Clarification regarding undetermined coefficient rule The Next CEO of Stack Overflowparticular solution of $4y''-y= sin(x)cdot cos(x/2)$Find a particular solution of a nonhomogenous equationSolve ODE using undetermined coefficientsVariation of Parameters: Why didn't multiply by $x$ in particular solution of $y''-2y'+y=frace^xx$?How to deal with repeating terms when solving differential equations using Method of Undetermined Coefficient?The method of undetermined coefficients: clarificationUndetermined Coefficients - Differential EquationsODE Initial Value Problem Help for Particular Solution of $y^(3)-2y''+y'=1+xe^x$Why do we multiply repeated terms by x in the method of undetermined coefficients ?Should I apply boundary conditions in the general solution before finding the particular solution?
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Clarification regarding undetermined coefficient rule
The Next CEO of Stack Overflowparticular solution of $4y''-y= sin(x)cdot cos(x/2)$Find a particular solution of a nonhomogenous equationSolve ODE using undetermined coefficientsVariation of Parameters: Why didn't multiply by $x$ in particular solution of $y''-2y'+y=frace^xx$?How to deal with repeating terms when solving differential equations using Method of Undetermined Coefficient?The method of undetermined coefficients: clarificationUndetermined Coefficients - Differential EquationsODE Initial Value Problem Help for Particular Solution of $y^(3)-2y''+y'=1+xe^x$Why do we multiply repeated terms by x in the method of undetermined coefficients ?Should I apply boundary conditions in the general solution before finding the particular solution?
$begingroup$
There is a rule in my book that states: If the particular solution contains terms that duplicate in $y_c$, multiply the particular solution with the lowest power of x such that no duplication remains.
So let's say I have a complementary solution
$y_c = c_1x $ and a particular solution
$y_p = Ax^2+Bx+C $.
Now I know that I will multiply the B term in the particular solution by x since it duplicates in the complementary solution. My question is that do I also multiply the A term and C term in the particular solution by x even though they don't have a duplicate in the complementary?
ordinary-differential-equations
asked Mar 19 at 10:38
JohnySmith12JohnySmith12
413
$endgroup$
add a comment |
$begingroup$
There is a rule in my book that states: If the particular solution contains terms that duplicate in $y_c$, multiply the particular solution with the lowest power of x such that no duplication remains.
So let's say I have a complementary solution
$y_c = c_1x $ and a particular solution
$y_p = Ax^2+Bx+C $.
Now I know that I will multiply the B term in the particular solution by x since it duplicates in the complementary solution. My question is that do I also multiply the A term and C term in the particular solution by x even though they don't have a duplicate in the complementary?
ordinary-differential-equations
asked Mar 19 at 10:38
JohnySmith12JohnySmith12
413
$endgroup$
$begingroup$
It says, "multiply the particular solution...." You write "a particular solution [is] $y_p=Ax^2+Bx+C$. Doesn't that answer your question?
$endgroup$
– Gerry Myerson
Mar 19 at 11:34
$begingroup$
It would be helpful if you could give the differential equation under consideration. From the available data it could be in the simplest case $y''=x^2$.
$endgroup$
– LutzL
Mar 19 at 16:19
$begingroup$
@Lutz, how is $y_p=Ax^2+Bx+C$ a solution of $y''=x^2$? And how would giving the differential equation help?
$endgroup$
– Gerry Myerson
Mar 19 at 22:17
$begingroup$
The concrete ODE would confirm that the method is applicable at all to your task. One could suspect that you have $xy'-y=x^2$ where the homogeneous solution is indeed $c_1x$ but the coefficients are non-constant. // $Ax^2+Bx+C$ is the first ingredient to the solution, after that you notice the double resonance and complete it to $y_p=x^2(Ax^2+Bx+C)$. In the given equation you could of course also just integrate twice to find $y_p=frac112x^4$.
$endgroup$
– LutzL
Mar 19 at 22:54
add a comment |
$begingroup$
There is a rule in my book that states: If the particular solution contains terms that duplicate in $y_c$, multiply the particular solution with the lowest power of x such that no duplication remains.
So let's say I have a complementary solution
$y_c = c_1x $ and a particular solution
$y_p = Ax^2+Bx+C $.
Now I know that I will multiply the B term in the particular solution by x since it duplicates in the complementary solution. My question is that do I also multiply the A term and C term in the particular solution by x even though they don't have a duplicate in the complementary?
ordinary-differential-equations
asked Mar 19 at 10:38
JohnySmith12JohnySmith12
413
$endgroup$
There is a rule in my book that states: If the particular solution contains terms that duplicate in $y_c$, multiply the particular solution with the lowest power of x such that no duplication remains.
So let's say I have a complementary solution
$y_c = c_1x $ and a particular solution
$y_p = Ax^2+Bx+C $.
Now I know that I will multiply the B term in the particular solution by x since it duplicates in the complementary solution. My question is that do I also multiply the A term and C term in the particular solution by x even though they don't have a duplicate in the complementary?
ordinary-differential-equations
ordinary-differential-equations
asked Mar 19 at 10:38
JohnySmith12JohnySmith12
413
asked Mar 19 at 10:38
JohnySmith12JohnySmith12
413
asked Mar 19 at 10:38
JohnySmith12JohnySmith12
413
asked Mar 19 at 10:38
JohnySmith12JohnySmith12
413
asked Mar 19 at 10:38
JohnySmith12JohnySmith12
413
413
$begingroup$
It says, "multiply the particular solution...." You write "a particular solution [is] $y_p=Ax^2+Bx+C$. Doesn't that answer your question?
$endgroup$
– Gerry Myerson
Mar 19 at 11:34
$begingroup$
It would be helpful if you could give the differential equation under consideration. From the available data it could be in the simplest case $y''=x^2$.
$endgroup$
– LutzL
Mar 19 at 16:19
$begingroup$
@Lutz, how is $y_p=Ax^2+Bx+C$ a solution of $y''=x^2$? And how would giving the differential equation help?
$endgroup$
– Gerry Myerson
Mar 19 at 22:17
$begingroup$
The concrete ODE would confirm that the method is applicable at all to your task. One could suspect that you have $xy'-y=x^2$ where the homogeneous solution is indeed $c_1x$ but the coefficients are non-constant. // $Ax^2+Bx+C$ is the first ingredient to the solution, after that you notice the double resonance and complete it to $y_p=x^2(Ax^2+Bx+C)$. In the given equation you could of course also just integrate twice to find $y_p=frac112x^4$.
$endgroup$
– LutzL
Mar 19 at 22:54
add a comment |
$begingroup$
It says, "multiply the particular solution...." You write "a particular solution [is] $y_p=Ax^2+Bx+C$. Doesn't that answer your question?
$endgroup$
– Gerry Myerson
Mar 19 at 11:34
$begingroup$
It would be helpful if you could give the differential equation under consideration. From the available data it could be in the simplest case $y''=x^2$.
$endgroup$
– LutzL
Mar 19 at 16:19
$begingroup$
@Lutz, how is $y_p=Ax^2+Bx+C$ a solution of $y''=x^2$? And how would giving the differential equation help?
$endgroup$
– Gerry Myerson
Mar 19 at 22:17
$begingroup$
The concrete ODE would confirm that the method is applicable at all to your task. One could suspect that you have $xy'-y=x^2$ where the homogeneous solution is indeed $c_1x$ but the coefficients are non-constant. // $Ax^2+Bx+C$ is the first ingredient to the solution, after that you notice the double resonance and complete it to $y_p=x^2(Ax^2+Bx+C)$. In the given equation you could of course also just integrate twice to find $y_p=frac112x^4$.
$endgroup$
– LutzL
Mar 19 at 22:54
$begingroup$
It says, "multiply the particular solution...." You write "a particular solution [is] $y_p=Ax^2+Bx+C$. Doesn't that answer your question?
$endgroup$
– Gerry Myerson
Mar 19 at 11:34
$begingroup$
It says, "multiply the particular solution...." You write "a particular solution [is] $y_p=Ax^2+Bx+C$. Doesn't that answer your question?
$endgroup$
– Gerry Myerson
Mar 19 at 11:34
$begingroup$
It would be helpful if you could give the differential equation under consideration. From the available data it could be in the simplest case $y''=x^2$.
$endgroup$
– LutzL
Mar 19 at 16:19
$begingroup$
It would be helpful if you could give the differential equation under consideration. From the available data it could be in the simplest case $y''=x^2$.
$endgroup$
– LutzL
Mar 19 at 16:19
$begingroup$
@Lutz, how is $y_p=Ax^2+Bx+C$ a solution of $y''=x^2$? And how would giving the differential equation help?
$endgroup$
– Gerry Myerson
Mar 19 at 22:17
$begingroup$
@Lutz, how is $y_p=Ax^2+Bx+C$ a solution of $y''=x^2$? And how would giving the differential equation help?
$endgroup$
– Gerry Myerson
Mar 19 at 22:17
$begingroup$
The concrete ODE would confirm that the method is applicable at all to your task. One could suspect that you have $xy'-y=x^2$ where the homogeneous solution is indeed $c_1x$ but the coefficients are non-constant. // $Ax^2+Bx+C$ is the first ingredient to the solution, after that you notice the double resonance and complete it to $y_p=x^2(Ax^2+Bx+C)$. In the given equation you could of course also just integrate twice to find $y_p=frac112x^4$.
$endgroup$
– LutzL
Mar 19 at 22:54
$begingroup$
The concrete ODE would confirm that the method is applicable at all to your task. One could suspect that you have $xy'-y=x^2$ where the homogeneous solution is indeed $c_1x$ but the coefficients are non-constant. // $Ax^2+Bx+C$ is the first ingredient to the solution, after that you notice the double resonance and complete it to $y_p=x^2(Ax^2+Bx+C)$. In the given equation you could of course also just integrate twice to find $y_p=frac112x^4$.
$endgroup$
– LutzL
Mar 19 at 22:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The general rule is that if the term on the right side that you are currently considering is in resonance with the linear differential operator, then you have to multiply with a monomial. More precisely, if the term is $$p(x)e^rx$$ and $r$ is a characteristic root of multiplicity $m$ of the left side, then the corresponding term in the particular solution $y_p$ is of the form $$x^mq(x)e^rx$$ where $q$ has the same degree as $p$ with as of yet undetermined coefficients.
In conclusion, for your concrete problem it means that you have to multiply the whole expression, to get $$x(Ax^2+Bx+C).$$
answered Mar 19 at 13:44
LutzLLutzL
60.1k42057
$endgroup$
$begingroup$
If I multiply the whole particular solution by x then the C term becomes $xC$. Now the C term duplicates the complementary solution, Would I need to multiply the entire particular solution by x again them?
$endgroup$
– JohnySmith12
Mar 19 at 15:45
$begingroup$
You are right, if $1$ and $x$ are basis solutions, then the multiplicity of $r=0$ is $m=2$ and thus you need to multiply with $x^2$.
$endgroup$
– LutzL
Mar 19 at 16:18
add a comment |
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1 Answer
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votes
$begingroup$
The general rule is that if the term on the right side that you are currently considering is in resonance with the linear differential operator, then you have to multiply with a monomial. More precisely, if the term is $$p(x)e^rx$$ and $r$ is a characteristic root of multiplicity $m$ of the left side, then the corresponding term in the particular solution $y_p$ is of the form $$x^mq(x)e^rx$$ where $q$ has the same degree as $p$ with as of yet undetermined coefficients.
In conclusion, for your concrete problem it means that you have to multiply the whole expression, to get $$x(Ax^2+Bx+C).$$
answered Mar 19 at 13:44
LutzLLutzL
60.1k42057
$endgroup$
$begingroup$
If I multiply the whole particular solution by x then the C term becomes $xC$. Now the C term duplicates the complementary solution, Would I need to multiply the entire particular solution by x again them?
$endgroup$
– JohnySmith12
Mar 19 at 15:45
$begingroup$
You are right, if $1$ and $x$ are basis solutions, then the multiplicity of $r=0$ is $m=2$ and thus you need to multiply with $x^2$.
$endgroup$
– LutzL
Mar 19 at 16:18
add a comment |
$begingroup$
The general rule is that if the term on the right side that you are currently considering is in resonance with the linear differential operator, then you have to multiply with a monomial. More precisely, if the term is $$p(x)e^rx$$ and $r$ is a characteristic root of multiplicity $m$ of the left side, then the corresponding term in the particular solution $y_p$ is of the form $$x^mq(x)e^rx$$ where $q$ has the same degree as $p$ with as of yet undetermined coefficients.
In conclusion, for your concrete problem it means that you have to multiply the whole expression, to get $$x(Ax^2+Bx+C).$$
answered Mar 19 at 13:44
LutzLLutzL
60.1k42057
$endgroup$
$begingroup$
If I multiply the whole particular solution by x then the C term becomes $xC$. Now the C term duplicates the complementary solution, Would I need to multiply the entire particular solution by x again them?
$endgroup$
– JohnySmith12
Mar 19 at 15:45
$begingroup$
You are right, if $1$ and $x$ are basis solutions, then the multiplicity of $r=0$ is $m=2$ and thus you need to multiply with $x^2$.
$endgroup$
– LutzL
Mar 19 at 16:18
add a comment |
$begingroup$
The general rule is that if the term on the right side that you are currently considering is in resonance with the linear differential operator, then you have to multiply with a monomial. More precisely, if the term is $$p(x)e^rx$$ and $r$ is a characteristic root of multiplicity $m$ of the left side, then the corresponding term in the particular solution $y_p$ is of the form $$x^mq(x)e^rx$$ where $q$ has the same degree as $p$ with as of yet undetermined coefficients.
In conclusion, for your concrete problem it means that you have to multiply the whole expression, to get $$x(Ax^2+Bx+C).$$
answered Mar 19 at 13:44
LutzLLutzL
60.1k42057
$endgroup$
The general rule is that if the term on the right side that you are currently considering is in resonance with the linear differential operator, then you have to multiply with a monomial. More precisely, if the term is $$p(x)e^rx$$ and $r$ is a characteristic root of multiplicity $m$ of the left side, then the corresponding term in the particular solution $y_p$ is of the form $$x^mq(x)e^rx$$ where $q$ has the same degree as $p$ with as of yet undetermined coefficients.
In conclusion, for your concrete problem it means that you have to multiply the whole expression, to get $$x(Ax^2+Bx+C).$$
answered Mar 19 at 13:44
LutzLLutzL
60.1k42057
answered Mar 19 at 13:44
LutzLLutzL
60.1k42057
answered Mar 19 at 13:44
LutzLLutzL
60.1k42057
answered Mar 19 at 13:44
LutzLLutzL
60.1k42057
60.1k42057
$begingroup$
If I multiply the whole particular solution by x then the C term becomes $xC$. Now the C term duplicates the complementary solution, Would I need to multiply the entire particular solution by x again them?
$endgroup$
– JohnySmith12
Mar 19 at 15:45
$begingroup$
You are right, if $1$ and $x$ are basis solutions, then the multiplicity of $r=0$ is $m=2$ and thus you need to multiply with $x^2$.
$endgroup$
– LutzL
Mar 19 at 16:18
add a comment |
$begingroup$
If I multiply the whole particular solution by x then the C term becomes $xC$. Now the C term duplicates the complementary solution, Would I need to multiply the entire particular solution by x again them?
$endgroup$
– JohnySmith12
Mar 19 at 15:45
$begingroup$
You are right, if $1$ and $x$ are basis solutions, then the multiplicity of $r=0$ is $m=2$ and thus you need to multiply with $x^2$.
$endgroup$
– LutzL
Mar 19 at 16:18
$begingroup$
If I multiply the whole particular solution by x then the C term becomes $xC$. Now the C term duplicates the complementary solution, Would I need to multiply the entire particular solution by x again them?
$endgroup$
– JohnySmith12
Mar 19 at 15:45
$begingroup$
If I multiply the whole particular solution by x then the C term becomes $xC$. Now the C term duplicates the complementary solution, Would I need to multiply the entire particular solution by x again them?
$endgroup$
– JohnySmith12
Mar 19 at 15:45
$begingroup$
You are right, if $1$ and $x$ are basis solutions, then the multiplicity of $r=0$ is $m=2$ and thus you need to multiply with $x^2$.
$endgroup$
– LutzL
Mar 19 at 16:18
$begingroup$
You are right, if $1$ and $x$ are basis solutions, then the multiplicity of $r=0$ is $m=2$ and thus you need to multiply with $x^2$.
$endgroup$
– LutzL
Mar 19 at 16:18
add a comment |
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$begingroup$
It says, "multiply the particular solution...." You write "a particular solution [is] $y_p=Ax^2+Bx+C$. Doesn't that answer your question?
$endgroup$
– Gerry Myerson
Mar 19 at 11:34
$begingroup$
It would be helpful if you could give the differential equation under consideration. From the available data it could be in the simplest case $y''=x^2$.
$endgroup$
– LutzL
Mar 19 at 16:19
$begingroup$
@Lutz, how is $y_p=Ax^2+Bx+C$ a solution of $y''=x^2$? And how would giving the differential equation help?
$endgroup$
– Gerry Myerson
Mar 19 at 22:17
$begingroup$
The concrete ODE would confirm that the method is applicable at all to your task. One could suspect that you have $xy'-y=x^2$ where the homogeneous solution is indeed $c_1x$ but the coefficients are non-constant. // $Ax^2+Bx+C$ is the first ingredient to the solution, after that you notice the double resonance and complete it to $y_p=x^2(Ax^2+Bx+C)$. In the given equation you could of course also just integrate twice to find $y_p=frac112x^4$.
$endgroup$
– LutzL
Mar 19 at 22:54