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June 20th, 2017, 02:22 PM   #11
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If y is set divisor then $\displaystyle y\neq 0 $
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June 20th, 2017, 03:38 PM   #12
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Quote:
Originally Posted by v8archie View Post
$$y=\begin{cases}x^2 &(x \lt 0) \\ -x^2 &(x \ge 0)\end{cases}$$is also a solution.

You've clearly misremembered something.
No, you have misremembered something:

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Originally Posted by zylo View Post
That answers all subsequent questions, except:

1) xy'=2y
dy/y=2dx/x
lny=2lnx+c
y=Cx$\displaystyle {^2}$
x=-2, y=4 -> C=1

y=x$\displaystyle {^2}$ sat's 1) and BC for all x, including 0 (0.y'=2.0).

Finished. End of story.
There is a theorem that says if you have a solution which satisfies ODE and BC it is the only solution (forgot all conditions but applies in this case).

If you divide 1) by x, x=0 is excluded, trivially and obviously.
Did you catch and BC (Boundary Condition) this time? Would you like to see it again?

Last edited by skipjack; June 20th, 2017 at 10:10 PM.
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June 20th, 2017, 10:13 PM   #13
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Quote:
Originally Posted by zylo View Post
There is a theorem that says if you have a solution which satisfies ODE and BC it is the only solution (forgot all conditions but applies in this case).
There are various conditions you might have come across, but they don't apply to the equation xy' = 2y.
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June 21st, 2017, 05:04 AM   #14
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Quote:
Originally Posted by zylo View Post
Did you catch and BC (Boundary Condition) this time? Would you like to see it again?
The example I gave satisfies your boundary condition. Would you like to try again?
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June 21st, 2017, 07:35 AM   #15
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Quote:
Originally Posted by v8archie View Post
The example I gave satisfies your boundary condition. Would you like to try again?
y=x$\displaystyle {^2}$
y=-x$\displaystyle {^2}$
BC: y(-2)=4
Your turn.

I sympathize slightly with the post-Romsek trend of the thread because when I took DE I understood nothing and hence remember nothing. It was just one method after the other when I was really looking for a start-from-scratch analysis of individual equations. I got an A when I deserved an F and I suspect the people who got an F or D deserved an A because they tried to understand things but under the circumstances couldn't because there was nothing to understand - somebody "just" thought of a method.

As for skipjack's claim that conditions for uniqueness don't apply to this equation, see, for example:

http://www2.fiu.edu/~aladrog/BasicTheoryLinear.pdf

"There exists a unique function f(x) defined on [a, b] such that f(x) is a solution of the linear differential equation and satisfies
f(x0) = c0, f’(x0) = c1, f’’(x0) = c2, … , f$\displaystyle ^{n-1}$(n-1)(x0) = cn-1."

Last edited by skipjack; June 21st, 2017 at 02:26 PM.
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June 21st, 2017, 08:04 AM   #16
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Quote:
Originally Posted by zylo View Post
y=x$\displaystyle {^2}$
y=-x$\displaystyle {^2}$
BC: y(-2)=4
Your turn.
That is neither what I wrote, nor does it communicate anything meaningful.

My function meets your boundary condition because $(-2)^2=4$.


Quote:
Originally Posted by zylo View Post
As for skipjack's claim that conditions for uniqueness don't apply to this equation, see, for example:

http://www2.fiu.edu/~aladrog/BasicTheoryLinear.pdf

"There exists a unique function f(x) defined on [a, b] such that f(x) is a solution of the linear differential equation and satisfies
f(x0) = c0, f’(x0) = c1, f’’(x0) = c2, … , f$\displaystyle ^{n-1}$(n-1)(x0) = cn-1."
Your quote is very mangled, we are not working on a closed interval here (although that may not matter depending on the definition of $a$ and $b$, and the linear equation is not specified, but most importantly we have a counter example so your theorem clearly doesn't apply.

Last edited by skipjack; June 21st, 2017 at 02:28 PM.
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June 21st, 2017, 08:25 AM   #17
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Your theorem (implicitly) assumes that $f(x)$ is analytic everywhere (in the given interval). There's no reason to assume that any derivative past the first exists in $xy'=2y$ and indeed, for my solution, $y''$ doesn't exist at $x=0$.

This is confirmed by the corollary which requires $y''(x_0)=0$ in the equation $xy'=2y$ to hold true.

You have to understand the theorem before quoting it blindly.

Last edited by v8archie; June 21st, 2017 at 08:47 AM.
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June 21st, 2017, 09:01 AM   #18
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Quote:
Originally Posted by zylo View Post

As for skipjack's claim that conditions for uniqueness don't apply to this equation, see, for example:

http://www2.fiu.edu/~aladrog/BasicTheoryLinear.pdf

"There exists a unique function f(x) defined on [a, b] such that f(x) is a solution of the linear differential equation and satisfies
f(x0) = c0, f’(x0) = c1, f’’(x0) = c2, … , f$\displaystyle ^{n-1}$(n-1)(x0) = cn-1."
xy'-y=0 and xy'-2y, y(-2)=4 satisfy the conditions of the theorem.
Obviously the derivatives have to exist to order n.

Let's assume everyone else understands whatever you are talking about and there is no point in trying to explain it to me.

Last edited by skipjack; June 21st, 2017 at 02:29 PM.
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June 21st, 2017, 09:32 AM   #19
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Fair enough. But the understanding you have is wrong.

In my solution, all derivatives up to $n=1$ exist. So the counterexample contradicts your understanding of the theorem.
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June 21st, 2017, 09:35 AM   #20
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As usual, don't know what you are talking about, so please don't bother explaining - remember, everyone else does so it's entirely unnecessary.

Last edited by skipjack; June 21st, 2017 at 02:31 PM.
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