Mixing
Solving differential equations of the form $f'(x)=f(x+1)$ [duplicate]
3
$begingroup$
This question already has an answer here:
Differential equations that are also functional
3 answers
When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$?
5 answers
What function satisfies $F'(x) = F(2x)$?
1 answer
How to solve differential equations of the following form:
$frac{df}{dx} = f(x+1)$
ordinary-differential-equations
edited Jan 11 at 7:54
Jepsilon
396216
asked Jan 11 at 7:40
Fayez Abdlrazaq DeabFayez Abdlrazaq Deab
30217
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marked as duplicate by Abcd, LutzL differential-equations
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Jan 11 at 7:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
3
$begingroup$
This question already has an answer here:
Differential equations that are also functional
3 answers
When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$?
5 answers
What function satisfies $F'(x) = F(2x)$?
1 answer
How to solve differential equations of the following form:
$frac{df}{dx} = f(x+1)$
ordinary-differential-equations
edited Jan 11 at 7:54
Jepsilon
396216
asked Jan 11 at 7:40
Fayez Abdlrazaq DeabFayez Abdlrazaq Deab
30217
$endgroup$
marked as duplicate by Abcd, LutzL differential-equations
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Jan 11 at 7:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Not really sure about solving because this isn't my field, but after a quick Google search, these are a specific type of differential equations known as Functional Differential Equations (FDEs) en.m.wikipedia.org/wiki/Functional_differential_equation
$endgroup$
– Jepsilon
Jan 11 at 7:50
1
$begingroup$
Because I cannor edit my comment: If you work with the assumption that $f$ is linear, i.e. $f(a+b)=f(a)+f(b)$, you can easily solve it normally as $f(x)=Ae^x-f(1)$ and substituting $x=1$ you get $f(1)=frac{Ae}{2}$. Then you find $A$ with some initial condition.
$endgroup$
– Jepsilon
Jan 11 at 8:06
1
$begingroup$
@Jepsilon This is also a delay differential equation whose solutions are well-known
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– Dylan
Jan 11 at 12:49
$begingroup$
Thank you for your helpful comments 🌹
$endgroup$
– Fayez Abdlrazaq Deab
Jan 22 at 3:20
add a comment |
3
3
3
1
$begingroup$
This question already has an answer here:
Differential equations that are also functional
3 answers
When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$?
5 answers
What function satisfies $F'(x) = F(2x)$?
1 answer
How to solve differential equations of the following form:
$frac{df}{dx} = f(x+1)$
ordinary-differential-equations
edited Jan 11 at 7:54
Jepsilon
396216
asked Jan 11 at 7:40
Fayez Abdlrazaq DeabFayez Abdlrazaq Deab
30217
$endgroup$
This question already has an answer here:
Differential equations that are also functional
3 answers
When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$?
5 answers
What function satisfies $F'(x) = F(2x)$?
1 answer
How to solve differential equations of the following form:
$frac{df}{dx} = f(x+1)$
This question already has an answer here:
Differential equations that are also functional
3 answers
When $f(x+1)-f(x)=f'(x)$, what are the solutions for $f(x)$?
5 answers
What function satisfies $F'(x) = F(2x)$?
1 answer
ordinary-differential-equations
ordinary-differential-equations
edited Jan 11 at 7:54
Jepsilon
396216
asked Jan 11 at 7:40
Fayez Abdlrazaq DeabFayez Abdlrazaq Deab
30217
edited Jan 11 at 7:54
Jepsilon
396216
asked Jan 11 at 7:40
Fayez Abdlrazaq DeabFayez Abdlrazaq Deab
30217
edited Jan 11 at 7:54
Jepsilon
396216
edited Jan 11 at 7:54
Jepsilon
396216
edited Jan 11 at 7:54
Jepsilon
396216
396216
asked Jan 11 at 7:40
Fayez Abdlrazaq DeabFayez Abdlrazaq Deab
30217
asked Jan 11 at 7:40
Fayez Abdlrazaq DeabFayez Abdlrazaq Deab
30217
asked Jan 11 at 7:40
Fayez Abdlrazaq DeabFayez Abdlrazaq Deab
30217
30217
marked as duplicate by Abcd, LutzL differential-equations
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Jan 11 at 7:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Abcd, LutzL differential-equations
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Jan 11 at 7:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Not really sure about solving because this isn't my field, but after a quick Google search, these are a specific type of differential equations known as Functional Differential Equations (FDEs) en.m.wikipedia.org/wiki/Functional_differential_equation
$endgroup$
– Jepsilon
Jan 11 at 7:50
1
$begingroup$
Because I cannor edit my comment: If you work with the assumption that $f$ is linear, i.e. $f(a+b)=f(a)+f(b)$, you can easily solve it normally as $f(x)=Ae^x-f(1)$ and substituting $x=1$ you get $f(1)=frac{Ae}{2}$. Then you find $A$ with some initial condition.
$endgroup$
– Jepsilon
Jan 11 at 8:06
1
$begingroup$
@Jepsilon This is also a delay differential equation whose solutions are well-known
$endgroup$
– Dylan
Jan 11 at 12:49
$begingroup$
Thank you for your helpful comments 🌹
$endgroup$
– Fayez Abdlrazaq Deab
Jan 22 at 3:20
add a comment |
1
$begingroup$
Not really sure about solving because this isn't my field, but after a quick Google search, these are a specific type of differential equations known as Functional Differential Equations (FDEs) en.m.wikipedia.org/wiki/Functional_differential_equation
$endgroup$
– Jepsilon
Jan 11 at 7:50
1
$begingroup$
Because I cannor edit my comment: If you work with the assumption that $f$ is linear, i.e. $f(a+b)=f(a)+f(b)$, you can easily solve it normally as $f(x)=Ae^x-f(1)$ and substituting $x=1$ you get $f(1)=frac{Ae}{2}$. Then you find $A$ with some initial condition.
$endgroup$
– Jepsilon
Jan 11 at 8:06
1
$begingroup$
@Jepsilon This is also a delay differential equation whose solutions are well-known
$endgroup$
– Dylan
Jan 11 at 12:49
$begingroup$
Thank you for your helpful comments 🌹
$endgroup$
– Fayez Abdlrazaq Deab
Jan 22 at 3:20
1
1
$begingroup$
Not really sure about solving because this isn't my field, but after a quick Google search, these are a specific type of differential equations known as Functional Differential Equations (FDEs) en.m.wikipedia.org/wiki/Functional_differential_equation
$endgroup$
– Jepsilon
Jan 11 at 7:50
$begingroup$
Not really sure about solving because this isn't my field, but after a quick Google search, these are a specific type of differential equations known as Functional Differential Equations (FDEs) en.m.wikipedia.org/wiki/Functional_differential_equation
$endgroup$
– Jepsilon
Jan 11 at 7:50
1
1
$begingroup$
Because I cannor edit my comment: If you work with the assumption that $f$ is linear, i.e. $f(a+b)=f(a)+f(b)$, you can easily solve it normally as $f(x)=Ae^x-f(1)$ and substituting $x=1$ you get $f(1)=frac{Ae}{2}$. Then you find $A$ with some initial condition.
$endgroup$
– Jepsilon
Jan 11 at 8:06
$begingroup$
Because I cannor edit my comment: If you work with the assumption that $f$ is linear, i.e. $f(a+b)=f(a)+f(b)$, you can easily solve it normally as $f(x)=Ae^x-f(1)$ and substituting $x=1$ you get $f(1)=frac{Ae}{2}$. Then you find $A$ with some initial condition.
$endgroup$
– Jepsilon
Jan 11 at 8:06
1
1
$begingroup$
@Jepsilon This is also a delay differential equation whose solutions are well-known
$endgroup$
– Dylan
Jan 11 at 12:49
$begingroup$
@Jepsilon This is also a delay differential equation whose solutions are well-known
$endgroup$
– Dylan
Jan 11 at 12:49
$begingroup$
Thank you for your helpful comments 🌹
$endgroup$
– Fayez Abdlrazaq Deab
Jan 22 at 3:20
$begingroup$
Thank you for your helpful comments 🌹
$endgroup$
– Fayez Abdlrazaq Deab
Jan 22 at 3:20
add a comment |
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1
$begingroup$
Not really sure about solving because this isn't my field, but after a quick Google search, these are a specific type of differential equations known as Functional Differential Equations (FDEs) en.m.wikipedia.org/wiki/Functional_differential_equation
$endgroup$
– Jepsilon
Jan 11 at 7:50
1
$begingroup$
Because I cannor edit my comment: If you work with the assumption that $f$ is linear, i.e. $f(a+b)=f(a)+f(b)$, you can easily solve it normally as $f(x)=Ae^x-f(1)$ and substituting $x=1$ you get $f(1)=frac{Ae}{2}$. Then you find $A$ with some initial condition.
$endgroup$
– Jepsilon
Jan 11 at 8:06
1
$begingroup$
@Jepsilon This is also a delay differential equation whose solutions are well-known
$endgroup$
– Dylan
Jan 11 at 12:49
$begingroup$
Thank you for your helpful comments 🌹
$endgroup$
– Fayez Abdlrazaq Deab
Jan 22 at 3:20