A difficulty in understanding a solution of a differential equation in Vinberg section 0.
0
$begingroup$
The differential equation is given in the following picture:
I believe that it is a separable differential equation, but because we differentiate with respect to you, I believe that the solution should be $$f(t) = C e^{ua}$$ instead of $$f(t) = C e^{ta},$$ could anyone clarify this for me please?
ordinary-differential-equations pde
asked Jan 14 at 7:03
hopefullyhopefully
251114
$endgroup$
add a comment |
0
$begingroup$
The differential equation is given in the following picture:
I believe that it is a separable differential equation, but because we differentiate with respect to you, I believe that the solution should be $$f(t) = C e^{ua}$$ instead of $$f(t) = C e^{ta},$$ could anyone clarify this for me please?
ordinary-differential-equations pde
asked Jan 14 at 7:03
hopefullyhopefully
251114
$endgroup$
$begingroup$
We differentiated the homomorphism identity with respect to $u$, but then set $u = 0$ so you end up with a differential equation in $t$, $f'(t) = f(t) a$, so the solution is a function of $t$.
$endgroup$
– Christopher A. Wong
Jan 14 at 7:09
$begingroup$
but $f '(t) = df/du$ correct?@ChristopherA.Wong
$endgroup$
– hopefully
Jan 14 at 7:13
$begingroup$
No. See my answer.
$endgroup$
– Fred
Jan 14 at 7:20
1
$begingroup$
@hopefully The function $f$ is a function of one variable. It does not matter what letter you call it. So $f'$ is the derivative of $f$ with respect to that variable. $f'(t)$ is that same derivative, evaluated at the point $t$.
$endgroup$
– Christopher A. Wong
Jan 14 at 8:05
add a comment |
0
0
0
$begingroup$
The differential equation is given in the following picture:
I believe that it is a separable differential equation, but because we differentiate with respect to you, I believe that the solution should be $$f(t) = C e^{ua}$$ instead of $$f(t) = C e^{ta},$$ could anyone clarify this for me please?
ordinary-differential-equations pde
asked Jan 14 at 7:03
hopefullyhopefully
251114
$endgroup$
The differential equation is given in the following picture:
I believe that it is a separable differential equation, but because we differentiate with respect to you, I believe that the solution should be $$f(t) = C e^{ua}$$ instead of $$f(t) = C e^{ta},$$ could anyone clarify this for me please?
ordinary-differential-equations pde
ordinary-differential-equations pde
asked Jan 14 at 7:03
hopefullyhopefully
251114
asked Jan 14 at 7:03
hopefullyhopefully
251114
asked Jan 14 at 7:03
hopefullyhopefully
251114
asked Jan 14 at 7:03
hopefullyhopefully
251114
asked Jan 14 at 7:03
hopefullyhopefully
251114
251114
$begingroup$
We differentiated the homomorphism identity with respect to $u$, but then set $u = 0$ so you end up with a differential equation in $t$, $f'(t) = f(t) a$, so the solution is a function of $t$.
$endgroup$
– Christopher A. Wong
Jan 14 at 7:09
$begingroup$
but $f '(t) = df/du$ correct?@ChristopherA.Wong
$endgroup$
– hopefully
Jan 14 at 7:13
$begingroup$
No. See my answer.
$endgroup$
– Fred
Jan 14 at 7:20
1
$begingroup$
@hopefully The function $f$ is a function of one variable. It does not matter what letter you call it. So $f'$ is the derivative of $f$ with respect to that variable. $f'(t)$ is that same derivative, evaluated at the point $t$.
$endgroup$
– Christopher A. Wong
Jan 14 at 8:05
add a comment |
$begingroup$
We differentiated the homomorphism identity with respect to $u$, but then set $u = 0$ so you end up with a differential equation in $t$, $f'(t) = f(t) a$, so the solution is a function of $t$.
$endgroup$
– Christopher A. Wong
Jan 14 at 7:09
$begingroup$
but $f '(t) = df/du$ correct?@ChristopherA.Wong
$endgroup$
– hopefully
Jan 14 at 7:13
$begingroup$
No. See my answer.
$endgroup$
– Fred
Jan 14 at 7:20
1
$begingroup$
@hopefully The function $f$ is a function of one variable. It does not matter what letter you call it. So $f'$ is the derivative of $f$ with respect to that variable. $f'(t)$ is that same derivative, evaluated at the point $t$.
$endgroup$
– Christopher A. Wong
Jan 14 at 8:05
$begingroup$
We differentiated the homomorphism identity with respect to $u$, but then set $u = 0$ so you end up with a differential equation in $t$, $f'(t) = f(t) a$, so the solution is a function of $t$.
$endgroup$
– Christopher A. Wong
Jan 14 at 7:09
$begingroup$
We differentiated the homomorphism identity with respect to $u$, but then set $u = 0$ so you end up with a differential equation in $t$, $f'(t) = f(t) a$, so the solution is a function of $t$.
$endgroup$
– Christopher A. Wong
Jan 14 at 7:09
$begingroup$
but $f '(t) = df/du$ correct?@ChristopherA.Wong
$endgroup$
– hopefully
Jan 14 at 7:13
$begingroup$
but $f '(t) = df/du$ correct?@ChristopherA.Wong
$endgroup$
– hopefully
Jan 14 at 7:13
$begingroup$
No. See my answer.
$endgroup$
– Fred
Jan 14 at 7:20
$begingroup$
No. See my answer.
$endgroup$
– Fred
Jan 14 at 7:20
1
1
$begingroup$
@hopefully The function $f$ is a function of one variable. It does not matter what letter you call it. So $f'$ is the derivative of $f$ with respect to that variable. $f'(t)$ is that same derivative, evaluated at the point $t$.
$endgroup$
– Christopher A. Wong
Jan 14 at 8:05
$begingroup$
@hopefully The function $f$ is a function of one variable. It does not matter what letter you call it. So $f'$ is the derivative of $f$ with respect to that variable. $f'(t)$ is that same derivative, evaluated at the point $t$.
$endgroup$
– Christopher A. Wong
Jan 14 at 8:05
add a comment |
1 Answer
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oldest
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$begingroup$
We have $ frac{d}{du}f(t+u)=frac{d}{du}f(t)f(u)$, hence $f'(t+u)=f(t)f'(u)$. With $u=0$ we get
$$f'(t)=f(t)f'(0).$$
answered Jan 14 at 7:19
FredFred
48.6k11849
$endgroup$
$begingroup$
what I am asking you are writing ..... in your writings$( ')$ is equivalent to differentiation with respect to u ... this is what you are writing
$endgroup$
– hopefully
Jan 14 at 7:25
$begingroup$
I think if you write the whole solution it may be clearer for me
$endgroup$
– hopefully
Jan 14 at 7:28
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
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1
$begingroup$
We have $ frac{d}{du}f(t+u)=frac{d}{du}f(t)f(u)$, hence $f'(t+u)=f(t)f'(u)$. With $u=0$ we get
$$f'(t)=f(t)f'(0).$$
answered Jan 14 at 7:19
FredFred
48.6k11849
$endgroup$
$begingroup$
what I am asking you are writing ..... in your writings$( ')$ is equivalent to differentiation with respect to u ... this is what you are writing
$endgroup$
– hopefully
Jan 14 at 7:25
$begingroup$
I think if you write the whole solution it may be clearer for me
$endgroup$
– hopefully
Jan 14 at 7:28
add a comment |
1
$begingroup$
We have $ frac{d}{du}f(t+u)=frac{d}{du}f(t)f(u)$, hence $f'(t+u)=f(t)f'(u)$. With $u=0$ we get
$$f'(t)=f(t)f'(0).$$
answered Jan 14 at 7:19
FredFred
48.6k11849
$endgroup$
$begingroup$
what I am asking you are writing ..... in your writings$( ')$ is equivalent to differentiation with respect to u ... this is what you are writing
$endgroup$
– hopefully
Jan 14 at 7:25
$begingroup$
I think if you write the whole solution it may be clearer for me
$endgroup$
– hopefully
Jan 14 at 7:28
add a comment |
1
1
1
$begingroup$
We have $ frac{d}{du}f(t+u)=frac{d}{du}f(t)f(u)$, hence $f'(t+u)=f(t)f'(u)$. With $u=0$ we get
$$f'(t)=f(t)f'(0).$$
answered Jan 14 at 7:19
FredFred
48.6k11849
$endgroup$
We have $ frac{d}{du}f(t+u)=frac{d}{du}f(t)f(u)$, hence $f'(t+u)=f(t)f'(u)$. With $u=0$ we get
$$f'(t)=f(t)f'(0).$$
answered Jan 14 at 7:19
FredFred
48.6k11849
answered Jan 14 at 7:19
FredFred
48.6k11849
answered Jan 14 at 7:19
FredFred
48.6k11849
answered Jan 14 at 7:19
FredFred
48.6k11849
48.6k11849
$begingroup$
what I am asking you are writing ..... in your writings$( ')$ is equivalent to differentiation with respect to u ... this is what you are writing
$endgroup$
– hopefully
Jan 14 at 7:25
$begingroup$
I think if you write the whole solution it may be clearer for me
$endgroup$
– hopefully
Jan 14 at 7:28
add a comment |
$begingroup$
what I am asking you are writing ..... in your writings$( ')$ is equivalent to differentiation with respect to u ... this is what you are writing
$endgroup$
– hopefully
Jan 14 at 7:25
$begingroup$
I think if you write the whole solution it may be clearer for me
$endgroup$
– hopefully
Jan 14 at 7:28
$begingroup$
what I am asking you are writing ..... in your writings$( ')$ is equivalent to differentiation with respect to u ... this is what you are writing
$endgroup$
– hopefully
Jan 14 at 7:25
$begingroup$
what I am asking you are writing ..... in your writings$( ')$ is equivalent to differentiation with respect to u ... this is what you are writing
$endgroup$
– hopefully
Jan 14 at 7:25
$begingroup$
I think if you write the whole solution it may be clearer for me
$endgroup$
– hopefully
Jan 14 at 7:28
$begingroup$
I think if you write the whole solution it may be clearer for me
$endgroup$
– hopefully
Jan 14 at 7:28
add a comment |
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$begingroup$
We differentiated the homomorphism identity with respect to $u$, but then set $u = 0$ so you end up with a differential equation in $t$, $f'(t) = f(t) a$, so the solution is a function of $t$.
$endgroup$
– Christopher A. Wong
Jan 14 at 7:09
$begingroup$
but $f '(t) = df/du$ correct?@ChristopherA.Wong
$endgroup$
– hopefully
Jan 14 at 7:13
$begingroup$
No. See my answer.
$endgroup$
– Fred
Jan 14 at 7:20
1
$begingroup$
@hopefully The function $f$ is a function of one variable. It does not matter what letter you call it. So $f'$ is the derivative of $f$ with respect to that variable. $f'(t)$ is that same derivative, evaluated at the point $t$.
$endgroup$
– Christopher A. Wong
Jan 14 at 8:05