constant evaluation when using differential equations.About the Legendre differential equationInhomogeneous...
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constant evaluation when using differential equations.
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constant evaluation when using differential equations.
About the Legendre differential equationInhomogeneous 2nd-order linear differential equationDetermining $y' = 1 - y^2$ generally excludes $y = pm 1$Solving a set of coupled first order differential equationsSolution of a Modified Bessel Differential Equation with Complex CoefficientFirst order nonlinear differential equation 2Should I include constants of integration while solving for the particular integral or not?“Trivial” differential equations solving, in Hilbert spaces
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$begingroup$
This is regards to constant evaluation when using differential equations.
- A solution is given to be:
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )-(c_1+1) e^x+(c_2-1) e^{2x}$$
- A simplified solution in an answer book is given as:
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )+(c_1 ) e^x+(c_2 ) e^{2x}$$
There is a change in sign of $c_1$ in the third term. $C_1$ is a constant and not specified to be positive or negative or is it supposed to be positive and that information is simply not specified. I never know how to interpret this kind of results. Can someone explain, please? Thank you.
Sincerely,
Mary A. Marion
ordinary-differential-equations
edited 9 hours ago
LutzL
67.7k4 gold badges22 silver badges61 bronze badges
asked 9 hours ago
Mary A. MarionMary A. Marion
136 bronze badges
$endgroup$
add a comment |
$begingroup$
This is regards to constant evaluation when using differential equations.
- A solution is given to be:
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )-(c_1+1) e^x+(c_2-1) e^{2x}$$
- A simplified solution in an answer book is given as:
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )+(c_1 ) e^x+(c_2 ) e^{2x}$$
There is a change in sign of $c_1$ in the third term. $C_1$ is a constant and not specified to be positive or negative or is it supposed to be positive and that information is simply not specified. I never know how to interpret this kind of results. Can someone explain, please? Thank you.
Sincerely,
Mary A. Marion
ordinary-differential-equations
edited 9 hours ago
LutzL
67.7k4 gold badges22 silver badges61 bronze badges
asked 9 hours ago
Mary A. MarionMary A. Marion
136 bronze badges
$endgroup$
add a comment |
$begingroup$
This is regards to constant evaluation when using differential equations.
- A solution is given to be:
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )-(c_1+1) e^x+(c_2-1) e^{2x}$$
- A simplified solution in an answer book is given as:
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )+(c_1 ) e^x+(c_2 ) e^{2x}$$
There is a change in sign of $c_1$ in the third term. $C_1$ is a constant and not specified to be positive or negative or is it supposed to be positive and that information is simply not specified. I never know how to interpret this kind of results. Can someone explain, please? Thank you.
Sincerely,
Mary A. Marion
ordinary-differential-equations
edited 9 hours ago
LutzL
67.7k4 gold badges22 silver badges61 bronze badges
asked 9 hours ago
Mary A. MarionMary A. Marion
136 bronze badges
$endgroup$
This is regards to constant evaluation when using differential equations.
- A solution is given to be:
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )-(c_1+1) e^x+(c_2-1) e^{2x}$$
- A simplified solution in an answer book is given as:
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )+(c_1 ) e^x+(c_2 ) e^{2x}$$
There is a change in sign of $c_1$ in the third term. $C_1$ is a constant and not specified to be positive or negative or is it supposed to be positive and that information is simply not specified. I never know how to interpret this kind of results. Can someone explain, please? Thank you.
Sincerely,
Mary A. Marion
ordinary-differential-equations
ordinary-differential-equations
edited 9 hours ago
LutzL
67.7k4 gold badges22 silver badges61 bronze badges
asked 9 hours ago
Mary A. MarionMary A. Marion
136 bronze badges
edited 9 hours ago
LutzL
67.7k4 gold badges22 silver badges61 bronze badges
asked 9 hours ago
Mary A. MarionMary A. Marion
136 bronze badges
edited 9 hours ago
LutzL
67.7k4 gold badges22 silver badges61 bronze badges
edited 9 hours ago
LutzL
67.7k4 gold badges22 silver badges61 bronze badges
edited 9 hours ago
LutzL
67.7k4 gold badges22 silver badges61 bronze badges
67.7k4 gold badges22 silver badges61 bronze badges
asked 9 hours ago
Mary A. MarionMary A. Marion
136 bronze badges
asked 9 hours ago
Mary A. MarionMary A. Marion
136 bronze badges
asked 9 hours ago
Mary A. MarionMary A. Marion
136 bronze badges
136 bronze badges
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Analysts have the lazy (but time-preserving) habit of using the same symbol for constants as they go along in such calculations. Those constants are different, but it's a slight abuse of notation that doesn't cause much trouble once you understand what's intended. It's like when they make such statements as $epsilon=frac{epsilon}{2},$ for example, since $epsilon>0$ is usually considered to be an arbitrarily small number anyway. Another place where we use such initially annoying (until you get used to it) abuse of notation is in the reindexing of series, where the same index is blithely used with different meanings.
Such are just conventions to shorten the work, and as you get used to them yourself, you'd begin to appreciate them -- and perhaps even prefer them to the more fastidious but tedious path.
answered 8 hours ago
AllawonderAllawonder
3,7578 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
As $c_1$ and $c_2$ are constants, we could define two other constants $tilde{c_1}=-(c_1+1)$ and $tilde{c_2}=c_2-1$ which differ from $c_1$ and $c_2$ by the value of one.
The answer key shows that
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )-(c_1+1) e^x+(c_2-1) e^{2x}tag{1}$$
can be simplified to
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )+(c_1) e^x+(c_2 ) e^{2x}tag{2}$$
where $(2)$ could be rewritten as
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )+(tilde{c_1} ) e^x+(tilde{c_2} ) e^{2x}tag{3}$$
provided that
$$tilde{c_1}=-(c_1+1)$$
$$tilde{c_2}=c_2-1$$
The author has chosen to omit rewriting $c_1$ and $c_2$ as new constants since it is implied that $c_1$ and $c_2$ refer to arbitrary constants in both $(1)$ and $(2)$.
answered 9 hours ago
Axion004Axion004
1,2906 silver badges17 bronze badges
$endgroup$
add a comment |
$begingroup$
Both solutions are correct and they are equivalent. The constants $C_1$ and $C_2$ are just place holders for numbers to be found from initial values and you may as well call them $-C_1-1$ or $C_2+1$
Once the initial values are given the constants are found and the final result is unique regardless of the notations for constants.
answered 9 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
51.1k4 gold badges27 silver badges72 bronze badges
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Analysts have the lazy (but time-preserving) habit of using the same symbol for constants as they go along in such calculations. Those constants are different, but it's a slight abuse of notation that doesn't cause much trouble once you understand what's intended. It's like when they make such statements as $epsilon=frac{epsilon}{2},$ for example, since $epsilon>0$ is usually considered to be an arbitrarily small number anyway. Another place where we use such initially annoying (until you get used to it) abuse of notation is in the reindexing of series, where the same index is blithely used with different meanings.
Such are just conventions to shorten the work, and as you get used to them yourself, you'd begin to appreciate them -- and perhaps even prefer them to the more fastidious but tedious path.
answered 8 hours ago
AllawonderAllawonder
3,7578 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
Analysts have the lazy (but time-preserving) habit of using the same symbol for constants as they go along in such calculations. Those constants are different, but it's a slight abuse of notation that doesn't cause much trouble once you understand what's intended. It's like when they make such statements as $epsilon=frac{epsilon}{2},$ for example, since $epsilon>0$ is usually considered to be an arbitrarily small number anyway. Another place where we use such initially annoying (until you get used to it) abuse of notation is in the reindexing of series, where the same index is blithely used with different meanings.
Such are just conventions to shorten the work, and as you get used to them yourself, you'd begin to appreciate them -- and perhaps even prefer them to the more fastidious but tedious path.
answered 8 hours ago
AllawonderAllawonder
3,7578 silver badges18 bronze badges
$endgroup$
add a comment |
$begingroup$
Analysts have the lazy (but time-preserving) habit of using the same symbol for constants as they go along in such calculations. Those constants are different, but it's a slight abuse of notation that doesn't cause much trouble once you understand what's intended. It's like when they make such statements as $epsilon=frac{epsilon}{2},$ for example, since $epsilon>0$ is usually considered to be an arbitrarily small number anyway. Another place where we use such initially annoying (until you get used to it) abuse of notation is in the reindexing of series, where the same index is blithely used with different meanings.
Such are just conventions to shorten the work, and as you get used to them yourself, you'd begin to appreciate them -- and perhaps even prefer them to the more fastidious but tedious path.
answered 8 hours ago
AllawonderAllawonder
3,7578 silver badges18 bronze badges
$endgroup$
Analysts have the lazy (but time-preserving) habit of using the same symbol for constants as they go along in such calculations. Those constants are different, but it's a slight abuse of notation that doesn't cause much trouble once you understand what's intended. It's like when they make such statements as $epsilon=frac{epsilon}{2},$ for example, since $epsilon>0$ is usually considered to be an arbitrarily small number anyway. Another place where we use such initially annoying (until you get used to it) abuse of notation is in the reindexing of series, where the same index is blithely used with different meanings.
Such are just conventions to shorten the work, and as you get used to them yourself, you'd begin to appreciate them -- and perhaps even prefer them to the more fastidious but tedious path.
answered 8 hours ago
AllawonderAllawonder
3,7578 silver badges18 bronze badges
answered 8 hours ago
AllawonderAllawonder
3,7578 silver badges18 bronze badges
answered 8 hours ago
AllawonderAllawonder
3,7578 silver badges18 bronze badges
answered 8 hours ago
AllawonderAllawonder
3,7578 silver badges18 bronze badges
3,7578 silver badges18 bronze badges
add a comment |
add a comment |
$begingroup$
As $c_1$ and $c_2$ are constants, we could define two other constants $tilde{c_1}=-(c_1+1)$ and $tilde{c_2}=c_2-1$ which differ from $c_1$ and $c_2$ by the value of one.
The answer key shows that
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )-(c_1+1) e^x+(c_2-1) e^{2x}tag{1}$$
can be simplified to
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )+(c_1) e^x+(c_2 ) e^{2x}tag{2}$$
where $(2)$ could be rewritten as
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )+(tilde{c_1} ) e^x+(tilde{c_2} ) e^{2x}tag{3}$$
provided that
$$tilde{c_1}=-(c_1+1)$$
$$tilde{c_2}=c_2-1$$
The author has chosen to omit rewriting $c_1$ and $c_2$ as new constants since it is implied that $c_1$ and $c_2$ refer to arbitrary constants in both $(1)$ and $(2)$.
answered 9 hours ago
Axion004Axion004
1,2906 silver badges17 bronze badges
$endgroup$
add a comment |
$begingroup$
As $c_1$ and $c_2$ are constants, we could define two other constants $tilde{c_1}=-(c_1+1)$ and $tilde{c_2}=c_2-1$ which differ from $c_1$ and $c_2$ by the value of one.
The answer key shows that
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )-(c_1+1) e^x+(c_2-1) e^{2x}tag{1}$$
can be simplified to
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )+(c_1) e^x+(c_2 ) e^{2x}tag{2}$$
where $(2)$ could be rewritten as
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )+(tilde{c_1} ) e^x+(tilde{c_2} ) e^{2x}tag{3}$$
provided that
$$tilde{c_1}=-(c_1+1)$$
$$tilde{c_2}=c_2-1$$
The author has chosen to omit rewriting $c_1$ and $c_2$ as new constants since it is implied that $c_1$ and $c_2$ refer to arbitrary constants in both $(1)$ and $(2)$.
answered 9 hours ago
Axion004Axion004
1,2906 silver badges17 bronze badges
$endgroup$
add a comment |
$begingroup$
As $c_1$ and $c_2$ are constants, we could define two other constants $tilde{c_1}=-(c_1+1)$ and $tilde{c_2}=c_2-1$ which differ from $c_1$ and $c_2$ by the value of one.
The answer key shows that
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )-(c_1+1) e^x+(c_2-1) e^{2x}tag{1}$$
can be simplified to
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )+(c_1) e^x+(c_2 ) e^{2x}tag{2}$$
where $(2)$ could be rewritten as
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )+(tilde{c_1} ) e^x+(tilde{c_2} ) e^{2x}tag{3}$$
provided that
$$tilde{c_1}=-(c_1+1)$$
$$tilde{c_2}=c_2-1$$
The author has chosen to omit rewriting $c_1$ and $c_2$ as new constants since it is implied that $c_1$ and $c_2$ refer to arbitrary constants in both $(1)$ and $(2)$.
answered 9 hours ago
Axion004Axion004
1,2906 silver badges17 bronze badges
$endgroup$
As $c_1$ and $c_2$ are constants, we could define two other constants $tilde{c_1}=-(c_1+1)$ and $tilde{c_2}=c_2-1$ which differ from $c_1$ and $c_2$ by the value of one.
The answer key shows that
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )-(c_1+1) e^x+(c_2-1) e^{2x}tag{1}$$
can be simplified to
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )+(c_1) e^x+(c_2 ) e^{2x}tag{2}$$
where $(2)$ could be rewritten as
$$y=(e^{2x}+e^x ) ln(1+e^{-x} )+(tilde{c_1} ) e^x+(tilde{c_2} ) e^{2x}tag{3}$$
provided that
$$tilde{c_1}=-(c_1+1)$$
$$tilde{c_2}=c_2-1$$
The author has chosen to omit rewriting $c_1$ and $c_2$ as new constants since it is implied that $c_1$ and $c_2$ refer to arbitrary constants in both $(1)$ and $(2)$.
answered 9 hours ago
Axion004Axion004
1,2906 silver badges17 bronze badges
edited 7 hours ago
answered 9 hours ago
Axion004Axion004
1,2906 silver badges17 bronze badges
answered 9 hours ago
Axion004Axion004
1,2906 silver badges17 bronze badges
answered 9 hours ago
Axion004Axion004
1,2906 silver badges17 bronze badges
1,2906 silver badges17 bronze badges
add a comment |
add a comment |
$begingroup$
Both solutions are correct and they are equivalent. The constants $C_1$ and $C_2$ are just place holders for numbers to be found from initial values and you may as well call them $-C_1-1$ or $C_2+1$
Once the initial values are given the constants are found and the final result is unique regardless of the notations for constants.
answered 9 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
51.1k4 gold badges27 silver badges72 bronze badges
$endgroup$
add a comment |
$begingroup$
Both solutions are correct and they are equivalent. The constants $C_1$ and $C_2$ are just place holders for numbers to be found from initial values and you may as well call them $-C_1-1$ or $C_2+1$
Once the initial values are given the constants are found and the final result is unique regardless of the notations for constants.
answered 9 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
51.1k4 gold badges27 silver badges72 bronze badges
$endgroup$
add a comment |
$begingroup$
Both solutions are correct and they are equivalent. The constants $C_1$ and $C_2$ are just place holders for numbers to be found from initial values and you may as well call them $-C_1-1$ or $C_2+1$
Once the initial values are given the constants are found and the final result is unique regardless of the notations for constants.
answered 9 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
51.1k4 gold badges27 silver badges72 bronze badges
$endgroup$
Both solutions are correct and they are equivalent. The constants $C_1$ and $C_2$ are just place holders for numbers to be found from initial values and you may as well call them $-C_1-1$ or $C_2+1$
Once the initial values are given the constants are found and the final result is unique regardless of the notations for constants.
answered 9 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
51.1k4 gold badges27 silver badges72 bronze badges
answered 9 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
51.1k4 gold badges27 silver badges72 bronze badges
answered 9 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
51.1k4 gold badges27 silver badges72 bronze badges
answered 9 hours ago
Mohammad Riazi-KermaniMohammad Riazi-Kermani
51.1k4 gold badges27 silver badges72 bronze badges
51.1k4 gold badges27 silver badges72 bronze badges
add a comment |
add a comment |
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