Simple differential equation example

Let f be a real valued differentiable function on R-- the set of all real numbers-- such that f of 1 is equal to 1. If the y-intercept of the tangent at any point P-- which has the coordinates x and y-- on the curve y equals f of x is equal to the cube of the abscissa-- I haven't seen the word abscissa in a while-- of P, then the value of f of negative 3 is equal to. So let's just decode what they're saying. So let me draw a potential f right here. So that's our y-axis. And then, let's call that our x-axis. And let me just draw a simple, potential f of x. So let's say it looks something like that. I don't know what it looks like. We don't know what it looks like. We just know it's some real valued differentiable function on R. So that is f of x. So what is the tangent going to look like? So let's say we have some point here, which is x, 0 and y, 0. So it's a particular x and y. What would the tangent there look like? Well, it would be a line that has the same slope as the function does at that point. So it would be a line. I can draw a better version than that. It would be a line that looks like that. This line right here would have the equation y is equal to mx plus b, where m is the slope and b is the y-intercept. Now, we know that the slope is going to be the same as the slope at this point in the function. And what is the slope at that point in the function? Well, the slope at that point in that function is the derivative of the function. So this is going to be f prime of x, 0. So we take the derivative of valuated x, 0. That will give us the slope right at that point. So this is going to be y is equal to f prime of x, 0 times x plus b. Now, if we wanted to solve for the y-intercept-- for b right there-- we just have to find out some point that's on this line. And we know a point that's on the line. x, 0 and y, 0 is on the line. So let's substitute x, 0 and y, 0 into this equation and then solve for b. So we have y, 0 is equal to f prime of x, 0-- that's just the slope-- times x, 0-- we're just substituting y, 0 and x, 0 into this equation-- plus b. So I just put x, 0 and y, 0 into this equation to solve for b. If you solve for b, you could just subtract this from both sides of the equation. We get b. b is equal to y, 0 minus f prime of x, 0 times x, 0. Now, that's the y-intercept of the tangent. They tell us the y-intercept of the tangent at any point P x, y on the curve y equals f of x is equal to the cube of the abscissa of P. An abscissa is just a fancy way of saying the x-coordinate. Clearly, they're just using the word to filter out people who don't know the word "abscissa." Abscissa just means the x-coordinate. If I tell you I have the point 2 comma 3, the abscissa of this is just 2. It's just the x-coordinate. So all this sentence is telling us is that the y-intercept is equal to the cube of the x-coordinate. So the y-intercept-- it's this-- is going to be equal to the cube of the x-coordinate. The x-coordinate here is x, 0. So it's going to be equal to x, 0 cubed. So for any x, 0 and y, 0 for any x, 0 and y, 0 that's on the function itself, we now get this expression. I'll do it in a new color. We get y, 0 minus f prime of x, 0 times x, 0 is equal to x, 0 to the third. Now, these are particular x's and y's. It's just for a case of example. It applies to all of them. So we could say it applies to all of the points on the curve. So we could say, for any point on the curve, we get y minus f prime of x times x is equal to x to the third. And we know that y is equal to f of x. Or another way of thinking about it is that y prime is equal to f prime of x. So we could write this. And what we're getting here is a little bit of a differential equation. We get y minus y prime times x is equal to x to the third. Let's see. We can could divide both sides of this equation by x. We get y over x minus y prime is equal to x squared-- just divided everything by x. And then we can subtract this from both sides and then add this to both sides-- kind of skipping a step right there. So we get y over x-- actually, let me write it this way. I don't want to skip steps. So let's subtract y over x from both sides. We get negative y prime is equal to negative y over x plus x squared. And then, multiply both sides times negative 1. We get y prime is equal to y over x minus x squared. And so a little bit of a differential equation here. And solving any differential equation is really a bit of an art. And so we have to think about it-- if we could separate the y's from the x's-- if we could put the y's all on this side-- then it would be pretty easy to solve, but it's not so obvious. And just to give an example-- just to inform our intuition-- if this was a simpler version, if this was just dy/dx-- which is the same thing as y prime-- if it was just equal to y over x, if it was just that, so if this negative x squared term wasn't there, then this would be a separable differential equation. You could divide both sides by y and multiply both sides times dx. So if you multiplied this times dx over y and you multiplied this times dx over y, these would cancel out. Those would cancel out. And you would get ydy is equal to x-- sorry. Not ydy. You'd get 1 over y. 1 over y dy is equal to 1 over x dx. You integrate both sides of that. You would get the natural log of y-- that's the antiderivative of 1 over y-- is equal to the natural log of x plus some constant. And we could write any constant as plus the natural log of some constant. It could just be any constant over there. And then this is the same thing as the natural log of y is equal to the natural log of-- when you add exponents like this, that's essentially what a logarithm is. It's essentially taking a product of these two terms. This is equal to cx or that y is equal to cx. Now, this is all just to inform our intuition here. If this negative x squared term wasn't here, then we would be able solve it pretty easily. y would be equal to cx. If this term wasn't here, taking the antiderivative of this is pretty straightforward. It would be negative 1/3 x to the third power. Our intuition here is-- maybe some combination of the antiderivatives of this and this-- and you can't just take them separately because this y is going to have to be equal to the antiderivative of the whole thing. It's not just the antiderivative of this part over here. But if this guy wasn't here, then it could be y is equal to cx. If this guy wasn't here, you would have something to the third power. So just to try things out-- and solving differential equations is always a bit of an art-- let's just say that y is equal to-- we know we're going to have to have something to the third power in there. So let's call it Ax to the third power. And we know we're going to have a x power in there. Let's call it plus Bx. And then, let's call it plus C. And that's our intuition-- that somehow that this will help you get to this term right over here. And then, this term-- when I take the derivative-- will help to get that term over there. So what would be the derivative of this? The derivative of this-- y prime, in this case-- would be 3Ax squared plus B. And now, what would y over x be? y over x would be equal to-- notice, I'm just saying y prime is this. y over x would be Ax squared plus B plus C over x. This differential equation here would reduce to-- and we just suspect that this might be a solution-- this y prime is the same thing as this. So we have 3Ax squared plus B is equal to y over x. It's equal to this thing over here-- Ax squared plus B plus C over x minus this x squared. So what does this simplify to? If we wanted to solve this equation, we could combine the x squared term. So let me write it this way. We have 3Ax squared plus B is equal to-- we have Ax squared minus x squared. So that's equal to A minus 1. This is equal to A minus 1 x squared. And then, I have plus B plus C over x. This is what the right-hand side simplifies to, and this is what the left-hand side is. In this, we have no constrain on B. B so far can be anything. B could be anything. There's no C term over here. And actually, more importantly, there's no 1 over x term over here. So C would have to be equal to 0. There's no C. C is equal to 0. This term does not exist. There is no C. And then, when you look at this, this coefficient-- the 3A-- the coefficient of the x squared on the left-hand side has to be equal to the coefficient of the x squared on the right-hand side in order for these 2 to be equal. So we'd get 3A is equal to A minus 1. Subtract A from both sides. 2A is equal to negative 1. A is equal to negative 1/2. So our hunch paid off. We were able to figure out A's, B's, and C's so that this function does satisfy our little differential equation. And our value we got is y is equal to-- I keep using that green color. I don't want to. We get y is equal to negative 1/2 x to the third. That's our value for A. Plus, B could be anything. Plus Bx. So this is our value for f of x. And they gave us some information. They told us in the problem that f of 1 is equal to 1. We can use that information to solve for B. f of 1 is equal to 1. So when x is equal to 1, y is equal to 1. So we get 1 is equal to negative 1/2 times 1 plus B times 1 is just B. So we can add 1/2 to both sides. And we get 3/2 is equal to B. So we're now able to fully describe our function. Our function is y is equal to negative 1/2 x to the third plus 3/2x. So that's our function, but that's not what they're asking us. They're not asking us for the function. They're asking us for the value of f of negative 3. f of negative 3 is equal to negative 1/2 times negative 3 to the third is negative 27 plus 3/2 times negative 3. So minus 9/2. This will simplify to-- well, everything's over 2. Negative 27 is positive 27. And then you have a minus 9. This is 27 over 2, 9 over 2. negative 9 over 2. So this is equal to 18 over 2, which is equal to 9. And we are done. This is equal to 9.