Theorem 2.3.1 If f is continuous on an open rectangle (a) that contains (xo yo) then the initial value problem f(a, y), y(o)yo has at least one solution on some open subinterval of (a, b) that contains ro (b) If both fand fy are continuous on R then (2.3.1) has a unique solution on some open subinterval of (a, b) that contains ro. In Exercises 1-13 find all (xo, Vo) for which Theorem 2.3.1 implies that the initial value problem...
Problem 1. In (a) and (b) below, consider the following intial-value problem. (x + 1)(x - 4)y" + (sin x) - (In xy = 0, y(1) = 3, (1) = -2 a. What is the largest interval on which the above initial-value problem has a unique solution? Explain why.
Solve the given initial-value problem. dy = x + 5y, y(0) = 3 y(x) = Give the largest interval I over which the solution is defined. (Enter your answer using interval notation.) I =
If f(x, y) is continuous in an open rectangle R = (a, b) x (c, d) in the xy-plane that contains the point (xo, Yo), then there exists a solution y(x) to the initial-value problem dy = f(x, y), y(xo) = yo, dx that is defined in an open interval I = (a, b) containing xo. In addition, if the partial derivative Ofjay is continuous in R, then the solution y(x) of the given equation is unique. For the initial-value...
Consider the initial value problem (t-2) y" + cot(t) y' +ty=e', y( 3 ) = 41/3, ' ( 3 ) =- T/ 4. Without solving the equation, what is the largest interval in which a unique solution is guaranteed to exist?
Differential equation
1. Chapter 4 covers differential equations of the form an(x)y("4a-,(x)ye-i) + +4(x)y'+4(x)-g(x) Subject to initial conditions y)oyy-Co) Consider the second order differential equation 2x2y" + 5xy, + y-r-x 2- The Existence of a Unique Solution Theorem says there will be a unique solution y(x) to the initial-value problem at x=而over any interval 1 for which the coefficient functions, ai (x) (0 S is n) and g(x) are continuous and a, (x)0. Are there any values of x for...
1. (5 points) Find an interval containing x 0for which the given initial value problem has a unique solution. y=x2 +2 +cos(x
1. (5 points) Find an interval containing x 0for which the given initial value problem has a unique solution. y=x2 +2 +cos(x
Find the solution of the given initial value problem: y′′′+y′=sec(t), y(0)=6, y′(0)=7, y′′(0)=−3 Enter ln|a| as ln(|a|). Enclose arguments of functions in parentheses. For example, sin(2x). y(t)=
Please give the solution (how
you get an answer) not only an answer.
10 points 2. Find the largest possible open rectangular box within tyy'-space in which the existence and uniqueness theorem applies to the nonlinear initial value problem and thereby infer that it has a unique local solution. ; y(0)-1 , y'(0) 2 2
10 points 2. Find the largest possible open rectangular box within tyy'-space in which the existence and uniqueness theorem applies to the nonlinear initial value...
dy 2x2 + 3y2 1. Solve the initial value problem y(-1) = -V3 by first making a dac ту substitution. Write your solution y(x) as an explicit function of x. Find the largest open interval on which the solution is valid.