Question

Differential Equations help?? Find two different solutions to initial value problem, why there is no contradiction of the Existence Uniqueness Theorem? dy/dx=2x(y-5)^(3/4) y(1)=5 Differential Equations help?? Find two different solutions to initial value problem, why there is no contradiction of the Existence Uniqueness Theorem? dy/dx=2x(y-5)^(3/4) y(1)=5 @Mathematics

Answer 1

I can solve it but I can't see how to get two answers...

Answer 2

I don't see how you can ever get two solutions with these types of questions

Answer 3

like another one is y'=y^(1/5)

Answer 4

unless it's like ln and you can use absolute?

Answer 5

Well I'm looking at it...
James knows though.

Answer 6

The other one
y'=y^(1/5)
is not an IVP without the initial value though, so it makes sense it has different answers depending on the value of C you get.

Answer 7

Hm. If y'=2x(y-5)^(3/4) then
\[ (y-5)^{-3/4} dy = 2x dx \]
\[ \implies (1/4)(y-5)^{1/4} = x^2 + C \]
\[ \implies y - 5 = [ 4(x^2 + C) ]^4 \]
\[ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 4^4 (x^2 + C)^4 \]

With the initial condition y(1) = 5, it must be that C = -1 hence

\[ y = 2^8 (x^2 - 1)^4 \]

So I guess I don't see how there are two solutions.

Answer 8

sorry ... final solution is
\[ y = 2^8 (x^2 -1)^4 + 5 \]

Answer 9

second initial is y(0)=0

Answer 10

I think it must exist since it ask why does these 2 solutions not contradict existence uniqueness

Answer 11

Well there you have it I guess. I can say James is usually right about this kind of thing.

Answer 12

Oh, I see.

The uniqueness theorem says that if f'(x) = g(x,y) and g is reasonably behaved, then given an initial condition y(a) = b, then the equation has a unique solution.

So saying the equation \( y'=2x(y-5)^{3/4} \) has a solution for one initial condition and a different solution for a different initial condition doesn't violate the uniqueness theorem, because ...

Answer 13

the hypotheses of the uniqueness theorem aren't satisfied. Instead, you're showing equation + initial condition A => solution A; and
equation + initial condition B => solution B
and these two solution aren't the same. Great, no problem.

Answer 14

What the unique theorem says is that if

equation + initial condition A => solution A
equation + initial condition A => solution B

then solution A = solution B

Answer 15

So we make up another not-given initial condition to get our other equation?

Answer 16

Oh I see so if you have two solutions then they are not related but if they are equal then the initial conditions should be the same

Answer 17

Yes.

Given the initial conditions
y(1) = 5, we get one solution (above)
y(0) = 0, we will get another solution (not yet written out)

Answer 18

Sweet, thanks.

Answer 19

I don't see how they are two different solutions though? I get that they are two different initial conditions

Answer 20

but they have the same value, oh so that goes along with the Existence Uniqueness theorem

Answer 21

By the way, it can be the case that

equation + initial condition A => solution S
equation + initial condition B => solution S

That's not a contradiction.

Answer 22

your equation + initial condition y(1) = 5 => \( y = 2^8 (x^2 - 1)^4 + 5 \)

That's a solution and indeed THE solution given the ODE and the IC.

Now with the ODE and the IC y(0) = 0 you will have a different solution. I haven't written it down. You write it down.

Answer 23

Yeah I did they both are 0

Answer 24

but yeah I get it now, they don't contradict b/c of the condition just wrote up top

Answer 25

So write down the function y(x) for me for y(0) = 0

Answer 26

Isn't it the same but with a different initial condition? Or am I doing this wrong. I just plugged in y(0)=0

Answer 27

And so what then is the solution of the
ODE: dy/dx=2x(y-5)^(3/4)
+ IC: y(0) = 0

Answer 28

what function satisfies that ODE and IC.

Answer 29

dy/dx=0? ==> 2(0)(0-5)^(3/4)=0

Answer 30

The function that satisfies
ODE: dy/dx=2x(y-5)^(3/4)
+ IC: y(1) = 5

is y(x) = 2^8 (x^2 - 1)^4 + 5

Answer 31

No.

Answer 32

Ohhhh I get it you have to integrate

Answer 33

that function is for both initial condition right?

Answer 34

I've already solved the equation in general: look above. The procedure is to take that general solution and then substitute the initial condition into that general solution to solve for the constant C introduced in the general solution.

Answer 35

Namely, the general solution of the ODE
dy/dx=2x(y-5)^(3/4)

is
\[ y(x) = 2^8 (x^2 + C)^4 + 5 \]

Answer 36

Now, for the IC y(1) = 5, I substituted in that condition to solve for C. That gave me THE solution the Initial Value Problem --- ODE + IC --- of \( y(x) = 2^8 (x^2 -1)^4 + 5 \).

Answer 37

Oh okay I gotcha then you sub in y(0)=0 and find C for that case

Answer 38

Yes.

Answer 39

and that gives two equations/solutions. Okay thanks for your help! Makes a lot of sense now

Answer 40

Actually, it looks to me as though there is no solution for y(0) = 0.

Write \( dy/dx = g(x,y) \) where \( g(x,y) = 2x(y-5)^{3/4} \). What's the domain of g?

\[ Dom(g) = \mathbb{R} \times [5,\infty) \]

Hence (0,0) isn't in the domain of g and we shouldn't expect to find a solution to the ODE with IC y(0) = 0.

Answer 41

Then how do you know what initial condition to use for the second solution? Aren't you trying to get an initial where dy/dx=2x(y-5)^(3/4) is equal to zero?

Answer 42

can I do like y(0)=5? should work as long as y> = 5