form this quotient here
would be equal to this quotient on the right,
which means another way to write this equation
in differential form would be y-squared dx,
and then we'd have plus the opposite
of the quantity x plus y dy
or just minus the quantity x plus y dy
equals zero.
So the main thing to notice here,
when writing a first-order DE in differential form,
the form is not unique, though it would be equivalent
to the equation in standard form.
Let's take a look at another example.
Notice this differential equation
is not given in standard form,
so the first step is going to be to write this in standard form
and then convert it to differential form.
So we'll have to start by solving this for y prime.
So we have y times the quantity y-squared y prime
minus two equals x.
I'm going to go ahead and distribute here,
so we'll have y to the third y prime
minus two y equals x.
Let's go ahead and add two y to both sides of the equation,
so we'll have y cubed y prime equals,
this will be x plus two y.
Last step, we'll divide both sides by y the third.
So this simplifies to y prime or dy/dx
equals this quotient here, x plus two y
divided by y the third,
and now we'll go ahead and write this in differential form.
Again, we'll go ahead and do this two different ways.
We're going to start with M of x comma y
divided by negative N of x comma y
equals dy/dx, which in this case is x plus two y
divided by y the third.
So, again, one way that always works
is to let M of x comma y
equal the entire quotient here,
and if we do this, then N of x comma y
is just going to be equal negative one.
So, again, writing this in differential form
or this form here.
We'll have our quotient, the quantity x plus two x
divided by y to the third
times dx plus negative one times dy
or just minus one dy or just minus dy equals zero.
But notice that's not the only way to do this.
If we go back and take a look at this proportion here
one more time,
we could let M of x comma y
equal just the numerator, x plus two y.
If we did this, then N of x comma y
would be the opposite of the denominator,
or negative y cubed.
So now if we write this in differential form,
we would have the quantity x plus two y times dx
plus negative y cubed times dy equals zero
or just minus y cubed dy
equals zero.
Okay, that's going to do it for this video.
I hope you found this helpful.