Mixing
8th order particular solution
Clash Royale CLAN TAG#URR8PPP
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I am given that $r^7 - 6r^6 + 20r^5 - 56r^4 + 112r^3 - 160r^2 + 192r - 128 = (r-2)^3(r^2+4)^2$.
Then I know that the characteristic equation is then $r*(r-2)^3(r^2+4)^2 = 0$
This gives me $r = 0,2,2,2,+2i,-2i,+2i,-2i$
So I know the complimentary solution will be $y(t) = c_1 + c_2e^2t + c_3te^2t + c_4t^2e^2t + c_5cos(2t) + c_6 sin(2t) + c_7tcos(2t) + c_8tsin(2t)$.
However, what can I assume for my Y(t) for the particular solution?
Is $Ate^2t + Bt + Csin(4t) + Dcos(4t)$ a good guess for $Y(t)$?
Are there more efficient ways of computing $Y^(8)(t)$ without a brute force approach of having to compute eight consecutive derivatives?
Thank You
differential-equations proof-verification computational-complexity homogeneous-equation
asked Jul 22 at 17:24
KhanMan
447
add a comment |Â
up vote
1
down vote
favorite
I am given that $r^7 - 6r^6 + 20r^5 - 56r^4 + 112r^3 - 160r^2 + 192r - 128 = (r-2)^3(r^2+4)^2$.
Then I know that the characteristic equation is then $r*(r-2)^3(r^2+4)^2 = 0$
This gives me $r = 0,2,2,2,+2i,-2i,+2i,-2i$
So I know the complimentary solution will be $y(t) = c_1 + c_2e^2t + c_3te^2t + c_4t^2e^2t + c_5cos(2t) + c_6 sin(2t) + c_7tcos(2t) + c_8tsin(2t)$.
However, what can I assume for my Y(t) for the particular solution?
Is $Ate^2t + Bt + Csin(4t) + Dcos(4t)$ a good guess for $Y(t)$?
Are there more efficient ways of computing $Y^(8)(t)$ without a brute force approach of having to compute eight consecutive derivatives?
Thank You
differential-equations proof-verification computational-complexity homogeneous-equation
asked Jul 22 at 17:24
KhanMan
447
Note that the RHS contains $t^colorred3 e^2t$.
– Math Lover
Jul 22 at 17:39
I guess you are missing the fact that the exponential is multiplied by $t^3$ and not by $t$ only. And as far as I remember this is important for the choose of the particular solution.
– mrtaurho
Jul 22 at 17:39
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am given that $r^7 - 6r^6 + 20r^5 - 56r^4 + 112r^3 - 160r^2 + 192r - 128 = (r-2)^3(r^2+4)^2$.
Then I know that the characteristic equation is then $r*(r-2)^3(r^2+4)^2 = 0$
This gives me $r = 0,2,2,2,+2i,-2i,+2i,-2i$
So I know the complimentary solution will be $y(t) = c_1 + c_2e^2t + c_3te^2t + c_4t^2e^2t + c_5cos(2t) + c_6 sin(2t) + c_7tcos(2t) + c_8tsin(2t)$.
However, what can I assume for my Y(t) for the particular solution?
Is $Ate^2t + Bt + Csin(4t) + Dcos(4t)$ a good guess for $Y(t)$?
Are there more efficient ways of computing $Y^(8)(t)$ without a brute force approach of having to compute eight consecutive derivatives?
Thank You
differential-equations proof-verification computational-complexity homogeneous-equation
asked Jul 22 at 17:24
KhanMan
447
I am given that $r^7 - 6r^6 + 20r^5 - 56r^4 + 112r^3 - 160r^2 + 192r - 128 = (r-2)^3(r^2+4)^2$.
Then I know that the characteristic equation is then $r*(r-2)^3(r^2+4)^2 = 0$
This gives me $r = 0,2,2,2,+2i,-2i,+2i,-2i$
So I know the complimentary solution will be $y(t) = c_1 + c_2e^2t + c_3te^2t + c_4t^2e^2t + c_5cos(2t) + c_6 sin(2t) + c_7tcos(2t) + c_8tsin(2t)$.
However, what can I assume for my Y(t) for the particular solution?
Is $Ate^2t + Bt + Csin(4t) + Dcos(4t)$ a good guess for $Y(t)$?
Are there more efficient ways of computing $Y^(8)(t)$ without a brute force approach of having to compute eight consecutive derivatives?
Thank You
differential-equations proof-verification computational-complexity homogeneous-equation
asked Jul 22 at 17:24
KhanMan
447
edited Jul 22 at 17:34
asked Jul 22 at 17:24
KhanMan
447
asked Jul 22 at 17:24
KhanMan
447
asked Jul 22 at 17:24
KhanMan
447
447
Note that the RHS contains $t^colorred3 e^2t$.
– Math Lover
Jul 22 at 17:39
I guess you are missing the fact that the exponential is multiplied by $t^3$ and not by $t$ only. And as far as I remember this is important for the choose of the particular solution.
– mrtaurho
Jul 22 at 17:39
add a comment |Â
Note that the RHS contains $t^colorred3 e^2t$.
– Math Lover
Jul 22 at 17:39
I guess you are missing the fact that the exponential is multiplied by $t^3$ and not by $t$ only. And as far as I remember this is important for the choose of the particular solution.
– mrtaurho
Jul 22 at 17:39
Note that the RHS contains $t^colorred3 e^2t$.
– Math Lover
Jul 22 at 17:39
Note that the RHS contains $t^colorred3 e^2t$.
– Math Lover
Jul 22 at 17:39
I guess you are missing the fact that the exponential is multiplied by $t^3$ and not by $t$ only. And as far as I remember this is important for the choose of the particular solution.
– mrtaurho
Jul 22 at 17:39
I guess you are missing the fact that the exponential is multiplied by $t^3$ and not by $t$ only. And as far as I remember this is important for the choose of the particular solution.
– mrtaurho
Jul 22 at 17:39
add a comment |Â
1 Answer
1
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0
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The R.H.S. consists of a polynomial, a trigonometric function and a product of an exponential with a cube. Respectively
$$t^3cdot e^2t + t + sin(4t)$$
Therefore the right attempt would be
$$Y_p(t)~=~At + B + e^2t(Ct^3 + Dt^2 +Et +F) + Gsin(4t) + Hcos(4t) $$
But I do not know a way to compute the constants without derivate eight times.
answered Jul 22 at 17:46
mrtaurho
740219
Perhaps, I am in the wrong. The question asks for a "suitable form for the particular solution Y(t)" if we were to apply the method of undetermined coefficients. You are correct. I think taking eight derivatives would be an unreasonable work exercise.
– KhanMan
Jul 22 at 17:52
Yeah, I think so too. But I guess my approach should be right and respects all of the given functions.
– mrtaurho
Jul 22 at 17:57
Thank you for your responses. A few question regarding your response: Where does the B come from in your particular solution? In what part of the RHS is B supposed to be matching. Likewise, why are there four coefficients attached to e^2t when we have repeated roots of 2 three times from the complimentary solution?
– KhanMan
Jul 22 at 17:59
Try it on the equation $y''+y=x^2$ with the guess $Y_p(x)=Ax^2+Bx+C$ and as contrast only with $Y_p(x)=Ax^2+Bx$. I cannot argue why the $C$ has to be there but you have to get it with you.
– mrtaurho
Jul 22 at 18:05
$2$ is a triple root, thus the term for $t^3e^2t$ gets an additional factor $t^3$, thus $t^3e^2t(Ct^3+Dt^2+Et+F)$.
– LutzL
Jul 22 at 18:24
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The R.H.S. consists of a polynomial, a trigonometric function and a product of an exponential with a cube. Respectively
$$t^3cdot e^2t + t + sin(4t)$$
Therefore the right attempt would be
$$Y_p(t)~=~At + B + e^2t(Ct^3 + Dt^2 +Et +F) + Gsin(4t) + Hcos(4t) $$
But I do not know a way to compute the constants without derivate eight times.
answered Jul 22 at 17:46
mrtaurho
740219
Perhaps, I am in the wrong. The question asks for a "suitable form for the particular solution Y(t)" if we were to apply the method of undetermined coefficients. You are correct. I think taking eight derivatives would be an unreasonable work exercise.
– KhanMan
Jul 22 at 17:52
Yeah, I think so too. But I guess my approach should be right and respects all of the given functions.
– mrtaurho
Jul 22 at 17:57
Thank you for your responses. A few question regarding your response: Where does the B come from in your particular solution? In what part of the RHS is B supposed to be matching. Likewise, why are there four coefficients attached to e^2t when we have repeated roots of 2 three times from the complimentary solution?
– KhanMan
Jul 22 at 17:59
Try it on the equation $y''+y=x^2$ with the guess $Y_p(x)=Ax^2+Bx+C$ and as contrast only with $Y_p(x)=Ax^2+Bx$. I cannot argue why the $C$ has to be there but you have to get it with you.
– mrtaurho
Jul 22 at 18:05
$2$ is a triple root, thus the term for $t^3e^2t$ gets an additional factor $t^3$, thus $t^3e^2t(Ct^3+Dt^2+Et+F)$.
– LutzL
Jul 22 at 18:24
add a comment |Â
up vote
0
down vote
accepted
The R.H.S. consists of a polynomial, a trigonometric function and a product of an exponential with a cube. Respectively
$$t^3cdot e^2t + t + sin(4t)$$
Therefore the right attempt would be
$$Y_p(t)~=~At + B + e^2t(Ct^3 + Dt^2 +Et +F) + Gsin(4t) + Hcos(4t) $$
But I do not know a way to compute the constants without derivate eight times.
answered Jul 22 at 17:46
mrtaurho
740219
Perhaps, I am in the wrong. The question asks for a "suitable form for the particular solution Y(t)" if we were to apply the method of undetermined coefficients. You are correct. I think taking eight derivatives would be an unreasonable work exercise.
– KhanMan
Jul 22 at 17:52
Yeah, I think so too. But I guess my approach should be right and respects all of the given functions.
– mrtaurho
Jul 22 at 17:57
Thank you for your responses. A few question regarding your response: Where does the B come from in your particular solution? In what part of the RHS is B supposed to be matching. Likewise, why are there four coefficients attached to e^2t when we have repeated roots of 2 three times from the complimentary solution?
– KhanMan
Jul 22 at 17:59
Try it on the equation $y''+y=x^2$ with the guess $Y_p(x)=Ax^2+Bx+C$ and as contrast only with $Y_p(x)=Ax^2+Bx$. I cannot argue why the $C$ has to be there but you have to get it with you.
– mrtaurho
Jul 22 at 18:05
$2$ is a triple root, thus the term for $t^3e^2t$ gets an additional factor $t^3$, thus $t^3e^2t(Ct^3+Dt^2+Et+F)$.
– LutzL
Jul 22 at 18:24
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The R.H.S. consists of a polynomial, a trigonometric function and a product of an exponential with a cube. Respectively
$$t^3cdot e^2t + t + sin(4t)$$
Therefore the right attempt would be
$$Y_p(t)~=~At + B + e^2t(Ct^3 + Dt^2 +Et +F) + Gsin(4t) + Hcos(4t) $$
But I do not know a way to compute the constants without derivate eight times.
answered Jul 22 at 17:46
mrtaurho
740219
The R.H.S. consists of a polynomial, a trigonometric function and a product of an exponential with a cube. Respectively
$$t^3cdot e^2t + t + sin(4t)$$
Therefore the right attempt would be
$$Y_p(t)~=~At + B + e^2t(Ct^3 + Dt^2 +Et +F) + Gsin(4t) + Hcos(4t) $$
But I do not know a way to compute the constants without derivate eight times.
answered Jul 22 at 17:46
mrtaurho
740219
answered Jul 22 at 17:46
mrtaurho
740219
answered Jul 22 at 17:46
mrtaurho
740219
answered Jul 22 at 17:46
mrtaurho
740219
740219
Perhaps, I am in the wrong. The question asks for a "suitable form for the particular solution Y(t)" if we were to apply the method of undetermined coefficients. You are correct. I think taking eight derivatives would be an unreasonable work exercise.
– KhanMan
Jul 22 at 17:52
Yeah, I think so too. But I guess my approach should be right and respects all of the given functions.
– mrtaurho
Jul 22 at 17:57
Thank you for your responses. A few question regarding your response: Where does the B come from in your particular solution? In what part of the RHS is B supposed to be matching. Likewise, why are there four coefficients attached to e^2t when we have repeated roots of 2 three times from the complimentary solution?
– KhanMan
Jul 22 at 17:59
Try it on the equation $y''+y=x^2$ with the guess $Y_p(x)=Ax^2+Bx+C$ and as contrast only with $Y_p(x)=Ax^2+Bx$. I cannot argue why the $C$ has to be there but you have to get it with you.
– mrtaurho
Jul 22 at 18:05
$2$ is a triple root, thus the term for $t^3e^2t$ gets an additional factor $t^3$, thus $t^3e^2t(Ct^3+Dt^2+Et+F)$.
– LutzL
Jul 22 at 18:24
add a comment |Â
Perhaps, I am in the wrong. The question asks for a "suitable form for the particular solution Y(t)" if we were to apply the method of undetermined coefficients. You are correct. I think taking eight derivatives would be an unreasonable work exercise.
– KhanMan
Jul 22 at 17:52
Yeah, I think so too. But I guess my approach should be right and respects all of the given functions.
– mrtaurho
Jul 22 at 17:57
Thank you for your responses. A few question regarding your response: Where does the B come from in your particular solution? In what part of the RHS is B supposed to be matching. Likewise, why are there four coefficients attached to e^2t when we have repeated roots of 2 three times from the complimentary solution?
– KhanMan
Jul 22 at 17:59
Try it on the equation $y''+y=x^2$ with the guess $Y_p(x)=Ax^2+Bx+C$ and as contrast only with $Y_p(x)=Ax^2+Bx$. I cannot argue why the $C$ has to be there but you have to get it with you.
– mrtaurho
Jul 22 at 18:05
$2$ is a triple root, thus the term for $t^3e^2t$ gets an additional factor $t^3$, thus $t^3e^2t(Ct^3+Dt^2+Et+F)$.
– LutzL
Jul 22 at 18:24
Perhaps, I am in the wrong. The question asks for a "suitable form for the particular solution Y(t)" if we were to apply the method of undetermined coefficients. You are correct. I think taking eight derivatives would be an unreasonable work exercise.
– KhanMan
Jul 22 at 17:52
Perhaps, I am in the wrong. The question asks for a "suitable form for the particular solution Y(t)" if we were to apply the method of undetermined coefficients. You are correct. I think taking eight derivatives would be an unreasonable work exercise.
– KhanMan
Jul 22 at 17:52
Yeah, I think so too. But I guess my approach should be right and respects all of the given functions.
– mrtaurho
Jul 22 at 17:57
Yeah, I think so too. But I guess my approach should be right and respects all of the given functions.
– mrtaurho
Jul 22 at 17:57
Thank you for your responses. A few question regarding your response: Where does the B come from in your particular solution? In what part of the RHS is B supposed to be matching. Likewise, why are there four coefficients attached to e^2t when we have repeated roots of 2 three times from the complimentary solution?
– KhanMan
Jul 22 at 17:59
Thank you for your responses. A few question regarding your response: Where does the B come from in your particular solution? In what part of the RHS is B supposed to be matching. Likewise, why are there four coefficients attached to e^2t when we have repeated roots of 2 three times from the complimentary solution?
– KhanMan
Jul 22 at 17:59
Try it on the equation $y''+y=x^2$ with the guess $Y_p(x)=Ax^2+Bx+C$ and as contrast only with $Y_p(x)=Ax^2+Bx$. I cannot argue why the $C$ has to be there but you have to get it with you.
– mrtaurho
Jul 22 at 18:05
Try it on the equation $y''+y=x^2$ with the guess $Y_p(x)=Ax^2+Bx+C$ and as contrast only with $Y_p(x)=Ax^2+Bx$. I cannot argue why the $C$ has to be there but you have to get it with you.
– mrtaurho
Jul 22 at 18:05
$2$ is a triple root, thus the term for $t^3e^2t$ gets an additional factor $t^3$, thus $t^3e^2t(Ct^3+Dt^2+Et+F)$.
– LutzL
Jul 22 at 18:24
$2$ is a triple root, thus the term for $t^3e^2t$ gets an additional factor $t^3$, thus $t^3e^2t(Ct^3+Dt^2+Et+F)$.
– LutzL
Jul 22 at 18:24
add a comment |Â
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Note that the RHS contains $t^colorred3 e^2t$.
– Math Lover
Jul 22 at 17:39
I guess you are missing the fact that the exponential is multiplied by $t^3$ and not by $t$ only. And as far as I remember this is important for the choose of the particular solution.
– mrtaurho
Jul 22 at 17:39