>> Welcome to the Cypress College Math Review
on First-Order Linear Differential Equations.
A first-order linear differential equation is one
that can be written in the form y prime plus some function
of x times y equals some other function
of x. Your first step is to write it in that form.
The second step is to obtain
and simplify the integrating factor mu of x. Mu of x is equal
to e to the power of the antiderivative of p of x.
So find mu of x, which is e to the power
of the antiderivative of p and simplify.
Leave off the constant of integration and in this case,
and simplify the result that you get for mu.
Third step, to solve the differential equation y times mu
is equal to the antiderivative of q times mu.
Let's first write its standard form by dividing by 3x minus 1,
so we have y prime minus 6 over the quantity 3x minus 1 times y,
is equal to negative 10 times and we have 3x minus 1 quantity
to the 1/3 power divided by 3x minus 1 to the 1st power,
which gives us that quantity to the negative 2/3 power.
We find our integrating factor, mu of x, so mu of x is equal
to e to the power of an antiderivative of the function p
of x. And the function p of x is this function here,
including the negative.
So we have e to the power of the antiderivative of negative 6
over the quantity 3x minus 1.
Pull out the negative 6, we're integrating 1 over 3x minus 1.
So let's multiply by 3.
And divide by 3.
So we're integrating 1 over udu, so we have natural log, yes.
So we have e to the power of negative 2 natural log
of the absolute value of what e was, which is 3x minus 1.
And we leave off the constant of integration
for the integrating factor.
Raise the negative 2 up for our natural log,
so we get the absolute value of 3x minus 1 raised
to the negative 2 power
and since we're squaring we can leave off the absolute value.
And we simply get quantity 3x minus 1 to the negative 2 power.
As our integrating factor.
So, we now know that mu times y is equal to the antiderivative
of mu times q with respect to x. So mu, which is 3x minus 1
on a quantity to the negative 2 power times y,
is equal to the antiderivative of mu,
which is the quantity 3x minus 1 to the negative 2 power,
times q which is this function here,
negative 10 times 3x minus 1 to the negative 2/3 power.
We're integrating that with respect to x.
Alright, so over here on the right hand side,
we pull out the negative 10 and we're going
to get integrating the quantity 3x minus 1
to the negative 8/3 power.
So that would be our u, u is 3x minus 1.
So it would be 3x minus 1 to the negative 5/3 power,
divided by 5/3, which is the same thing
as multiplying by negative 3/5.
Because u is 3x minus 1, we would have to divide by 3
and then we add a negative 10 out there.
You don't leave off your constant of integration
in general, that was only for the integrating factor.
These 3 cancel, so do the negatives, and the 10
and the 5 reduce down to just 2.
So, now we solve for y, so we have y is equal to 3x minus 1
to the positive 2 power, times 2, times quantity 3x minus 1
to the negative 5/3 power plus c. And distribute
to get our answer of 2 times the quantity 3x minus 1
to the 1/3 power, plus c times quantity 3x minus 1
to the 2nd power.
Let's write it in standard form, so we divide
by 1 plus x squared each term.
So we have y prime minus 2x over 1 plus x squared times y.
And a profact routed x square we're left
with x squared plus 1.
And the denominators the same, x squared plus 1.
So those cancel and simply give us x squared.
Alright, we have the standard form.
Let's the find the integrating factor mu of x as e
to the power of the function p
So we have e to the power of, don't forget the negative,
to the antiderivative of negative 2x
over 1 plus x squared dx.
So a u substitution would help us here.
U equals 1 plus x squared.
So du is equal to 2x dx, ok.
So we pull out the negative, so mu of x is equal to e
to the power of negative
and then I would simply have 1 over u du.
So we have e to the power
of negative natural log absolute value of 1 plus x squared.
We leave off the constant of integration.
Since 1 plus x squared is positive,
we can leave off the absent value.
We bring the negative 1 up, because of the natural log
and we end up with quantity 1 plus x squared
to the negative 1 power.
There's our integrating factor.
Now we apply it.
So mu times y is equal to the antiderivative
of mu times q integrated with respect to x.
So 1 plus x squared quantity
to the negative 1 power times y is equal
to the antiderivative of that same thing.
Times q which is simply x squared.
What I'm integrating is x squared over x squared plus 1.
To integrate that, I would add and subtract 1.
Just a trick for, actually for dividing.
Just a short hand way
for actually doing the long division.
So I get 1 minus 1 over x squared plus 1.
So we have x minus arc tangent of x plus c
and then we multiply both sides by 1 plus x squared to solve
for y. So y is equal to the quantity 1 plus x squared
to the positive 1 power times the right hand side,
which is x minus arc tangent of x plus c
and that gives us our answer.
This equation is already in standard form, so we straight
to our integrating factor.
So mu of x is equal to e to the power of the antiderivative of p
with respect to x. So we have e to the power
of the antiderivative of the function tangent.
And the antiderivative of tangent is negative natural log
of the absolute value of cosign.
We leave off the constant of integration.
Bring the negative 1 up and we have the absolute value
of cosign raised to the negative 1 power,
which of course you're going to write as secant.
Now what about the absolute value?
You're actually going to drop that also,
so mu of x is simply secant.
Why can you leave off the absolute value?
Because when you apply an integrating factor,
you're actually multiplying both sides by it.
When you multiply both sides by a positive
or multiply both sides by a negative, it's really not going
to make any difference is it?
So if you have an absolute value on your integrating factor,
you can leave that off also.
Alright, we have our integrating factor, let's apply it.
So we know that mu times y is equal to the antiderivative
of mu times q integrated with respect to x. So we have secant
of x times y is equal to the antiderivative of secant of x,
times cosign of x, integrated with respect to x.
So in the right hand side, I'm only integrating 1.
When I integrate 1 I get x plus c. To solve for y,
I multiply both sides by cosign.
And there's our answer.
This is already in standard form, so we go straight
to our integrating factor.
Mu of x is equal to e to the power of the antiderivative of p
of x with respect to x. This time that is 3x plus 1,
quantity all over x. You can split up the numerator
but not a denominator.
So that would be 3x over x which is just 3 plus 1 over x.
And we integrate that.
We get 3x plus natural log of the absolute value
of x. Leave off the plus c. So I have e to the power of a sum.
So I can rewrite that as e to the 3x times e to the power
of natural log of the absolute value of x,
because of this property.
And then the second factor can be rewritten as absolute value
of x and like we said in the last problem,
we leave off the absolute value.
I'll put that out front.
And our integrating factor is x times e to the 3x.
Now let's apply it.
So we know that mu times y is going
to equal the antiderivative of mu times q integrated
with respect to x. So x times e
to the 3x times y will equal the antiderivative
of that same quantity times q, which was, e to the negative 3x
over x, integrated with respect to x. Once again,
all we're integrating on the right hand side is just 1.
So we get, once again x plus c. So y is equal
to x plus c all over, x e to the 3x.
Which is our answer.
Here's some extra practice I encourage you
to try these on your own.
Pause the video, do these problems.
And then once you've completed the problems, restart the video
and check your answers.
Here are the answers to the two practice problems.