I posted this before but I could not really understand the hand writing.
I posted this before but I could not really understand the hand writing. Existence and Uniqueness...
given ivp y' = (2y)/x, y(x0) = y0 using the existence and uniqueness theorem show that a unique solution exists on any interval where x0 does not equal 0, no solution exists if y(0) = y0 does not equal 0, and and infinite number of solutions exist if y(0) = 0
Determine whether the Existence and Uniqueness of Solution Theorem implies that the given initial value problem has a unique solution. 2 dy Select the correct choice below and fill in the answer box(es) to complete your choice. The theorem implies the existence of a unique solution because a rectangle containing the point Type an ordered pair.) The theorem does not imply the existence of a unique solution becauseis not continuous in any rectangle containing the point Type an ordered pair.)...
2y (9 points) Given the initial value problem y' => y (xo) = yo. Use the existence and uniqueness theorem to show that a) a unique solution exists on any interval where xo + 0, b) no solution exists if y (0) = yo # 0, and c) an infinite number of solutions exist if y (0) = 0.
x (9 points) Given the initial value problem y' 2y 29, 2014 ,y (xo) = yo. Use the existence and uniqueness theorem to show that a) a unique solution exists on any interval where Xo 70, b) no solution exists if y (0) = yo #0, and c) an infinite number of solutions exist if y (0) = 0.
2y 1. (9 points) Given the initial value problem y' = y (xo) = yo. Use the existence and uniqueness theorem to show that a) a unique solution exists on any interval where x, 60, b) no solution exists if y(0) = % 70, and c) an infinite number of solutions exist if y(0) = 0.
For each initial value problem, does Picards's theorem apply? If
so, determine if it guarantees that a solutio exists and is
unique.
Theorem (Picard). Consider the initial value problem dy = f(t,y), dt (IVP) y(to) = Yo- (a) Existence: If f(t,y) is continuous in an open rectangle R = {(t,y) |a<t < b, c < y < d} and (to, Yo) belongs in R, then there exist h > 0 and a solution y = y(t) of (IVP) defined in...
Problem 2. (a) Solve the initial value problem I y' + 2y = g(t), 1 y(0) = 0, where where | 1 if t < 1, g(t) = { 10 if t > 1 (t) = { for all t. Is this solution unique for all time? Is it unique for any time? Does this contradict the existence and uniqueness theorem? Explain. (b) If the initial condition y(0) = 0 were replaced with y(1) = 0, would there necessarily be...
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...
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...
I Do We Have the Complete Solution Set? A differential operator in R[D] has order n can be written out in the form o(n-1) with the last coefficient cn (at least) not equal to zero. The key to determining the dimension of these solution spaces is the following existence and uniqueness theorem for initial value problems. 'So it can be efficiently described by giving a basis. ethciently described by giving a basis Theorem 1 (Existence and Unique ness Theorem for...