## What is exactness condition of second-order?

## What is exactness condition of second-order?

Because the quantity dS = d′Qmax/ Higher-order equations are also called exact if they are the result of differentiating a lower-order equation. For example, the second-order equation p(x)y″ + q(x)y′ + r(x)y = 0 is exact if there is a first-order expression p(x)y′ + s(x)y such that its derivative is the given equation.

## What does the existence and uniqueness theorem say about the correspond ing solution?

The existence theorem is used to check whether there exists a solution for an ODE, while the uniqueness theorem is used to check whether there is one solution or infinitely many solutions.

**How do you differentiate a second-order differential equation?**

A linear second order differential equation is written as y” + p(x)y’ + q(x)y = f(x), where the power of the second derivative y” is equal to one which makes the equation linear. Some of its examples are y” + 6x = 5, y” + xy’ + y = 0, etc.

**What is the second order differential equation?**

Definition A second-order ordinary differential equation is an ordinary differential equation that may be written in the form. x”(t) = F(t, x(t), x'(t)) for some function F of three variables.

### What is the existence and uniqueness theorem for second-order differential equations?

The simplest Existence and Uniqueness Theorem (EUT) for second-order differential equations is one that is a natural extension of the result we saw in Section 2.8. Suppose we have a second-order IVP d2y dt2 = f(t, y, ˙ y), with y(t0) = y0 and ˙ y(t0) = ˙ y0.

### Is there a local existence and uniqueness theorem for the SPP?

A local existence and uniqueness theorem for the SPP can be found in Ebin and Marsden paper [20]: if h and I are sufficiently close in a sufficiently high order Sobolev norm, then there is a unique shortest path. In large, uniqueness can fail for the SPP.

**What is the existence and uniqueness theorem for a rectangle?**

In terms of the Existence and Uniqueness Theorem, we have. f ( x, y) = − P ( x) y + Q ( x) and ∂ f ∂ y ( x, y) = − P ( x). But both P ( x) and Q ( x) are assumed continuous on the rectangle R = { ( x, y) | a ≤ x ≤ b, c ≤ y ≤ d } for any values of c and d, and f ( x, y) is a combination of continuous functions.

**What is the existence theorem in math?**

This is an existence theorem, which means that if the right conditions are satisfied, you can find a solution, but you are not told how to find it. In particular, you may not be able to describe the interval I without actually solving the differential equation.