Question

solve the differential equation

dy/dx = x^2-y^2

Answer 1

hmm, can we let vx = y ?

Answer 2

v'x + v = x^2 - (vx)^2

v'x + v = x^2(1 - v^2)

prolly not ...

Answer 3

this appeared in the section before substitution, so I feel like an idiot not knowing the answer.

Answer 4

its not a linear setup, and its not seperable that i can tell

Answer 5

the answer in the book says you can guarantee the existence of a solution and the uniqueness of a solution

Answer 6

y' = x^2-y^2

y' + y^2 = x^2

subbing in z = y^2
z' = 2yy' doesnt help since there is no y term

Answer 7

theorem 1 says if dy/dx and Dy(dy/dx) is continuous on some interval near a, it guarantees uniqueness.

so here, it seems its continuous everywhere. but for a solution to exist, don't you have to explicitly solve for y(x)?

Answer 8

you dont need to explicitly solve for it no

Answer 9

http://www.wolframalpha.com/input/?i=y%27%2By%5E2%3Dx%5E2

the solution is a mess for sure

Answer 10

but the examples in the book always solve it to show existence. am I misunderstanding?

Answer 11

and how does the book present the question?

Answer 12

x^2 - y^2 is a continuous function, it has no bad inputs to restrict

Fy = -2yy' is just as good .....

so the thrm holds

Answer 13

determine if theorem 1 guarantees existence. if it does, determine if theorem 1 guarantees the solution is unique.

Answer 14

so, by thrm 1 .... we see that f(x,y) is good, and Dy f(x,y) is good

Answer 15

havent you previously proved the thrm?

Answer 16

I see the existence part. But does it prove uniqueness?

Answer 17

you can always make the rectangle "domain" small enough so that f(x,y) is unique given these conditions

Answer 18

for dy/dx = y^1/3 the answer says it proves existence of y(0)=0 but not uniqueness. if you take f(x,y) = y^1/3 then df/dy = 1/(3y^2/3) but thats not continuous at y=0. so how does that work?

Answer 19

since Dy is not continuous at x=0, then the thrm is not wrong is it?

Answer 20

IF f(x,y) and Dy are continuous at a, then we can be assured that there exists a unique solution for some y(a)=b

Answer 21

that what I'm confused about

Answer 22

spose we have 2 solutions ...

they can be defined as existing at their common point, and they can be defined in a unique fashion at every other point. It is only at the intersection that they are simply not uniquely definable