Existence and Uniqueness For ODE

[ver_mathstats]

Homework Statement:

1. For which initial points exists a solution in some interval containing x-naught?
2. For which initial points exists a unique solution in some interval containing x-naught?

Relevant Equations:

Existence and Uniqueness

I'm new to learning about ODE's and I just want to make sure I am on the right track and understanding everything properly.

We have our ODE which is y' = 6x³(y-1)^1/6 with y(x₀)=y₀.

I know that existence means that if f is continuous on an open rectangle that contains (x₀, y₀) then the IVP has at least one solution on some open subinterval of (a,b) that contains x₀. Uniqueness is when f and f_y are continuous on the rectangle then we will have a unique solution on some open subinterval of (a,b) that contains x₀.

Here is my attempt at a solution for the questions:

1. y' = 6x³(y-1)^1/6 is continuous for all x∈R and all y∈R, thus there is a solution for all (x₀, y₀) according to the theorem of existence.

2. Since f_y = x³/(y-1)^5/6 is continuous for all x∈R and all y∈R except for y = 1 we see there is a unique solution on some open interval containing x₀ for all (x₀, y₀) except when y = 1.

Could someone please check this over to see if I have the right idea and if this is correct? Thank you.

 

[haruspex]

if f is continuous

What is f in this context?
Some of your post reads as though you are considering some function f of x and y (continuous on a rectangle, etc.) but in the question there is only that y is a function of x.

 

[ver_mathstats]

Sorry I should've been more clear y' = 6x³(y-1)^1/6 and f(x,y) = 6x³(y-1)^1/6 and then f_y is the partial derivative. So "if f is continuous" pertains to f(x,y).

 

[haruspex]

Sorry I should've been more clear y' = 6x³(y-1)^1/6 and f(x,y) = 6x³(y-1)^1/6 and then f_y is the partial derivative. So "if f is continuous" pertains to f(x,y).

Then all looks ok except that the ODE cannot apply for y<1.

 

[ver_mathstats]

Then all looks ok except that the ODE cannot apply for y<1.

Sorry this is for uniqueness right? Thank you though.

 

[haruspex]

Sorry this is for uniqueness right? Thank you though.

No, for existence.
The ODE is only defined for y≥1, so you cannot really say that a solution exists for the point (0,0).

 

[ver_mathstats]

No, for existence.
The ODE is only defined for y≥1, so you cannot really say that a solution exists for the point (0,0).

Oh okay, sorry I got a bit confused, I understand now. Thank you for the help.