Basic existence/uniqueness question

[milloak]

Homework Statement

Does the solution exist, and is it unique?

a ) dy/dx = y^(1/3) ; y(0) = 0

b) y * dy/dx = x - 1 ; y(1) = 0

Homework Equations

Theorem: If a function and its partial derivative (w/ y) are continuous in some area that has a point (a,b) in it, then on some interval with (a,0), dy/dx has one and only one solution y(a) = b on the interval.

The Attempt at a Solution

Existence is found by continuity of f(x,y) and its partial derivative in some interval. If one of these is not continuous, a solution either does not exist or there is more than one solution. I'm not sure how to do this. Do you just find plug in the initial conditions and check for continuity? And if the partial derivative is discontinuous, then it has more than one solution?

 

[hunt_mat]

Both equations are seperable and can be integrated.

 

[HallsofIvy]

Homework Statement

Does the solution exist, and is it unique?

a ) dy/dx = y^(1/3) ; y(0) = 0

b) y * dy/dx = x - 1 ; y(1) = 0

Homework Equations

Theorem: If a function and its partial derivative (w/ y) are continuous in some area that has a point (a,b) in it, then on some interval with (a,0), dy/dx has one and only one solution y(a) = b on the interval.

The Attempt at a Solution

Existence is found by continuity of f(x,y) and its partial derivative in some interval. If one of
these is not continuous, a solution either does not exist or there is more than one solution. I'm not sure how to do this. Do you just find plug in the initial conditions and check for continuity? And if the partial derivative is discontinuous, then it has more than one solution?

For the first equation, y^(1/3) is continuous in a neighborhod of (0, 0). What about the derivative, with respect to y, which is (1/3)y^{-2/3}?

For the first equation, (x- 1)/y is NOT continuous in a neighborhood of (0, 0). However, the theorem you cite does NOT say what happens if those are not true.