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Suppose you have an equation (I don't see a latex editor or know how to use math tags here)

f'(h(t))*h'(t) = f(h(t)+ \alpha)

where f' is differentiated with respect to t, following from the chain rule on f(h(t)). Is there a substitution that will transform this differential equation into the form of

 f'(w) = f(w+\alpha)

? It seems reasonable but I am not finding an easy way to do it.

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It looks like w=h(t).

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Not quite because then you have an extra factor of h'(h-inverse(w)).

Edited by Strange
Edited (t - > w) at request of op

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Suppose you have an equation (I don't see a latex editor or know how to use math tags here)


f'(h(t))*h'(t) = f(h(t)+ \alpha)

where f' is differentiated with respect to t, following from the chain rule on f(h(t)). Is there a substitution that will transform this differential equation into the form of


f'(w) = f(w+\alpha)

? It seems reasonable but I am not finding an easy way to do it.

Is this the equation you mean ?

$\frac{d}{{dt}}\left( {h\frac{{dh}}{{dt}}} \right) = h + \alpha$

where α is a constant?

If not I have left the square brackets of the math tags, can you revise the mathml stament and reinstate the square brackets?

Note you may (will) have to refresh the page after posting to view the mathml  - there is a fault with prsenting math on this forum.

math\frac{d}{{dt}}\left( {h\frac{{dh}}{{dt}}} \right) = h + \alpha /math]

Edited by studiot

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That isn't the equation I meant unfortunately, unless possibly you are proposing that as a substitution. In the original equation, f(g(t)) is a composite function, and I represented its derivative under the chain rule as f'(h(t))*h'(t). Although alpha is a constant, it seems like you are missing the extra function on the outside of h+alpha to make f(h(t)+alpha).

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That isn't the equation I meant unfortunately, unless possibly you are proposing that as a substitution. In the original equation, f(g(t)) is a composite function, and I represented its derivative under the chain rule as f'(h(t))*h'(t). Although alpha is a constant, it seems like you are missing the extra function on the outside of h+alpha to make f(h(t)+alpha).

Pity, but I did post the tex for you to amend so that we could all be sure of the equation you do mean.

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Again there's no edit button for some reason so I have to waste another post saying f(g(t)) should have been f(h(t)).

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It can't be too much to ask you to write your equation in conventional format.

I puzzled over your own format and came to the conclusion your notation may have introduced a circular definition.

All you have to do is copy/paste my mathml statement and then change it to your taste.

You can only come back and edit posts for 1 hour after posting here.

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I don't know what "mathml" is, I use latex. It says there is a latex editor here but it didn't work the first time, so...

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I don't know what "mathml" is, I use latex. It says there is a latex editor here but it didn't work the first time, so...

Write your script as though it was LaTex.

But instead of Tex tags, use the tags $at the beginning and$ at the end.

I tried to show you this by taking off the square brackets from the script that produced the differential equation I wrote.

Thus

13 hours ago, studiot said:

math\frac{d}{{dt}}\left( {h\frac{{dh}}{{dt}}} \right) = h + \alpha /math]

All you have to do is to put them back, try it and see.

Then you can write your own equation.

Since the parser here goes to another site to read and present the script you need to activate it after writing by using the refresh page button in your browser.
Then you will be able to view what you have written.

Many members test their Tex Math markup language whatever here.

Edited by studiot

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Okay, but is your name a hybrid of student-idiot, or stud-idiot, or stud-iota, or student-iota, or did you misspell studious, or something else entirely?

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