WEBVTT

00:00:01.316 --> 00:00:03.896 A:middle
>> Welcome to the Cypress
College Math Review

00:00:04.246 --> 00:00:08.206 A:middle
on Second Order Differential
Equations, Special Types.

00:00:09.286 --> 00:00:13.126 A:middle
In general a 2nd order
differential equation can be

00:00:13.126 --> 00:00:16.936 A:middle
written as a function of
the independent variable,

00:00:17.596 --> 00:00:20.876 A:middle
the dependent variable and the
first and second derivatives.

00:00:22.006 --> 00:00:26.126 A:middle
There are two special types of
2nd order differential equations

00:00:27.046 --> 00:00:29.606 A:middle
where a substitution
reduces them

00:00:29.966 --> 00:00:32.186 A:middle
to a 1st order differential
equation

00:00:32.516 --> 00:00:33.736 A:middle
that we're able to solve.

00:00:35.066 --> 00:00:41.246 A:middle
The first type has no dependent
variable or no x in this case.

00:00:43.346 --> 00:00:47.346 A:middle
By substituting v equal dx
dt or the first derivative,

00:00:47.896 --> 00:00:51.776 A:middle
the second derivative
becomes the derivative of v

00:00:51.776 --> 00:00:53.476 A:middle
with respect to t or v prime.

00:00:54.516 --> 00:01:00.076 A:middle
[ Pause ]

00:01:00.576 --> 00:01:04.416 A:middle
In this first example t is
the independent variable

00:01:04.986 --> 00:01:09.456 A:middle
and x is the dependent
variable and x is missing.

00:01:11.566 --> 00:01:13.886 A:middle
This is a 2nd order
differential equation.

00:01:14.806 --> 00:01:21.146 A:middle
So we let v equal dx dt which
means that the second derivative

00:01:21.146 --> 00:01:26.356 A:middle
of x with respect to t will
equal dv dt or v prime.

00:01:27.256 --> 00:01:28.806 A:middle
We make these substitutions.

00:01:31.276 --> 00:01:39.636 A:middle
So t squared times v prime
plus v squared is equal

00:01:39.636 --> 00:01:47.516 A:middle
to 2t times v. Interchanging
the last two terms.

00:01:48.516 --> 00:01:56.146 A:middle
[ Pause ]

00:01:56.646 --> 00:01:58.456 A:middle
Divide each term by t squared.

00:01:59.516 --> 00:02:11.846 A:middle
[ Pause ]

00:02:12.346 --> 00:02:15.616 A:middle
The left hand side looks like
we have a 1st order linear

00:02:15.616 --> 00:02:16.886 A:middle
differential equation.

00:02:17.836 --> 00:02:22.356 A:middle
But we look at the right hand
side and we recognize that no,

00:02:22.456 --> 00:02:25.726 A:middle
we have instead a Bernoulli
differential equation.

00:02:26.936 --> 00:02:29.606 A:middle
Here's the form for a
Bernoulli differential equation.

00:02:30.636 --> 00:02:34.376 A:middle
So the first step to solve a
Bernoulli differential equation

00:02:34.746 --> 00:02:37.016 A:middle
would be to divide
by y to the nth

00:02:37.446 --> 00:02:39.486 A:middle
which in this case is v squared.

00:02:40.666 --> 00:02:49.506 A:middle
So we have v to the negative
2, v prime minus 2 over t,

00:02:50.776 --> 00:02:57.416 A:middle
v to the negative 1 is equal
to negative 1 over t squared.

00:02:57.996 --> 00:03:00.886 A:middle
We make our substitution.

00:03:01.596 --> 00:03:06.306 A:middle
We let u equal y to
the 1 minus n power.

00:03:06.906 --> 00:03:13.706 A:middle
Well in this case u will be
v to the 1 minus 2 power.

00:03:14.636 --> 00:03:17.336 A:middle
So that's v to the
negative 1 power.

00:03:18.426 --> 00:03:21.046 A:middle
We also have to find
out what u prime is,

00:03:21.276 --> 00:03:24.916 A:middle
so u prime would be negative v

00:03:24.916 --> 00:03:28.066 A:middle
to the negative 2 times v
prime by the chain rule.

00:03:30.076 --> 00:03:34.986 A:middle
So we replace v to
the negative 2 v prime

00:03:36.056 --> 00:03:38.786 A:middle
with the opposite of u prime.

00:03:40.446 --> 00:03:45.856 A:middle
We replace v to the
negative 1 with u.

00:03:46.516 --> 00:03:54.676 A:middle
[ Pause ]

00:03:55.176 --> 00:03:58.346 A:middle
Multiply both sides of this
equation by negative 1.

00:03:59.516 --> 00:04:12.926 A:middle
[ Pause ]

00:04:13.426 --> 00:04:16.976 A:middle
Now we have a 1st order linear
differential equation that's

00:04:16.976 --> 00:04:17.976 A:middle
in standard form.

00:04:18.516 --> 00:04:28.366 A:middle
[ Pause ]

00:04:28.866 --> 00:04:32.086 A:middle
Now we solve this 1st order
linear differential equation.

00:04:34.776 --> 00:04:36.046 A:middle
Here's our format.

00:04:38.036 --> 00:04:40.346 A:middle
We need to find the
integrating factor mu.

00:04:42.256 --> 00:04:44.736 A:middle
There's our formula for mu.

00:04:45.286 --> 00:04:48.376 A:middle
P of x in this case
is positive 2

00:04:48.376 --> 00:04:53.296 A:middle
over t. We simplify
to get t squared.

00:04:53.846 --> 00:04:59.876 A:middle
So when we solve a 1st order
linear differential equation we

00:04:59.876 --> 00:05:04.486 A:middle
get mu times y equals
the antiderivative

00:05:04.656 --> 00:05:06.556 A:middle
of the product of q with mu.

00:05:09.076 --> 00:05:13.656 A:middle
So in this case we'll have
the integrating factor mu

00:05:13.656 --> 00:05:20.556 A:middle
which is t squared times
our function u is equal

00:05:20.556 --> 00:05:22.416 A:middle
to the antiderivative of mu

00:05:22.416 --> 00:05:26.716 A:middle
which is t squared times
the function q which is 1

00:05:26.716 --> 00:05:31.256 A:middle
over t squared, we're
integrating that with respect

00:05:31.256 --> 00:05:38.416 A:middle
to t. So we get t plus the
constant of integration.

00:05:41.006 --> 00:05:47.146 A:middle
So u is equal to t plus
c sub 1 over t squared.

00:05:48.666 --> 00:05:54.156 A:middle
And now we substitute back, u is
equal it v the negative 1 power

00:05:54.866 --> 00:05:58.976 A:middle
so 1 over v is equal to this.

00:05:59.516 --> 00:06:12.786 A:middle
[ Pause ]

00:06:13.286 --> 00:06:15.036 A:middle
So let's take reciprocals.

00:06:16.736 --> 00:06:22.376 A:middle
So v is equal to t
squared over t plus c sub 1

00:06:23.736 --> 00:06:24.996 A:middle
and we substitute back.

00:06:25.796 --> 00:06:27.816 A:middle
Recall that v is equal to dx dt.

00:06:28.516 --> 00:06:35.216 A:middle
[ Pause ]

00:06:35.716 --> 00:06:37.146 A:middle
We're going to need
to integrate that.

00:06:37.656 --> 00:06:40.316 A:middle
So since the degree on top
is greater than or equal

00:06:40.316 --> 00:06:43.876 A:middle
to the degree on the bottom,
we need to do long division.

00:06:44.516 --> 00:06:50.406 A:middle
[ Pause ]

00:06:50.906 --> 00:06:56.846 A:middle
So t squared divided by t is
t. So t times t is t squared,

00:06:57.336 --> 00:07:05.466 A:middle
c sub 1 times t, change the
signs, divide negative c sub 1 t

00:07:05.716 --> 00:07:07.876 A:middle
by t and I get negative c sub 1.

00:07:10.246 --> 00:07:13.046 A:middle
Multiply negative c
sub 1 times 9 divisor

00:07:13.556 --> 00:07:18.366 A:middle
and I get negative c sub
1t minus c sub 1 squared.

00:07:19.496 --> 00:07:21.976 A:middle
Change the signs and
I get c sub 1 squared.

00:07:24.426 --> 00:07:30.296 A:middle
So I'm integrating 1dx equals
and when I'm integrating

00:07:30.296 --> 00:07:32.806 A:middle
over here on the right
side is my quotient

00:07:33.626 --> 00:07:38.176 A:middle
which is t minus c sub
1 plus my remainder

00:07:38.176 --> 00:07:44.426 A:middle
which is c sub 1 squared over my
divisor which is t plus c sub 1.

00:07:45.386 --> 00:07:47.116 A:middle
I'm integrating with
respect to t.

00:07:49.066 --> 00:07:53.376 A:middle
So on the left hand
side I get x,

00:07:54.666 --> 00:07:56.996 A:middle
on the right hand
side I get t squared

00:07:56.996 --> 00:08:07.736 A:middle
over 2 minus c sub 1 times t
plus c sub 1 squared times the

00:08:07.736 --> 00:08:10.586 A:middle
natural log of the absent
value of my denominator

00:08:11.446 --> 00:08:16.836 A:middle
which t plus c sub 1 plus a
new constant of integration.

00:08:17.746 --> 00:08:22.336 A:middle
And notice that x equals another
constant is also a solution.

00:08:23.516 --> 00:08:44.426 A:middle
[ Pause ]

00:08:44.926 --> 00:08:47.636 A:middle
Here's another 2nd order
differential equation.

00:08:49.046 --> 00:08:50.816 A:middle
The independent variable is t,

00:08:51.576 --> 00:08:56.566 A:middle
the dependent variable
is x and x is missing.

00:08:59.036 --> 00:09:03.056 A:middle
When you look at this term here
that's not the 2nd derivative,

00:09:04.096 --> 00:09:07.336 A:middle
that's the second power
of the 1st derivative.

00:09:09.106 --> 00:09:10.916 A:middle
So we make our substitutions.

00:09:12.396 --> 00:09:20.056 A:middle
V is equal to dx dt and the 2nd
derivative is dv dt or v prime.

00:09:21.396 --> 00:09:31.626 A:middle
So we get 2t times v prime
equals v squared minus 1.

00:09:33.246 --> 00:09:35.216 A:middle
This is separable
differential equation.

00:09:36.616 --> 00:09:45.646 A:middle
So we have 1 over the quantity
v squared minus 1 dv equals 1

00:09:45.646 --> 00:09:47.236 A:middle
over the quantity 2t dt.

00:09:48.446 --> 00:09:53.036 A:middle
To be able to integrate
that we have do a partial

00:09:53.036 --> 00:09:54.436 A:middle
fraction decomposition.

00:09:56.576 --> 00:10:05.296 A:middle
So we have 1 over the product
of v plus 1 with v minus 1.

00:10:06.386 --> 00:10:14.316 A:middle
To set that up we would
have a over b plus 1 plus b

00:10:14.986 --> 00:10:18.156 A:middle
over v minus 1, multiply
both sides

00:10:18.216 --> 00:10:20.196 A:middle
by the least common denominator

00:10:20.196 --> 00:10:21.816 A:middle
which is the original
denominator.

00:10:22.176 --> 00:10:31.486 A:middle
So we get 1 equals a times v
minus 1 plus b times v plus 1,

00:10:35.216 --> 00:10:44.706 A:middle
let v equal 1 and we get
that 1 is equal to 2 times b

00:10:45.556 --> 00:10:49.396 A:middle
which gives that
b is equal to 1/2.

00:10:51.026 --> 00:10:56.666 A:middle
If we then let v equal negative
1 we get that 1 is equal

00:10:56.666 --> 00:11:02.246 A:middle
to negative 2a which tells us
that a is equal to negative 1/2.

00:11:05.426 --> 00:11:09.406 A:middle
So now the integral on the
left hand side becomes a

00:11:09.466 --> 00:11:20.006 A:middle
which is negative 1/2
over v plus 1 plus b

00:11:20.456 --> 00:11:23.676 A:middle
which is 1/2 over v minus 1.

00:11:25.926 --> 00:11:32.896 A:middle
On the right hand side we're
integrating 1/2 over t dt,

00:11:34.216 --> 00:11:43.426 A:middle
multiply by 2 and we
get negative natural log

00:11:43.426 --> 00:11:49.226 A:middle
of the absolute value of the
denominator plus natural log

00:11:49.226 --> 00:11:54.506 A:middle
of the absolute value of the
denominator equals natural log

00:11:54.506 --> 00:11:55.376 A:middle
of the absolute value

00:11:55.376 --> 00:12:00.636 A:middle
of the denominator plus our
constant of integration.

00:12:01.456 --> 00:12:03.416 A:middle
We're going to go ahead
and call our constant

00:12:03.416 --> 00:12:07.086 A:middle
of integration the natural
log of the absolute value of c

00:12:07.416 --> 00:12:11.016 A:middle
so that we're able to
combine these logarithms.

00:12:11.736 --> 00:12:13.266 A:middle
That's a trick that you can use

00:12:13.616 --> 00:12:15.096 A:middle
that makes things
easier for you.

00:12:16.516 --> 00:12:24.016 A:middle
[ Pause ]

00:12:24.516 --> 00:12:26.256 A:middle
Let's combine the logarithms.

00:12:27.516 --> 00:12:38.856 A:middle
[ Pause ]

00:12:39.356 --> 00:12:40.966 A:middle
Exponentiate both sides.

00:12:41.516 --> 00:12:49.006 A:middle
[ Pause ]

00:12:49.506 --> 00:12:54.466 A:middle
So now we have this
quotient equals plus

00:12:54.466 --> 00:12:58.716 A:middle
or minus c t. We can
go ahead and call

00:12:58.716 --> 00:13:05.006 A:middle
that a new constant c sub 1
times t. Now let's solve for v.

00:13:06.066 --> 00:13:16.086 A:middle
So we cross multiply and we have
v minus 1 equals c sub 1 tv plus

00:13:16.396 --> 00:13:28.856 A:middle
c sub 1 t. So v minus c sub
1 tv equals c sub 1 t plus 1.

00:13:29.646 --> 00:13:30.486 A:middle
Factor out a v.

00:13:31.516 --> 00:13:37.916 A:middle
[ Pause ]

00:13:38.416 --> 00:13:44.636 A:middle
So v is equal to c
sub 1 t plus 1 divided

00:13:44.636 --> 00:13:55.036 A:middle
by 1 minus c sub 1 times
t. Let's substitute back.

00:13:55.516 --> 00:14:01.746 A:middle
V is equal to dx dt, separate.

00:14:02.516 --> 00:14:12.246 A:middle
[ Pause ]

00:14:12.746 --> 00:14:15.326 A:middle
And since the degree on top
is greater than or equal

00:14:15.406 --> 00:14:18.146 A:middle
to the degree on the bottom
we have to do long division.

00:14:19.516 --> 00:14:27.936 A:middle
[ Pause ]

00:14:28.436 --> 00:14:32.036 A:middle
Divide c sub 1t by
negative c sub 1t

00:14:32.036 --> 00:14:33.566 A:middle
and we simply get negative 1.

00:14:34.866 --> 00:14:37.246 A:middle
Multiply negative
1 times my divisor

00:14:37.556 --> 00:14:43.886 A:middle
and we get c sub 1 times t
minus 1, change the signs

00:14:44.536 --> 00:14:46.306 A:middle
and we get 2 as our remainder.

00:14:46.886 --> 00:14:54.606 A:middle
So on the right hand side
we're integrating negative 1

00:14:54.606 --> 00:14:57.806 A:middle
which is our quotient
plus or remainder

00:14:57.806 --> 00:15:00.216 A:middle
which is 2 over our divisor.

00:15:06.576 --> 00:15:16.736 A:middle
So our answer x is equal to
negative t plus and we need

00:15:16.736 --> 00:15:19.626 A:middle
to do a u substitution here
to be able to integrate this.

00:15:22.086 --> 00:15:32.586 A:middle
U equals negative c sub 1t plus
1, so du is negative c sub 1 dt.

00:15:37.206 --> 00:15:45.726 A:middle
So our integral turns
out to be negative 2

00:15:45.996 --> 00:15:51.176 A:middle
over c sub 1 times natural
log of the absolute value

00:15:51.176 --> 00:15:59.236 A:middle
of 1 minus c sub 1 times t plus
the constant of integration.

00:15:59.786 --> 00:16:05.046 A:middle
That's the answer to our
differential equation.

00:16:06.516 --> 00:16:19.456 A:middle
[ Pause ]

00:16:19.956 --> 00:16:23.416 A:middle
Our second type of 2nd
order differential equation,

00:16:23.416 --> 00:16:27.156 A:middle
a special type has no
independent variable,

00:16:27.676 --> 00:16:28.486 A:middle
t is missing.

00:16:30.226 --> 00:16:36.766 A:middle
In this case we replace dx dt
with v and our goal here is

00:16:36.766 --> 00:16:39.086 A:middle
to make t completely disappear.

00:16:39.996 --> 00:16:42.436 A:middle
So we have x in our equation.

00:16:43.196 --> 00:16:47.306 A:middle
We're replacing dx dt
with v. Now we need

00:16:47.306 --> 00:16:49.886 A:middle
to replace the 2nd
derivative with something

00:16:49.886 --> 00:16:52.046 A:middle
that has no t's in it at all.

00:16:53.276 --> 00:17:00.836 A:middle
So the 2nd derivative is equal
to dv dt but we need that t

00:17:00.836 --> 00:17:02.386 A:middle
to be completely gone.

00:17:02.686 --> 00:17:05.136 A:middle
We can't just call it v prime

00:17:05.446 --> 00:17:07.816 A:middle
because v prime is
the same as dv dt.

00:17:07.956 --> 00:17:12.516 A:middle
We need that t to be out
of the equation completely.

00:17:13.726 --> 00:17:19.666 A:middle
By the chain rule dv dt is
equal dv dx times dx dt.

00:17:19.916 --> 00:17:25.096 A:middle
Dx dt recall is equal to v.
So we make that substitution.

00:17:25.886 --> 00:17:30.836 A:middle
So the 2nd derivative can
be written as dv dx times v.

00:17:31.376 --> 00:17:36.586 A:middle
If we make that substitution
now our equation can be written

00:17:36.586 --> 00:17:45.316 A:middle
in terms of x v and dv dx
times v. No t's at all.

00:17:46.866 --> 00:17:48.766 A:middle
Now we can solve the
differential equation.

00:17:49.936 --> 00:17:51.386 A:middle
Here's our first example.

00:17:52.486 --> 00:17:56.586 A:middle
It's a 2nd order differential
equation and t is missing.

00:18:00.116 --> 00:18:01.776 A:middle
So we make our substitutions.

00:18:03.146 --> 00:18:07.536 A:middle
We're going to replace dx dt
with v and then we're going

00:18:07.536 --> 00:18:11.766 A:middle
to replace the 2nd
derivative with dv dx times v.

00:18:13.706 --> 00:18:26.226 A:middle
So x squared plus 1 times dv
dx times v equals 2x times

00:18:26.226 --> 00:18:26.886 A:middle
v squared.

00:18:28.316 --> 00:18:30.426 A:middle
This is a separable
differential equation.

00:18:31.686 --> 00:18:37.576 A:middle
I get 1 over v dv equals 2x

00:18:38.186 --> 00:18:41.176 A:middle
over the quantity x
squared plus 1 dx.

00:18:42.966 --> 00:18:48.556 A:middle
So natural log of the absolute
value of v equals natural log

00:18:48.556 --> 00:18:53.626 A:middle
of the absolute value of the
quantity x squared plus 1 plus

00:18:55.316 --> 00:19:01.136 A:middle
and then I chose the form of
my constant of integration

00:19:01.716 --> 00:19:04.566 A:middle
to be natural log of
the absolute value of c

00:19:05.336 --> 00:19:08.236 A:middle
because all the other
terms are natural logs.

00:19:09.326 --> 00:19:12.626 A:middle
So this is equal to natural
log of the absolute value

00:19:12.626 --> 00:19:15.446 A:middle
of c times the quantity
x squared plus 1.

00:19:16.506 --> 00:19:19.676 A:middle
I combine those logarithms
on the right hand side,

00:19:20.156 --> 00:19:22.036 A:middle
exponentiate both sides.

00:19:23.826 --> 00:19:31.896 A:middle
So I have absolute value
of v equals absolute value

00:19:31.896 --> 00:19:36.576 A:middle
of c times the quantity
x squared plus 1.

00:19:37.516 --> 00:19:46.766 A:middle
[ Pause ]

00:19:47.266 --> 00:19:52.506 A:middle
So v is equal to plus or minus
c times the quantity x squared

00:19:52.566 --> 00:19:53.226 A:middle
plus 1.

00:19:55.276 --> 00:19:59.336 A:middle
So v is equal to a new
constant times that quantity.

00:20:00.716 --> 00:20:03.626 A:middle
Replace v to what it's
equal to which is dx dt.

00:20:04.516 --> 00:20:09.216 A:middle
[ Pause ]

00:20:09.716 --> 00:20:14.986 A:middle
And is a separable differential
equation so we have 1

00:20:14.986 --> 00:20:17.196 A:middle
over the quantity x
squared plus 1 dx.

00:20:17.196 --> 00:20:21.616 A:middle
And the right hand side we're
just integrating the constant.

00:20:24.426 --> 00:20:26.596 A:middle
So on the left hand
side we get arctangent,

00:20:30.456 --> 00:20:35.156 A:middle
the right hand side we get c sub
1 times t plus a new constant

00:20:35.156 --> 00:20:35.956 A:middle
of integration.

00:20:37.596 --> 00:20:41.646 A:middle
So x equals tangent
of this quantity.

00:20:46.736 --> 00:20:50.036 A:middle
X equals a constant
is also a solution

00:20:50.036 --> 00:20:51.316 A:middle
to this differential equation.

00:20:52.516 --> 00:20:58.876 A:middle
[ Pause ]

00:20:59.376 --> 00:21:02.296 A:middle
And that gives us the solutions
to this differential equation.

00:21:03.516 --> 00:21:13.076 A:middle
[ Pause ]

00:21:13.576 --> 00:21:16.606 A:middle
On this next example we don't
even see what our independent

00:21:16.606 --> 00:21:17.526 A:middle
variable would be.

00:21:18.426 --> 00:21:20.526 A:middle
Let's write it in terms
of [inaudible] notation

00:21:20.526 --> 00:21:22.226 A:middle
and just choose an
independent variable.

00:21:23.366 --> 00:21:27.606 A:middle
So it would be the
2nd derivative of y,

00:21:28.046 --> 00:21:38.466 A:middle
let's say with respect to t plus
y times the 1st derivative cubed

00:21:39.746 --> 00:21:40.496 A:middle
equals zero.

00:21:41.476 --> 00:21:45.266 A:middle
So the independent
variable t is missing,

00:21:46.146 --> 00:21:47.586 A:middle
we're going to make
our substitution.

00:21:48.576 --> 00:21:54.176 A:middle
So v is equal to dy dt and
the 2nd derivative is equal

00:21:54.176 --> 00:22:00.406 A:middle
to dv dy times v. So we
replace the 2nd derivative

00:22:00.406 --> 00:22:03.066 A:middle
with v times dv dy.

00:22:05.016 --> 00:22:12.626 A:middle
And then we have plus y times
and the derivative gets replaced

00:22:12.626 --> 00:22:15.816 A:middle
with v, so we have
the derivative cubed

00:22:15.816 --> 00:22:18.796 A:middle
so we have v cubed and
that's equal to zero.

00:22:20.366 --> 00:22:26.066 A:middle
This is a separable differential
equation separating the terms.

00:22:26.866 --> 00:22:39.616 A:middle
We have v over v cubed
dv equals negative y dy.

00:22:40.696 --> 00:22:43.736 A:middle
On the left hand side we're
integrating v to the negative 2.

00:22:46.286 --> 00:22:48.816 A:middle
On the right hand side we're
integrating negative y.

00:22:51.396 --> 00:22:53.236 A:middle
So we get negative v

00:22:53.236 --> 00:22:57.426 A:middle
to the negative 1
equals negative y squared

00:22:57.426 --> 00:23:00.776 A:middle
over 2 plus c sub 1.

00:23:01.516 --> 00:23:10.446 A:middle
[ Pause ]

00:23:10.946 --> 00:23:23.436 A:middle
So we get 1 over v is equal
to, to multiply by negative 1

00:23:23.436 --> 00:23:25.686 A:middle
but let's also get a
common denominator here.

00:23:29.136 --> 00:23:31.946 A:middle
So 1 over v is equal

00:23:31.946 --> 00:23:38.646 A:middle
to y squared plus
new constant over 2.

00:23:45.396 --> 00:23:58.036 A:middle
So v is equal to 2 over
y squared plus c sub 2.

00:24:02.486 --> 00:24:03.766 A:middle
Now we substitute back.

00:24:05.796 --> 00:24:08.116 A:middle
So v is equal to dy dt.

00:24:09.516 --> 00:24:19.466 A:middle
[ Pause ]

00:24:19.966 --> 00:24:23.016 A:middle
So we integrate the
left hand side,

00:24:23.456 --> 00:24:29.656 A:middle
we're integrating y squared
plus c sub 2 with respect to y.

00:24:29.656 --> 00:24:34.746 A:middle
And the right hand side
we're integrating just 2.

00:24:41.686 --> 00:24:54.396 A:middle
So we have y cubed over 3 plus
c sub 2 times y equals 2t plus

00:24:54.626 --> 00:24:57.496 A:middle
another constant c sub 3 or
a constant of integration.

00:24:59.216 --> 00:25:07.886 A:middle
If I multiply 3 by 3 we have
y cubed plus 3 times cu sub 2,

00:25:07.886 --> 00:25:10.096 A:middle
so we have another
constant, a new constant,

00:25:10.626 --> 00:25:19.476 A:middle
we'll call it c sub 4 times y
equals 6t plus 3 times our old

00:25:19.476 --> 00:25:22.176 A:middle
constant c sub 3, we'll
call that c sub 5.

00:25:23.636 --> 00:25:27.196 A:middle
So here's our solution to
the differential equation.

00:25:29.446 --> 00:25:32.846 A:middle
Now you do need to note that
you have two different arbitrary

00:25:32.876 --> 00:25:33.686 A:middle
constants here.

00:25:33.736 --> 00:25:35.886 A:middle
I call them c sub 4 and c sub 5.

00:25:36.606 --> 00:25:39.176 A:middle
Your numbering scheme will
probably be different.

00:25:39.906 --> 00:25:42.616 A:middle
But you do have to note that you
do have two different arbitrary

00:25:42.616 --> 00:25:43.286 A:middle
constants here.

00:25:44.516 --> 00:25:53.466 A:middle
[ Pause ]

00:25:53.966 --> 00:25:56.746 A:middle
Here's some additional
problems to try on your own.

00:25:57.596 --> 00:26:02.486 A:middle
I encourage you to pause
the video, do these problems

00:26:03.286 --> 00:26:05.956 A:middle
and then check with
the solutions that are

00:26:05.956 --> 00:26:07.296 A:middle
at the end of the video.

00:26:08.516 --> 00:26:16.426 A:middle
[ Pause ]

00:26:16.926 --> 00:26:19.346 A:middle
Here are the solutions
to the practice problems.

00:26:20.516 --> 00:26:27.500 A:middle
[ Pause ]