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AP®︎/College Calculus BC
Course: AP®︎/College Calculus BC > Unit 7
Lesson 2: Verifying solutions for differential equationsVerifying solutions to differential equations
AP.CALC:
FUN‑7 (EU)
, FUN‑7.B (LO)
, FUN‑7.B.1 (EK)
, FUN‑7.B.2 (EK)
We can check whether a potential solution to a differential equation is indeed a solution. What we need to do is differentiate and substitute both the solution and the derivative into the equation.
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How to find a solution or set of solutions of differential equations?(14 votes)
That's the main question of this entire field. All the videos and questions are here to answer it.(12 votes)
What does it mean for f(x) to be a solution of f'(x)? Is there a graphical way to describe this?(4 votes)
It would mean that taking the derivative of f(x) results int he same equation. The only functiont hat does that though is e^x
If you have other constraints though, like in f'(x)=f(x)-x there is more than just one function to satisfy the condition.
Maybe a better way of explaining it is saying if you have some f(x) and find f'(x) is there some other way to get it to its derivative other than taking it's derivative.
So let's say f(x) = 5x^2 so then f'(x)=10x. Now what can we do to 5x^2 without calculus to make it 10x? We could multiply it by 2/x. so we could say f'(x) = f(x)(2/x) and 5x^2 would be a solution to that.
So I guess you could say it's graph transformations to get to the graph of f'(x)(6 votes)
- can i use a differential equation to solve world hunger(4 votes)
- Are units (meters, seconds, m/s) can be used in differential equations? Or is it just the number that can be used?(2 votes)
It depends: in word problems it is often the case that the solution is looking for a rate (meters/sec, liters/sec, etc.). However, when faced with a problem such as y'' - 2y' + y = 0 the solution will be a function y = Ae^x + Bxe^x, where A & B are real values. No units, no measurements, just a good ol' fashioned function.(3 votes)
Why does Sal use y=4x to get y=x^4, and how does he get y=4x to get y=x^4(2 votes)- Why does Sal use y=4x to get y=x^4, and how does he get y=4x to get y=x^4?(2 votes)
- Why we find solution of differential equation ?(2 votes)
- Observation that I make (don't know if it's true or not):
if we are given 1st order differential equation: y' = y + cx^n , we can guess that the y is an unknown polynomial with degree of n(1 vote)
Yes, that's right.
For example with 𝑛 = 3,
we can let 𝑦 = 𝐴𝑥³ + 𝐵𝑥² + 𝐶𝑥 + 𝐷 ⇒ 𝑦' = 3𝐴𝑥² + 2𝐵𝑥 + 𝐶
𝑦' = 𝑦 + 𝑐𝑥³
⇒ 3𝐴𝑥² + 2𝐵𝑥 + 𝐶 = (𝐴 + 𝑐)𝑥³ + 𝐵𝑥² + 𝐶𝑥 + 𝐷
⇒
𝐴 = −𝑐
𝐵 = 3𝐴 = −3𝑐
𝐶 = 2𝐵 = −6𝑐
𝐷 = 𝐶 = −6𝑐
The leading coefficient is −𝑐, which then determines all the other coefficients.
This is of course not a formal proof for the general case, but I think you can imagine that this must be what happens for any whole number 𝑛.(2 votes)
at4:02, I understand we can just check whether the function is a solution of the differential equation by substituting the function with its derivative in the equation, but why does Sal emphasize it needs to work for any x?(1 vote)
Generally when you are asked to solve a differential equation it is implicitly assumed it is to be solved for the whole domain.(2 votes)
- at5:19of the video why does f'(x)= e^x + 1, specifically why the e^x part??(1 vote)
f'(x) = d/dx [f(x)] = d/dx [e^x + x + 1] =
d/dx[e^x] + d/dx [x] + d/dx [1] = e^x + 1 + 0.(1 vote)
4:02Video transcript
- [Instructor] So let's write
down a differential equation, the derivative of y with respect to x is equal to four y over x. And what we'll see in
this video is the solution to a differential equation isn't a value or a set of values. It's a function or a set of functions. But before we go about
actually trying to solve this or figure out all of the solutions, let's test whether certain equations, certain functions, are solutions to this
differential equation. So for example, if I have y is equal to four x, is this a solution to this
differential equation? Pause the video and see
if you can figure it out. Well to see if this is a solution, what we have to do is figure
out the derivative of y with respect to x and
see is that truly equal to four times y over x. And I'm gonna try to express
everything in terms of x to see if I really have an equality there. So first let's figure
out the derivative of y with respect to x. Well that's just going
to be equal to four. We've seen that many times before. And so what we need to test is, is four, the derivative
of y with respect to x, equal to four times, I could write y, but instead of y let's write four x. I'm gonna put everything in terms of x. So y is equal to four x, so instead of four y I could
write four times four x, all of that over x. Is this true? Well that x cancels with
that and I'm gonna get four is equal to 16,
which it clearly is not. And so this is not a solution. Not a solution to our differential equation. Let's look at another equation. What about y is equal to
x to the fourth power? Pause this video and see
if this is a solution to our original differential equation. Well we're going to do the same thing. What's the derivative
of y with respect to x? This is equal to, just
using the power rule, four x to the third power. And so what we have to test is, is four x to the third power, that's the derivative
of y with respect to x, equal to four times y, instead of writing a y I'm gonna
write it all in terms of x, so is that equal to four
times x to the fourth, because x to the fourth
is the same thing as y, divided by x? And so let's see, x to
the fourth divided by x, that is going to be x to the third. And so you will indeed
get four x to the third is equal to four x to the third. So check, this is a solution. So is a solution. It's not necessarily the only solution, but it is a solution to
that differential equation. Let's look at another
differential equation. Let's say that I had, and I'm gonna write it
with different notation, f prime of x is equal to f of x minus x. And the first function that I wanna test, let's say I have f of x is equal to two x. Is this a solution to this
differential equation? Pause the video again and
see if you can figure it out. Well to figure that out, you have to say well what is f prime of x? f prime of x is just
going to be equal to two. And then test the equality. Is two, is f prime of x, equal to f of x, which is two x, minus x, minus x? And so let's see we are going
to get two is equal to x. So you might be tempted to say oh hey I just solved for x or
something like that. But this would tell you
that this is not a solution because this needs to be true for any x that is in the domain of this function. And so this is, I'll just put an x there, or I'll put a incorrect there to say not, not a, not a solution. Just to be clear again, this needs, in order for a
function to be a solution of this differential equation, it needs to work for any x that you can put into the function. Let's look at another one. Let's say that we have f of
x is equal to x plus one. Pause the video and
see, is this a solution to our differential equation? Well same drill. f prime of x is going to be equal to one. And so we have to see is f prime of x, which is equal to one,
is it equal to f of x, which is x plus one, x plus one, minus x? And so here, you see no matter what x is, this equation is going to be true. So this is a solution, is a solution. Let's do a few more of these. Let me scroll down little bit
so I have a little bit more, a little bit more space, but make sure we see our
original differential equation. Let's test whether, I'm
gonna do it in a red color, let's test whether f of x equals e to the x plus x plus one is a solution to this
differential equation. Pause the video again and
see if you can figure it out. All right well let's figure
out the derivative here. f prime of x is going to be equal to, derivative of e to the x with
respect to x is e to the x, which I always find amazing. And so and then plus one
and the derivative of this with respect to x is just zero. And then let's substitute this into our original differential equation. So f prime of x is e to the x plus one. Is that equal to f of x, which is e to the x plus x
plus one, minus x, minus x? And if that x cancels out with that x, it is indeed, they are indeed equal. So this is also a solution. So this, this is a solution. And we're done.