Finding the roots of a characteristic polynomial
up vote
2
down vote
favorite
Main aim is to find the lowest order equation with the solution:
$$y(x)= 2 cosh(x) + 3 e^{-2x} sin(x)$$
Now, I am trying to find the roots to form the characteristic polynomial from which I get the lowest order equation.
However, am stuck with the second expression as the first can be easily expressed as $e^x - e^{-x}$ so I deduce $lambda_{1}=1,lambda_{1} = -1 $ but the other expression I am not quite sure whether it is $-2pm i $ or something else as there is an exponential and a trigonometric function at the same time ?
Any advice greatly appreciated!
differential-equations reduction-of-order-ode
edited Nov 15 at 19:49
Leucippus
19.6k102870
asked Nov 15 at 18:46
JKM
5415
add a comment |
up vote
2
down vote
favorite
Main aim is to find the lowest order equation with the solution:
$$y(x)= 2 cosh(x) + 3 e^{-2x} sin(x)$$
Now, I am trying to find the roots to form the characteristic polynomial from which I get the lowest order equation.
However, am stuck with the second expression as the first can be easily expressed as $e^x - e^{-x}$ so I deduce $lambda_{1}=1,lambda_{1} = -1 $ but the other expression I am not quite sure whether it is $-2pm i $ or something else as there is an exponential and a trigonometric function at the same time ?
Any advice greatly appreciated!
differential-equations reduction-of-order-ode
edited Nov 15 at 19:49
Leucippus
19.6k102870
asked Nov 15 at 18:46
JKM
5415
3
Can you try with $sin x=dfrac{e^{ix}-e^{-ix}}{2i}$.
– Yadati Kiran
Nov 15 at 18:48
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Main aim is to find the lowest order equation with the solution:
$$y(x)= 2 cosh(x) + 3 e^{-2x} sin(x)$$
Now, I am trying to find the roots to form the characteristic polynomial from which I get the lowest order equation.
However, am stuck with the second expression as the first can be easily expressed as $e^x - e^{-x}$ so I deduce $lambda_{1}=1,lambda_{1} = -1 $ but the other expression I am not quite sure whether it is $-2pm i $ or something else as there is an exponential and a trigonometric function at the same time ?
Any advice greatly appreciated!
differential-equations reduction-of-order-ode
edited Nov 15 at 19:49
Leucippus
19.6k102870
asked Nov 15 at 18:46
JKM
5415
Main aim is to find the lowest order equation with the solution:
$$y(x)= 2 cosh(x) + 3 e^{-2x} sin(x)$$
Now, I am trying to find the roots to form the characteristic polynomial from which I get the lowest order equation.
However, am stuck with the second expression as the first can be easily expressed as $e^x - e^{-x}$ so I deduce $lambda_{1}=1,lambda_{1} = -1 $ but the other expression I am not quite sure whether it is $-2pm i $ or something else as there is an exponential and a trigonometric function at the same time ?
Any advice greatly appreciated!
differential-equations reduction-of-order-ode
differential-equations reduction-of-order-ode
edited Nov 15 at 19:49
Leucippus
19.6k102870
asked Nov 15 at 18:46
JKM
5415
edited Nov 15 at 19:49
Leucippus
19.6k102870
asked Nov 15 at 18:46
JKM
5415
edited Nov 15 at 19:49
Leucippus
19.6k102870
edited Nov 15 at 19:49
Leucippus
19.6k102870
edited Nov 15 at 19:49
Leucippus
19.6k102870
19.6k102870
asked Nov 15 at 18:46
JKM
5415
asked Nov 15 at 18:46
JKM
5415
asked Nov 15 at 18:46
JKM
5415
5415
3
Can you try with $sin x=dfrac{e^{ix}-e^{-ix}}{2i}$.
– Yadati Kiran
Nov 15 at 18:48
add a comment |
3
Can you try with $sin x=dfrac{e^{ix}-e^{-ix}}{2i}$.
– Yadati Kiran
Nov 15 at 18:48
3
3
Can you try with $sin x=dfrac{e^{ix}-e^{-ix}}{2i}$.
– Yadati Kiran
Nov 15 at 18:48
Can you try with $sin x=dfrac{e^{ix}-e^{-ix}}{2i}$.
– Yadati Kiran
Nov 15 at 18:48
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
$$2ie^{-2x}sin x=e^{-2x}(e^{ix}-e^{-ix})=e^{(-2+i)x}-e^{(-2-i)x}.$$
These are indeed complex exponentials.
answered Nov 15 at 20:23
Yves Daoust
122k668217
add a comment |
up vote
2
down vote
$$ (lambda -1)(lambda +1)(lambda +2-i)(lambda +2+i)$$
$$= (lambda ^2 -1)((lambda+2)^2+1))$$
$$=(lambda ^2 -1)(lambda^2+4lambda +5)$$
Multiply out and get your equation out of it.
answered Nov 15 at 20:28
Mohammad Riazi-Kermani
40.3k41958
The OP is not asking the polynomial but how to handle the second term.
– Yves Daoust
Nov 15 at 20:35
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$$2ie^{-2x}sin x=e^{-2x}(e^{ix}-e^{-ix})=e^{(-2+i)x}-e^{(-2-i)x}.$$
These are indeed complex exponentials.
answered Nov 15 at 20:23
Yves Daoust
122k668217
add a comment |
up vote
1
down vote
accepted
$$2ie^{-2x}sin x=e^{-2x}(e^{ix}-e^{-ix})=e^{(-2+i)x}-e^{(-2-i)x}.$$
These are indeed complex exponentials.
answered Nov 15 at 20:23
Yves Daoust
122k668217
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$$2ie^{-2x}sin x=e^{-2x}(e^{ix}-e^{-ix})=e^{(-2+i)x}-e^{(-2-i)x}.$$
These are indeed complex exponentials.
answered Nov 15 at 20:23
Yves Daoust
122k668217
$$2ie^{-2x}sin x=e^{-2x}(e^{ix}-e^{-ix})=e^{(-2+i)x}-e^{(-2-i)x}.$$
These are indeed complex exponentials.
answered Nov 15 at 20:23
Yves Daoust
122k668217
edited Nov 15 at 20:41
answered Nov 15 at 20:23
Yves Daoust
122k668217
answered Nov 15 at 20:23
Yves Daoust
122k668217
answered Nov 15 at 20:23
Yves Daoust
122k668217
122k668217
add a comment |
add a comment |
up vote
2
down vote
$$ (lambda -1)(lambda +1)(lambda +2-i)(lambda +2+i)$$
$$= (lambda ^2 -1)((lambda+2)^2+1))$$
$$=(lambda ^2 -1)(lambda^2+4lambda +5)$$
Multiply out and get your equation out of it.
answered Nov 15 at 20:28
Mohammad Riazi-Kermani
40.3k41958
The OP is not asking the polynomial but how to handle the second term.
– Yves Daoust
Nov 15 at 20:35
add a comment |
up vote
2
down vote
$$ (lambda -1)(lambda +1)(lambda +2-i)(lambda +2+i)$$
$$= (lambda ^2 -1)((lambda+2)^2+1))$$
$$=(lambda ^2 -1)(lambda^2+4lambda +5)$$
Multiply out and get your equation out of it.
answered Nov 15 at 20:28
Mohammad Riazi-Kermani
40.3k41958
The OP is not asking the polynomial but how to handle the second term.
– Yves Daoust
Nov 15 at 20:35
add a comment |
up vote
2
down vote
up vote
2
down vote
$$ (lambda -1)(lambda +1)(lambda +2-i)(lambda +2+i)$$
$$= (lambda ^2 -1)((lambda+2)^2+1))$$
$$=(lambda ^2 -1)(lambda^2+4lambda +5)$$
Multiply out and get your equation out of it.
answered Nov 15 at 20:28
Mohammad Riazi-Kermani
40.3k41958
$$ (lambda -1)(lambda +1)(lambda +2-i)(lambda +2+i)$$
$$= (lambda ^2 -1)((lambda+2)^2+1))$$
$$=(lambda ^2 -1)(lambda^2+4lambda +5)$$
Multiply out and get your equation out of it.
answered Nov 15 at 20:28
Mohammad Riazi-Kermani
40.3k41958
answered Nov 15 at 20:28
Mohammad Riazi-Kermani
40.3k41958
answered Nov 15 at 20:28
Mohammad Riazi-Kermani
40.3k41958
answered Nov 15 at 20:28
Mohammad Riazi-Kermani
40.3k41958
40.3k41958
The OP is not asking the polynomial but how to handle the second term.
– Yves Daoust
Nov 15 at 20:35
add a comment |
The OP is not asking the polynomial but how to handle the second term.
– Yves Daoust
Nov 15 at 20:35
The OP is not asking the polynomial but how to handle the second term.
– Yves Daoust
Nov 15 at 20:35
The OP is not asking the polynomial but how to handle the second term.
– Yves Daoust
Nov 15 at 20:35
add a comment |
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3
Can you try with $sin x=dfrac{e^{ix}-e^{-ix}}{2i}$.
– Yadati Kiran
Nov 15 at 18:48