A First Course in Differential Equations with Modeling Applications (11th Edition) Edit edition Solutions for Chapter 1.2 Problem 27E: In Problems 25 –28 determine whether Theorem 1.2.1 guarantees that the differential equation possesses a unique solution through the given point.
11.09.2021 · For example by rewriting Equation \ref{eq:10.2.6} as an equivalent linear system it can be shown that Theorem 10.2.1 implies Theorem 9.1.1 (Exercise 10.2.12). Back to top; 10.1E: Introduction to Systems of Differential Equations (Exercises) 10.2E: Linear Systems of Differential Equations (Exercises)
Section 1.6 Existence and Uniqueness of Solutions. If \(x' = f(t, x)\) and \(x(t_0) = x_0\) is a linear differential equation, we have already shown that a solution exists and is unique. We will now take up the question of existence and uniqueness of solutions for all first-order differential equations. The existence and uniqueness of solutions will prove to be very important—even …
1 guarantees that the differential equation possesses a unique solution through the given point. Reference : Theorem 1.2.1” is broken down into a number of easy ...
1 guarantees that the differential equation y' = squareroot y^2 - 9 possesses a unique solution through the given point. 25. (1, 4) 26. (5, 3) 27. (2, -3) 28.
1 = A + iB is a root of the characteristic equation ar2 + br + c = 0 and (r r 1)k divides ar2 + br + c. Valid also for sin(Bx) when B > 0. Always, B . For second order, only k = 1;2 are possible. Euler’s theorem is valid for any order differential equation: replace the equation by a ny(n) + + 0y = 0 and the characteristic equation by a nrn ...
VIDEO ANSWER: Okay, so we need to consider the theorem one .2.1. ... Determine whether Theorem 1.2 .1 guarantees that the differential equation y′=√y2−9 ...
Okay, we're going to consider if theorem 1.2.1 guarantees a unique solution. So what we need to think about is inside that square root that y squared minus nine. We do need to be greater than zero. So that means that why squared is going to have to be greater than nine and the absolute value of why it's going to have to be greater or equal to ...
Note that a solution to a differential equation is not necessarily unique, primarily because the derivative of a constant is zero. For example, y = x 2 + 4 y = x 2 + 4 is also a solution to the first differential equation in Table 4.1.We will return to this idea a little bit later in this section.
DIFFERENTIAL EQUATIONS: EXISTENCE AND UNIQUENESS OF SOLUTIONS IN ECONOMIC MODELS5 P 1 ˙P 2 ˙P 3::: where P nis the n-pattern of u k n. Now, using the equicontinuity assumption in the statement of the proof, we know that for any given >0 there exists an Nlarge such that for any u2F varies at most by with intervals of length 2 NjIj.
rems for first-order differential equations because the criteria of continuity of f(x, y) and f0 y are relatively easy to check. The geometry of Theorem 1.2.1 is illustrated in Figure 1.2.6. EXAMPLE 5 Example 4 Revisited We saw in Example 4 that the differential equation dy dx xy1/2 possesses at least two solutions whose graphs pass through (0 ...
1 Theorem 2.4.1 (Existence and Uniqueness of solutions of 1st order linear differential equations). For the 1st order differential equation , if and are continuous on an open interval containing the point , then there exists a unqiue function that satisfies the differential equation ...
In Problems 25-28 determine whether Theorem 1.2 .1 guarantees that the differential equation y^′=sqrt y^2-9 possesses a unique solution through the given ...
Determine whether Theorem 1.2.1 guarantees that the differential equation y' = ˆš(y 2 - 9) possesses a unique solution through the given point.. 1. (1, 4) 2. (5, 3) 3. (2, 3) 4. (1, 1) Theorem 1.2.1. Let R be a rectangular region in the xy-plane defined by a ‰¤ x ‰¤ b, c ‰¤ y ‰¤ d that contains the point (x 0, y 0) in its interior.If f (x, y) and Ï‘f /Ï‘y are continuous on ...
FOR FIRST ORDER DIFFERENTIAL EQUATIONS I. Statement of the theorem. We consider the initial value problem ... existence and uniqueness theorem for (1.1) we just have to establish that the equation (3.1) has a unique solution in [x0 −h,x0 …
hypotheses of Theorem 1.2.1 do not hold on the line y = 0 for the differential equation dv/dx = xy02, so it is not surprising, as we saw in Example 4 of this section, that there are two solutions defined on a common interval —h <x < h satisfying y(0) = 0. On the other hand, the hypotheses of Theorem 1.2.1 do