Intro to differential equations
Simple Differential Equations 3 basic differential equations that can be solved by taking the antiderivatives of both sides.
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- Welcome to the video on the introduction to
- differential equations.
- What's a differential?
- Well, we've been doing things like if I said that y is
- equal to, I don't know, 3x.
- If we took the derivative of both sides of this equation and
- we said -- one of the notations we used is we said dy
- over dx is equal to 3.
- We took the derivative of both sides of that.
- But this you could almost use differential notation.
- dy is a differential and dx is a differential.
- What does a differential mean?
- A differential just means an infinitely small change in y,
- in this case, so that's a differential in y, or an
- infinitely small change in x.
- So a differential it's like a difference -- when we learned
- slope, we said slope is change in y over change in x.
- Or you could say the difference in y over difference in x.
- But once we started taking the slope of curves we had to make
- this difference and this difference approach 0.
- That's what a differential essentially is.
- A differential is a really, really, really, really
- infinitely small difference.
- So what is a differential equation?
- Well it's an equation that involves differentials.
- So, for example, this is a differential equation.
- This will be the first example that we solve.
- dy over dx is equal to x squared plus 1.
- So how do we solve this differential equation?
- Well, there's a couple of ways you could say it.
- You say oh, well this is the same thing as f prime of x.
- You know, y to the function of x and you just take the
- antiderivative of both sides.
- But let's solve this properly as a differential equation.
- So you can actually manipulate these differentials the same
- way that you would manipulate numbers for the most part.
- And so what could we do here?
- Let's multiply both sides of this equation by dx.
- So you get, for a very small change in y, dy, you get x
- squared plus 1 times a very small change in the x.
- Or another way to say it is for a very small change in x, if
- you want to figure out how much does y change, you multiply
- times x squared plus 1 wherever you are on the curve.
- So what does that do for us?
- Well let's take the antiderivative of both
- sides or let's take the integral of both sides.
- And if you watched all of the videos on integration and the
- definite integral and area and a curve, you realize that an
- integral is essentially a sum, it's kind of an infinite
- sum of a bunch of these infinitely small dy's.
- So if you take the sum of all of the changes in dy's,
- you're left with a y on this side of this equation.
- You might want to re-watch all the videos that we
- did on integration.
- Then what is this?
- This is essentially just -- this is the indefinite
- integral of x squared plus 1.
- So it's the antiderivative of that.
- And what's the antiderivative of x squared plus 1?
- Well it's x to the third over 3.
- We're just taking the derivative backwards.
- And once again, watch the videos on the antiderivative.
- There's I think eight or nine of them.
- Plus x and then plus c.
- And where does the c come from?
- Well we know that when we take the derivative of a
- constant it goes to 0.
- So when you take the antiderivative, we're like
- well, there could have been a constant there.
- And that's where that plus c comes from.
- So this is the general solution to this differential equation.
- And that's something interesting.
- So with traditional equations, the solution tended to
- be a number, right?
- If I just told you y is equal to 2y minus 1.
- Then you would get minus y is equal to minus 1, you'd
- get y is equal to 1.
- So this is a traditional equation and your solution
- was just a value -- you solved for the variable.
- Differential equations are something different.
- The solution is actually a function.
- You're saying what function satisfies this
- differential equation.
- So that's something to keep in mind.
- Right now we're doing very basic differential equations,
- but that's something to keep in mind the whole time you learn
- differential equations.
- I think I'll eventually do a play list on essentially
- an introductory course on differential equations that
- you would take at college.
- And that applies even when you start doing partial
- differential equations, et cetera, et cetera.
- The solution to a differential equation is not a number,
- it is a function.
- So anyway, this was the general solution to this
- differential equation.
- And if you want the particular solution, people normally give
- you initial conditions or they give you points on the function
- and then you can substitute back.
- In this problem they said that dy dx equals x squared plus 1.
- And they said that -- let me switch colors -- they said that
- y is equal to 1 at x is equal to 1, or y is equal to 1
- when x is equal to 1.
- So we can use this information now to solve for c.
- How do you do that?
- Well it says y is equal to 1.
- So 1 is equal to when x is equal to 1, is equal to 1/3,
- right, 1 over 3, 1 to the third power over 3, plus 1 plus c.
- And let's see what I can do.
- Subtract one from both sides -- this comes 0.
- Subtract 1/3 from both sides and you get c is
- equal to minus 1/3.
- So using these conditions, a point where this function
- crosses through, we can now give you the particular
- solution to this differential equation.
- And that is y is equal to x to the third over 3 plus x minus
- 1/3, and then we just solve for the c.
- And if you don't believe me, take this expression and
- substitute it here and you will see that it equals, if you were
- to take the derivative of y with respect to x, you would
- see that it equals x squared plus 1.
- Let's do another one.
- So it tells us -- so this is a little bit more interesting --
- it says dy dx is equal to x over y, and it has
- the same conditions.
- It says y equals 1 at x equals 1.
- Or when x equals 1, y equals 1.
- So let's find the function that satisfies this equation.
- This one's interesting because we have an x and a y on
- the right hand side of the equation.
- So this kind of looks like something we got after we
- did some type of implicit differentiation.
- But let's see where we go.
- So let's do the same thing.
- Let's multiply both sides of this equation by dx until you
- get dy is equal to x over y dx.
- And let's get this y over onto the dy side, because
- it will be easy to take the antiderivative then.
- So we get y dy is equal to x dx.
- Now we can take the integral of both sides of this equation, or
- take the antiderivative on this side with respect to y on
- the side with respect to x.
- I don't know why I did it in brown.
- So what's the antiderivative here?
- Well, it is y squared over 2.
- And I could say plus c, some constant.
- Let me do that.
- There could be a constant here.
- I'll call it c1.
- We don't know what that constant is.
- And that equals x squared over 2 plus c2, right?
- Or some other constant.
- Maybe it's the same number, I don't know.
- But we don't know what either of these are.
- I could re-write it as -- let's see, what could I do?
- Let me take the x out of that side.
- So I could have y squared over 2 minus x squared over 2 is
- equal to -- and let me subtract this from that side so I get
- the constants all on the right hand side of the equation.
- c2 minus c1.
- I just took a c1, put it on the right hand side, took the x
- squared over 2, put it on the left hand side, that's
- why it's negative.
- And we didn't know.
- We said this could be any constant and this
- could be any constant.
- So the difference between two arbitrary constants could just
- be a third arbitrary constant.
- So I'll just re-write that as c.
- So the general solution to this differential equation is y
- squared over 2 minus x squared over 2 is equal to c.
- Actually let's do something else just to clean it up a
- little bit, because once again this could be any constant.
- So let's multiply both sides of this equation by 2 and
- you get y squared minus x squared equal to 2c.
- Well now this is still any constant number, so we could
- still write this as a c.
- So we have y squared minus x squared is equal to c.
- Now let's use our initial conditions to see what c is.
- So when y is 1, so 1 squared minus when x is 1, 1 squared is
- equal to c, this is 1 minus 1, so it's 0, right?
- c is equal to 0.
- So what is the particular solution to this
- differential equation?
- I'll do it in green.
- It is y squared minus x squared is equal to 0.
- Or we could add x squared to both sides of that.
- We could also write it as y squared is equal to x squared.
- Now you might be tempted to take the square root of both
- sides of this and say that y is equal to x.
- The reason why this would not be accurate is because here x
- could be minus 2 and y could be plus 2 or vice versa.
- So this would satisfy this equation, but it would not
- satisfy this equation.
- So be careful when you take that square root.
- You have to worry about the plus or minus.
- Let's do one more.
- So this one says the derivative of y with respect to x is equal
- to -- OK, so this is even a little bit more interesting.
- This is equal to the square root of x over y.
- And it says that y is equal to 4 when is equal to 1.
- Yup, I think I'm reading that right.
- We could do this one very similarly.
- So let's just do the same step.
- So multiply both sides of the equation times dx.
- So you get dy is equal to -- and just to skip a step, I'm
- going to re-write square root of x over y is the square root
- of x over the square root of y, and I multiply both
- sides of that times dx.
- Now let's multiply both sides of this equation
- by the square root of y.
- And I'm just going to re-write it as y to the 1/2 power.
- That's the same thing as the square root of y.
- dy is equal to x to the 1/2 power dx.
- I just multiplied that there and re-wrote
- it as y to the 1/2.
- So what is the antiderivative of y to the 1/2 power?
- Well it's just this plus 1, so it's y the 3/2 power, and then
- times the inverse of this.
- So times 2/3.
- Or you could have said divided by this, either way.
- And if you're not sure about this, because sometimes it is
- confusing with the fractions and whatever else,
- take the derivative.
- 3/2 times 2/3 is 1, and then you subtract this 1 from
- the exponent and you get y to the 1/2.
- So that works.
- And of course, we kind of go through the same drill, plus c1
- is equal to -- it's going to be the same thing on this side
- -- 2/3 x to the 3/2 plus c2, and what can we do?
- Well let's take the x to the left hand side of this
- equation, so we get 2/3 y to the 3/2 minus 2/3 x to the 3/2
- is equal to c2 minus c1, which we can just re-write as c.
- And let's multiply both sides of this equation by 3/2, so
- these two will become 1.
- So you get y to 3/2 minus x to the 3/2 is equal to -- well
- what's 3/2 times some constant c.
- But we didn't even know what it was, we haven't solved for
- it, so we can still write c.
- I hope that doesn't confuse you.
- 3/2 c, we didn't know what it was.
- We could call this c3 and now this is c4.
- It's a different constant, but we still have to solve for it.
- And now let's use our initial conditions.
- So the initial conditions tell us that 4 -- and they don't
- have to be initial conditions.
- You can kind of say conditions or points where we know that
- the particular solution to this differential equation
- are satisfied.
- So 4 to the 3/2 minus 1 to the 3/2 is equal to c.
- What's 4 to the 3/2?
- So 4 to the 1/2 is 2, and then that to the 3 power, that's 8.
- Then 1 to any power, but especially in this case, 1 to
- the square root of 1 is 1.
- Then to the third power is 1.
- So 8 minus 1 is equal to c.
- And so c is equal to 7.
- So the particular solution of this differential equation I
- will do in a different color, and it is y to the 3/2 minus
- x to the 3/2 is equal to 7.
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