Finding particular solutions to differential equations: exponential function
About this video
Given f'(x) = 5e^x and f(7) = 40 + 5e^7, the video finds f(0). The antiderivative is f(x) = 5e^x + C; substituting x = 7 gives C = 40, so f(x) = 5e^x + 40 and f(0) = 45.
You'll be able to
- find the antiderivative of an exponential like 5e^x.
- use a given function value to solve for the constant of integration.
- evaluate the resulting function at a new point.
Key words
- antiderivative
- A function whose derivative is the given function.
- constant of integration
- The +C in an antiderivative, found using extra information like f(7).
- e^x
- The exponential function that is its own derivative and its own antiderivative.
Transcript
- [Voiceover] We're told that F of seven is equal to 40 plus five, E to the seventh power, and F prime of X is equal to five, E to the X. What is F of zero? So to evaluate F of zero, let's take the anti-derivative of F prime of X, and then we're going to have a constant of integration there, so we can use the information that they gave us up here that F of seven is equal to this.
This might look like an expression. Well, it is an expression, but it's really just a number. There's no variables in this, and so we can use that to solve for our constant of integration, and then we will have fully known what F of X is, and we can use that to evaluate F of zero, so let's just do it.
So if F prime of X is equal to five, E to the X, then F of X is going to be equal to the anti-derivative of F prime of X, or the anti-derivative of five, E to the X, DX, and this is the thing that I always find amazing about exponentials, and actually, let me just take a step.
I'll take that five out of the integral so it becomes a little bit more obvious. And so the anti-derivative of E to the X, well, that's just E to the X because the derivative of E to the X is E to the X, which I find amazing every time I have to manipulate or take the derivative or anti-derivative of E to the X.
So this is gonna be five, E to the X, plus C, and you can verify. Take the derivative of five, E to the X, plus C. The derivative of five, E to the X, well, that's five, E to the X, so that works out. Well, and the derivative of C is zero, so you wouldn't see it over here.
So now, let's use this information to figure out what C is so that we know exactly what F of X is, and then we can evaluate F of zero. So we know that F of seven, so when X is equal to seven, this expression is going to evaluate to this thing, 40 plus five, E to the seven.
So, five times E to the seventh power plus C is equal to 40, plus five, E to the seventh power. And notice, all I did is say, okay, F of seven. Well, if this is F of X-- Let me write this down. So, if this is F of seven, if this is F of X, I just replaced the X with a seven to find F of seven, and we know that F of seven is also going to be equal to that.
They gave us that information, but when you just look at this, it's pretty easy to figure out what C is going to be. You can subtract five, E to the seven from both sides, and you see that C is equal to 40. And so we can rewrite F of X. We can say that F of X is equal to five, E to the X, plus C, which is 40.
And so now, from that, we can evaluate F of zero. F of zero is going to be five times E to the zero power, plus 40. E to the zero is one, so it's gonna be five times one, which is just five, plus 40, which is equal to 45, and we're done.
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