Method of undetermined coefficients - when do I need to multiply with x?
0
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I have a book that suggests me to use a few classical approaches to solving inhomogenous differential equations, like $y=C*e^{kx}$ for an inhomogeneous part of $i=A*e^{kx}$. If $k$ is a zero of the charakteristic polynomial though, I need to multiply this with t. That said, what if I can not solve the homogeneous with this method, and instead use the separation of variables? How can I then determine if I need to multiply with zero or not?
ordinary-differential-equations
asked Dec 29 '18 at 0:40
ExperimentatornomExperimentatornom
1
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show 1 more comment
0
$begingroup$
I have a book that suggests me to use a few classical approaches to solving inhomogenous differential equations, like $y=C*e^{kx}$ for an inhomogeneous part of $i=A*e^{kx}$. If $k$ is a zero of the charakteristic polynomial though, I need to multiply this with t. That said, what if I can not solve the homogeneous with this method, and instead use the separation of variables? How can I then determine if I need to multiply with zero or not?
ordinary-differential-equations
asked Dec 29 '18 at 0:40
ExperimentatornomExperimentatornom
1
$endgroup$
$begingroup$
@NobleMushtak If k is a multiple zero though, and for a differential equation of fifth order there can be 5 multiple zeroes, then I need to multiple with $x^5$ though. So is using $x^{Order of equation}e{kx}$ a reasonable thing to do?
$endgroup$
– Experimentatornom
Dec 29 '18 at 0:49
$begingroup$
Sorry, ignore everything I just said. I just messed up and confused variation of parameters with separation of variables for some reason. I actually don't know that much about separation of variables, so you should probably ask someone else for help. Again, sorry for the confusion.
$endgroup$
– Noble Mushtak
Dec 29 '18 at 1:09
1
$begingroup$
Not every differential equation can be solved with a separation of variables. That being said once you solve the homogeneous equation, you need to avoid any of those solutions to obtain a particular solution to the non-homogeneous equation. Multiplying by a power of t allows for that to happen.
$endgroup$
– Joel Pereira
Dec 29 '18 at 1:28
$begingroup$
@JoelPereira That makes sense, but is there a way to know if I need to multiply before I try to solve it? Or will I only notice the need for it once I have tried solving unsucessfully?
$endgroup$
– Experimentatornom
Dec 29 '18 at 1:35
1
$begingroup$
Only after finding the homogeneous solutions is that possible. The question is does the forcing function look like the homogeneous solution.
$endgroup$
– Joel Pereira
Dec 29 '18 at 3:24
|
show 1 more comment
0
0
0
$begingroup$
I have a book that suggests me to use a few classical approaches to solving inhomogenous differential equations, like $y=C*e^{kx}$ for an inhomogeneous part of $i=A*e^{kx}$. If $k$ is a zero of the charakteristic polynomial though, I need to multiply this with t. That said, what if I can not solve the homogeneous with this method, and instead use the separation of variables? How can I then determine if I need to multiply with zero or not?
ordinary-differential-equations
asked Dec 29 '18 at 0:40
ExperimentatornomExperimentatornom
1
$endgroup$
I have a book that suggests me to use a few classical approaches to solving inhomogenous differential equations, like $y=C*e^{kx}$ for an inhomogeneous part of $i=A*e^{kx}$. If $k$ is a zero of the charakteristic polynomial though, I need to multiply this with t. That said, what if I can not solve the homogeneous with this method, and instead use the separation of variables? How can I then determine if I need to multiply with zero or not?
ordinary-differential-equations
ordinary-differential-equations
asked Dec 29 '18 at 0:40
ExperimentatornomExperimentatornom
1
asked Dec 29 '18 at 0:40
ExperimentatornomExperimentatornom
1
asked Dec 29 '18 at 0:40
ExperimentatornomExperimentatornom
1
asked Dec 29 '18 at 0:40
ExperimentatornomExperimentatornom
1
asked Dec 29 '18 at 0:40
ExperimentatornomExperimentatornom
1
1
$begingroup$
@NobleMushtak If k is a multiple zero though, and for a differential equation of fifth order there can be 5 multiple zeroes, then I need to multiple with $x^5$ though. So is using $x^{Order of equation}e{kx}$ a reasonable thing to do?
$endgroup$
– Experimentatornom
Dec 29 '18 at 0:49
$begingroup$
Sorry, ignore everything I just said. I just messed up and confused variation of parameters with separation of variables for some reason. I actually don't know that much about separation of variables, so you should probably ask someone else for help. Again, sorry for the confusion.
$endgroup$
– Noble Mushtak
Dec 29 '18 at 1:09
1
$begingroup$
Not every differential equation can be solved with a separation of variables. That being said once you solve the homogeneous equation, you need to avoid any of those solutions to obtain a particular solution to the non-homogeneous equation. Multiplying by a power of t allows for that to happen.
$endgroup$
– Joel Pereira
Dec 29 '18 at 1:28
$begingroup$
@JoelPereira That makes sense, but is there a way to know if I need to multiply before I try to solve it? Or will I only notice the need for it once I have tried solving unsucessfully?
$endgroup$
– Experimentatornom
Dec 29 '18 at 1:35
1
$begingroup$
Only after finding the homogeneous solutions is that possible. The question is does the forcing function look like the homogeneous solution.
$endgroup$
– Joel Pereira
Dec 29 '18 at 3:24
|
show 1 more comment
$begingroup$
@NobleMushtak If k is a multiple zero though, and for a differential equation of fifth order there can be 5 multiple zeroes, then I need to multiple with $x^5$ though. So is using $x^{Order of equation}e{kx}$ a reasonable thing to do?
$endgroup$
– Experimentatornom
Dec 29 '18 at 0:49
$begingroup$
Sorry, ignore everything I just said. I just messed up and confused variation of parameters with separation of variables for some reason. I actually don't know that much about separation of variables, so you should probably ask someone else for help. Again, sorry for the confusion.
$endgroup$
– Noble Mushtak
Dec 29 '18 at 1:09
1
$begingroup$
Not every differential equation can be solved with a separation of variables. That being said once you solve the homogeneous equation, you need to avoid any of those solutions to obtain a particular solution to the non-homogeneous equation. Multiplying by a power of t allows for that to happen.
$endgroup$
– Joel Pereira
Dec 29 '18 at 1:28
$begingroup$
@JoelPereira That makes sense, but is there a way to know if I need to multiply before I try to solve it? Or will I only notice the need for it once I have tried solving unsucessfully?
$endgroup$
– Experimentatornom
Dec 29 '18 at 1:35
1
$begingroup$
Only after finding the homogeneous solutions is that possible. The question is does the forcing function look like the homogeneous solution.
$endgroup$
– Joel Pereira
Dec 29 '18 at 3:24
$begingroup$
@NobleMushtak If k is a multiple zero though, and for a differential equation of fifth order there can be 5 multiple zeroes, then I need to multiple with $x^5$ though. So is using $x^{Order of equation}e{kx}$ a reasonable thing to do?
$endgroup$
– Experimentatornom
Dec 29 '18 at 0:49
$begingroup$
@NobleMushtak If k is a multiple zero though, and for a differential equation of fifth order there can be 5 multiple zeroes, then I need to multiple with $x^5$ though. So is using $x^{Order of equation}e{kx}$ a reasonable thing to do?
$endgroup$
– Experimentatornom
Dec 29 '18 at 0:49
$begingroup$
Sorry, ignore everything I just said. I just messed up and confused variation of parameters with separation of variables for some reason. I actually don't know that much about separation of variables, so you should probably ask someone else for help. Again, sorry for the confusion.
$endgroup$
– Noble Mushtak
Dec 29 '18 at 1:09
$begingroup$
Sorry, ignore everything I just said. I just messed up and confused variation of parameters with separation of variables for some reason. I actually don't know that much about separation of variables, so you should probably ask someone else for help. Again, sorry for the confusion.
$endgroup$
– Noble Mushtak
Dec 29 '18 at 1:09
1
1
$begingroup$
Not every differential equation can be solved with a separation of variables. That being said once you solve the homogeneous equation, you need to avoid any of those solutions to obtain a particular solution to the non-homogeneous equation. Multiplying by a power of t allows for that to happen.
$endgroup$
– Joel Pereira
Dec 29 '18 at 1:28
$begingroup$
Not every differential equation can be solved with a separation of variables. That being said once you solve the homogeneous equation, you need to avoid any of those solutions to obtain a particular solution to the non-homogeneous equation. Multiplying by a power of t allows for that to happen.
$endgroup$
– Joel Pereira
Dec 29 '18 at 1:28
$begingroup$
@JoelPereira That makes sense, but is there a way to know if I need to multiply before I try to solve it? Or will I only notice the need for it once I have tried solving unsucessfully?
$endgroup$
– Experimentatornom
Dec 29 '18 at 1:35
$begingroup$
@JoelPereira That makes sense, but is there a way to know if I need to multiply before I try to solve it? Or will I only notice the need for it once I have tried solving unsucessfully?
$endgroup$
– Experimentatornom
Dec 29 '18 at 1:35
1
1
$begingroup$
Only after finding the homogeneous solutions is that possible. The question is does the forcing function look like the homogeneous solution.
$endgroup$
– Joel Pereira
Dec 29 '18 at 3:24
$begingroup$
Only after finding the homogeneous solutions is that possible. The question is does the forcing function look like the homogeneous solution.
$endgroup$
– Joel Pereira
Dec 29 '18 at 3:24
|
show 1 more comment
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$begingroup$
@NobleMushtak If k is a multiple zero though, and for a differential equation of fifth order there can be 5 multiple zeroes, then I need to multiple with $x^5$ though. So is using $x^{Order of equation}e{kx}$ a reasonable thing to do?
$endgroup$
– Experimentatornom
Dec 29 '18 at 0:49
$begingroup$
Sorry, ignore everything I just said. I just messed up and confused variation of parameters with separation of variables for some reason. I actually don't know that much about separation of variables, so you should probably ask someone else for help. Again, sorry for the confusion.
$endgroup$
– Noble Mushtak
Dec 29 '18 at 1:09
1
$begingroup$
Not every differential equation can be solved with a separation of variables. That being said once you solve the homogeneous equation, you need to avoid any of those solutions to obtain a particular solution to the non-homogeneous equation. Multiplying by a power of t allows for that to happen.
$endgroup$
– Joel Pereira
Dec 29 '18 at 1:28
$begingroup$
@JoelPereira That makes sense, but is there a way to know if I need to multiply before I try to solve it? Or will I only notice the need for it once I have tried solving unsucessfully?
$endgroup$
– Experimentatornom
Dec 29 '18 at 1:35
1
$begingroup$
Only after finding the homogeneous solutions is that possible. The question is does the forcing function look like the homogeneous solution.
$endgroup$
– Joel Pereira
Dec 29 '18 at 3:24