# 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 fy 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 fy = 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.

Homework Helper
Gold Member
2022 Award
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 fy is the partial derivative. So "if f is continuous" pertains to f(x,y).

Homework Helper
Gold Member
2022 Award
Sorry I should've been more clear y' = 6x3(y-1)1/6 and f(x,y) = 6x3(y-1)1/6 and then fy 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
ver_mathstats
Then all looks ok except that the ODE cannot apply for y<1.
Sorry this is for uniqueness right? Thank you though.

Homework Helper
Gold Member
2022 Award
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
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.