Integral Equations
In Exercises 7–12, write an equivalen...

Question

Integral Equations

In Exercises 7–12, write an equivalent first-order differential equation

and initial condition for y.

y = ln x + ∫ₓᵉ √ (t² + (y(t))²) dt

Accepted Answer

Welcome back everyone. Determine the equivalent first order ODE and the initial value for the integral equation. Y of x is equal to e to the power of x plus integral from 0 to x of y tint dt. So for this problem, let's begin by finding the required first order differential equation. We can seek the derivative of both sides. In particular, on the left hand side, we will have the derivative of y with respect to x, which is the y divided by dx or simply y. And on the right hand side we're going to take the derivative of e to the power of x, and we're going to add the derivative of the integral. Let's write it as d divided by dx of the integral from 0 to x of y of tsin t dt. And to evaluate each derivative, let's begin by taking the derivative of e to the power of x, which is e to the power of x itself. And for the derivative of the integral, let's apply the fundamental theorem of calculus part 1. Since the lower limit of integration is 0 and the upper one is x, we simply evaluate our function at x, which gives us y of x. C becomes x, right? Multiplied by sin of x again c becomes x. So now we have the required equation. We can show that y is equal to e to the power of x plus y multiplied by sin of x. What we want to do is identify the initial condition, and the easiest way is to evaluate the integral at the, at the lower limit of integration, right? So what we're going to do is notice that the lower limit of integration is 0. So let's go ahead and evaluate y at 0, which is according to the original equation e to the power of 0 plus integral from 0 to 0. Of y of Tint DT. Notice that the integral is going to become 0, right? Because the lower limit of integration is equal to the upper limit of integration. And we get y at 0 equals e to the power of 0, which is 1. This is how we get the initial condition y at 0 equals 1. Let's label our final answer and thank you for watching.

Integral EquationsIn Exercises 7–12, write an equivalent first-order differential equation and initial condition for y.y = ln x + ∫ₓᵉ √ (t² + (y(t))²) dt

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Integral Equations
In Exercises 7–12, write an equivalent first-order differential equation
and initial condition for y.

y = ln x + ∫ₓᵉ √ (t² + (y(t))²) dt