BAM!
Mr. Tarrou
In this lesson, we are gonna take a look at two examples
of solving differential equations of one particular type
of differential equation
A separable differential equation.
Now there's a whole semester course on this in college,
It can be quite a complex type scenario,
but we're looking at just this one type
and then we're gonna use it in the next section that I'm teaching
uh, in my calculus course
were we solve differential equations involving exponential growth and decay.
...or direct in inverse variation.
Uh, with that exponential growth and decay model,
there's a rule that we can use as a shortcut,
to like quicken the process.
..but to see where that comes from, we gotta understand
how to solve differential equations.
like these separable differential equations.
A differential equation of the form dy/ dx, is equal to f(y)*g(x)
and this is called a SEPARABLE differential equation
we'll separate the variables by writing in the form of
1 over f(y) dy= g(x) dx
In other words we're going to write our equation
in this equation we have variables of x and y
but we're gonna separate those to opposite sides of the equation
and then we're going to find the solution by anti-differentiating each side
with respect to its isolated variable
Now we have a division of f(y) here, this may or may not look like one over some kind of function.
It probably will. And we have with our notation that we are using in our calculus here.
It is not really the division of d(x), but it is set up on purpose to allow us easily manipulate these symbols
and more easily see our way through the calculus techniques and steps that we have to do.
So we are going to undo that division of d(x), moving it over to the other side with what appears to be
multiplication.
Our first example we are going to take a look at, of two examples, is we want to solve the differential
equation y prime is equal to 7x over y. And this one is going to prove to be a fairly simple example.
We are going to rewrite that y prime in terms of this notation here, dy over dx.
Which is equal to 7x over y. We are going to move the dx over to the other side, and it appears to be divison.
So we are going to move it over with what appears to be multiplication. We will have dy is equal to 7x over y, dx.
So I have my variables of x on one side, but I am not done here, because I have not finished with y.
I am going to multiply both sides by y. And I get y, dy is equal to 7x, dx.
Now this is a fairly straight forward example. The next one will be quite a bit longer.
But you see here, when I have separated my variables of x and y, and I do not have in this particular situation,
one over something with a variable of y. That is just how we have our definition written.
Though you will see that pattern quite often.
Now we are going to take the integral of both sides of this, with respect to the isolated variable.
And of course this is pretty straight forward. We have y to the first, so we are going to make that y to the second.
Then divide by the power that we have increased by one, with is effectively dividing by 2.
Or as I am going to write it here, multiplying by one over 2.
And that is equal to...now, I am doing indefinite integration here and I do not have enough information
to find a particular solution. We are finding a general solution. And why did I not write a "plus C" on the left?
Because I am going to write a "plus C" on the right. I do not need to write a "plus C sub one" on the left,
and a "plus C sub 2" on the right. We would just subtract both sides by C sub one and make a third constant,
again, unknown.
I am just going to write the "plus C" on the right here.
This is x to the first, so we are going to raise that power by one and have 7 over 2 (we divide by that power rasied
by one) and x squared, plus C.
Depending on the format your teacher wants your answer in, if they are not concerned about you
simplifying or anything like that, like I have in my notes, that is my answer.
This is the soltution to this differential equation.
That was quick! Let's get on to the last example, which again, will be quite a bit harder.
Second and last example!
Solve: 3y-prime plus y-prime times x-cubed, minus 6x-squared times y is equal to zero.
This looks, like I said, a lot more complicated than our last one. But, this is the only technique and type
of differentiable equations we are going to look at.
So we must be able to separate these variables.
We have an x factor in this term, so let's take that minus and move it over to the other side.
And we have 3y-prime, plus y-prime x-cubed, is equal to 6x-squared times y.
Now we want to get this into the same form that the previous board showed. That rule which was dy over
dx equals f(y) times g(x). In other words, I need that y-prime alone. That is what I am trying to solve for.
So these two terms have a y-prime, or dy/dx, we are going to factor that out.
Divide both sides by 3 plus x-cubed. And I am also going to write this as dy over dx.
Okay.
Let's move what appears to be a denominator of dx over to the other side.
And of course, I am showing every step to help you follow my work and understand how to do this
yourself. You might already have this written with these variables separated.
Because now I have all my x's over here along with my dx, but I have a factor in the numerator of y.
So we are going to move that over to the other side by multiplying both sides by one over y, or we divide both
sides by y.
Okay.
So my variables are separated and I am going to take the integral of both sides.
Or antidifferentiate both sides of this equation with respect to the isolated variable.
Well, hopefully we know integration rules. We know that the integral of one over y, dy, or one over x, dx,
you many remember this of course as u-prime over u, or du over u.
Is going to be the natural log. So we have here, the natural log of the absolute value of y,
because you cannot take the natural log of negative numbers. So we have to put the absolute value around
that y to make sure that we are showing that work. And then over here...
Well, I should have given myself more room. Because really, the left side is not going to take a lot of work,
but we have to recognize that something is going on here.
Our denominator has a degree which is larger than the numerator by one. And when we take the derivative,
using the power rule, we do decrease those exponents by one. So it seems like that maybe the numerator
is in some form a derivative of the denominator.
So that is why I was bringing up that u-prime over u idea.
Memorizing these rules, of course, that we have been working on.
Let's just let u equal 3 plus x-cubed. Then du is equal to 3x-squared dx.
Usually I do my u-substitution to get my problem to exactly match the rules...
this time I am going to do a pattern recognition.
So I've got a 3x-squared dx and a 6x-squared dx.
I could pull a 2 out front...well I could pull the whole 6 out front, right?
But I would have a u, and a u-prime on the numerator if that was just simply 3x-squared.
6 is 3 times 2, so let's pull that out front.
I got all dressed up, by the way, to make this video.
Now we have, if you will, a u and u-prime in the numerator.
That means that we actually have a natural log function on both sides of this equation. The integral of one over
y, dy is again, the natural log of the absolute value of y
And this is going to be 2 times the natural log of the absolute value of again, with this being
u prime, and this being u...we've got the natural log of the absolute value of 3 plus x-cubed.
And now we have to remember our constant of integration.
Why am I wrapping this up in absolute value signs?
Because I cannot take the natural log of a negative number.
If that was an x-squared, or any even power, then regardless of what the value of x was, if raised to
an even power, would become positive. And with plus another positive number, it would we guaranteed that
expression would give us a positive number, and we would not need the absolute value sign.
Here we do, because of course that is not an even power, but rather, odd.
Now, cleaning this up a little bit, unlike what I did with the last one...
we have the multiplication of 2. That two can come up here as a power.
And we get the natural log of the absolute value of y is equal to the natural log of the absolute value of
3 plus x-cubed, squared.
Now, when you square something it always becomes positive. So these absolute value symbols are not
really needed. It is going to guarantee that what we are plugging into the natural log function is indeed
positive.
Now, I do have natural log fuctions in here. This is actually going to lead us into the next video where we
solve or work with exponential and decay models. But I can make the sides here an exponent of e.
Go through the process of removing those natural log functions, because natural log is log base e.
And that is the inverse of exponential functions, base e.
And so this base e, and log base e are going to cancel out.
And give us y is equal to... here it is not just e with an exponent of natural log, or log base e.
You also have a plus constant.
So to separate that...when you add exponents, right? Because this is an exponent of e, you add exponents
when you multiply like bases. So this becomes e to the c power, times e to the natual log of 3 plus x-cubed, squared.
This e to a constant just becomes another constant. I am going to write C sub-one just to differentiate this
constant with that one. But it is still an unknown constant.
Then base e and log base e, or natural log base e, are going to cancel out.
We have y is equal to C sub-one times 3 plus x-cubed to the second power.
And that is the solution to this differential equation.
Now I could just call it done and pop out and this video is over, because I am only doing 2 examples on this
video. Shorter than most of them. But I did just make a video, or if you are watching them as a class, where we
were given solutions to differential equations and checked them. Right?
If we were to find the derivative of this equation, or y-prime, or dy/dx, if you will.
And plug it into this equation...and if we take the derivative of this equation, plug it into here,
and get the fact that it is equal to zero, then we have shown and we have checked our solution.
We do not have to just get an answer and go, "Man, I hope that's right!"
We can actually check them! Let's do that right now!
As you can see, after finding our derivative and plugging it back into the original the original function
everywhere their was a y-prime. We plugged in our derivative and our last factor of y, which I wasn't
sure if we were going to plug that in, but we needed it, to get all this in terms of x. And ultimately simplify it
to where zero is equal to zero.
So, our solution checks out.
I'm Mr. Tarrou!
BAM!
Go do your homework!