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An integral equation is an equation that contains an unknown function $ y(x) $ and an integral that involves $ y(x). $ Solve the given integral equation. [Hint: Use an initial condition obtained from the integral equation.]$ y(x) = 4 + \int^x_0 2t \sqrt {y(t)} dt $

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Calculus 2 / BC

Chapter 9

Differential Equations

Section 3

Separable Equations

Campbell University

Baylor University

Idaho State University

Lectures

13:37

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.

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01:49

An integral equation is an…

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Find $d y / d x$$$…

Okay, So for this problem or given an integral equation, uh, Y of X equals four plus integral from zero to ex, uh, to tee times a squared of y f t d t. And at first glance, this might see kind of intimidating, I guess, Uh, since you have X and T and inter girls and all that. But actually, we can make this really simple using some fundamental theorem of calculus. So there's fundamental rules that calculus always follows. Um, so you start with the first thing we do is we take the derivative of both sides. So this would become do you the derivative y of X? With respect to X, you can just write as a dy DX equals. So during this would be 00 plus, um, and then you have the derivative with respect to X of this entire integral to D squared of Iot G. And basically, this is the important part here because the first fundamental theorem of calculus first rule basically says that the derivative of any integral right from a to B or eight x of a function, it's basically going to be equal to this dysfunction here. So what that's also telling us is that the derivative f prime of X to differentiate this side? If I integrate this side, then I'll get obviously capital F, which is one step higher. But the derivative would just be the normal thing. So from what we can tell from this is we can take this and we can apply that directly to here. And you can say that dy dx it's going to be equal to two X. So the derivative of this is just equal to the function and the function hit here on the inside is to t, uh, squared of y of t So we just plug X in there and we'll get to x times the square root of y of X uh, D X right. We don't need the d. X because the derivative and here will cancel out. So you'll have just two t times the square root of y of X, uh, two x Times Square to buy back to my back. Um, so now what this is telling us is, uh y of X. We can just rewrite as just why? Because in this case, that's all it is. So you have two x times word of y And now we can rewrite this, separating them right, applying the separation rules. So we take the dy DX and separate them so you can divide this side by squared of y and multiply this side by DX And what do that you'll get dy over the square root of y equals two x dx. And, uh, now, finally, the last thing I have to do is simply integrate both sides. So you integrate this side, uh, with respect to the y and you integrate this side with respect to the X Uh so another way you can be right. This is square, uh, integral of why two negative one half dy equals and an integral of two x is really, really simple. So you can just go ahead and write X squared. Plus c don't forget that constant of integration. Um, so yeah, so you have X squared. Plus, C uh, you have actually received there and then on this side for the why it's a negative one half. Uh, you just simply do the normal power rules, right? X two negative, one half plus one and then on the bottom, you're going to put negative one half plus one as well. Uh, so that will basically give you equals X squared plus c and then the plus See the integral integral in the constant of integration you get from this integral, uh, basically just add onto here. So in reality, you have, like, a C one and C two here, but it's all just a constant. So C one plus C two is implied in this plot. One single plus c over here. So you don't need to worry about both constant of integration. You can just combine them. Uh, and after all this, you'll end up with basically this will go to the top two times y to the one half mhm equals X squared plus c. And why did the one half is just a fancy way of saying square root of why? So you can go out and say that. And now if we separate out why, then we can simply say, uh, why it would be equal to X squared plus c over two. And then this whole thing squared Mhm. Okay, but now we can actually substitute X equals zero to our initial function here, which was this so if you say X equals zero, then because we need to find this plus see here. So we say X equals zero. So are we. Just rewrite real quick. Our original, uh, function here it was to t times the square of y f T d t. And if we plug in Y equals zero, then you'll get four plus interval from 0 to 0. That's really key. Uh, of two tee Times Square, the Via T d. T. Now, if you remember the rule of integration where if you're going from 0 to 0, then your integral, which is basically zero. So why have zero, which is equal to four? So now that we know why zero is equal to four, uh, we can use this. We just found out why here. So we can plug y of zero is equal to four, which is equal to zero squared, which is zero. And then you'll see over to see over two square. Uh, so we take the square root of both sides of this little thing here, and you're left with two equals C over two. And that means C is equal to four. So now that you know that you can finally go in and plug your C back into your final equation, which will change colors here for, uh, red. Why is equal to mhm? Was it X squared? Plus X squared. Plus 4/2 squared. So that would be your final answer. Uh, for evaluating this integral.

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