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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) = 2 + \int^x_2 [t - ty(t)] dt $

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Calculus 2 / BC

Chapter 9

Differential Equations

Section 3

Separable Equations

Missouri State University

Campbell University

Harvey Mudd College

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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An integral equation is an…

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for this problem, we are asked to solve the integral equation. Why of X equals two plus the integral from 22, X of t minus T Y D t. Uh the first thing that we can do here is not that for that integral, we know that whatever that is integrated out to be, it will come out to capital ffx minus capital F evaluated at two, depending on where the derivative of capital F is going to be t minus 50. So having that if we differentiate the integral equation, we'll get that Y prime of X is going to be capital F prime of x minus capital F prime of two. But if we evaluate capital F prime at two, it's just going to be a constant. So we get that why prime of X is going to be equal to capital F prime of X, which means that we have now are new differential equation or our differential equation for him is going to be x minus x. Y. Where we do actually get an initial condition from this integral equation. Specifically note that if we evaluate uh y of two, then we'd get two plus an integral from 2 to 2 of stuff. And if we integrate that, it's always going to go to zero. So we have an initial condition of Y of two equals two. So now we have a full initial value problem, kind of, you have a value problem of a kind and we can see that we have that. It is separable. Specifically we can factor out the exes from the right hand side. So we get y prime equals X Times one -Y. Which means that why prime over one minus Y is going to equal X. So integrating the right hand side, we get x squared over two plus C. And integrating the left hand side, one moment here It's going to be oh yeah, uh integrating the left hand side with a bit of substitution is going to give us negative lawn of one -Y. Oh we need to note we have a plus C over there, which then means that we have lawn of one minus Y is going to be equal to X squared over two plus or negative X squared over two. And we can leave that as plus C because C is arbitrary, Which means then that we have one -Y. It's going to be E to the power of negative X squared over two plus C. Or that's the equivalent of just putting a constant out in front of that C. So we have lastly that why is going to equal one? We can write it as one minus. It doesn't really make a huge difference. One minus C. E. To the power of negative X squared over two. Having all of that. Now we just need to evaluate this at X equals two. We have, we have two is equal to one minus C times E. To the power of negative X squared over two. So that would be negative for over two or 2002, Intern needs to equal to So we can see then, let's see here uh ad ce to power of negative two to both sides, subtract two from both sides. So we'll have negative one equal ce 2000 negative too. Which then means that C. Is going to equal negative one divided by E. To the power of negative two or just negative E. To the power of to. So our solution is going to be five X equals one. Now we have a double negative, so it would be one plus E. To the power of negative X squared over two Plus two. When we bring that exponent from the e into the exponent there.

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