Transient Terms in a General Solution
$begingroup$
Find the general solution of the given differential equation:
$$ (x^2-4)left(frac{dy}{dx}right) +4y = (x+2)^2 $$
I found the general solution of the D.E and I got the following correct solution:
$$ y = (x+C)left(frac{x+2}{x-2}right) $$
I know that this is the correct solution and the next part of the question says to determine whether there are any transient terms in the general solution. (Enter the transient terms as a comma-separated list; if there are none, enter NONE.)
The definition of a transient term from my understanding is a term that approaches zero as x goes to infinity.
Ive tried many answers and do not know how to find the transient terms. I thought there were none in this solution.
Can anyone guide me in understanding the problem?
Thanks.
ordinary-differential-equations
edited Dec 7 '18 at 12:19
Larry
2,1762828
asked Feb 9 '16 at 3:58
RichardRichard
6112
$endgroup$
add a comment |
$begingroup$
Find the general solution of the given differential equation:
$$ (x^2-4)left(frac{dy}{dx}right) +4y = (x+2)^2 $$
I found the general solution of the D.E and I got the following correct solution:
$$ y = (x+C)left(frac{x+2}{x-2}right) $$
I know that this is the correct solution and the next part of the question says to determine whether there are any transient terms in the general solution. (Enter the transient terms as a comma-separated list; if there are none, enter NONE.)
The definition of a transient term from my understanding is a term that approaches zero as x goes to infinity.
Ive tried many answers and do not know how to find the transient terms. I thought there were none in this solution.
Can anyone guide me in understanding the problem?
Thanks.
ordinary-differential-equations
edited Dec 7 '18 at 12:19
Larry
2,1762828
asked Feb 9 '16 at 3:58
RichardRichard
6112
$endgroup$
$begingroup$
$$lim_{x to infty} frac{1}{x-2} to 0$$ if that helps.
$endgroup$
– Mattos
Feb 9 '16 at 4:28
$begingroup$
Hello, I am from 2018
$endgroup$
– Aayush Paurana
Jul 5 '18 at 6:45
add a comment |
$begingroup$
Find the general solution of the given differential equation:
$$ (x^2-4)left(frac{dy}{dx}right) +4y = (x+2)^2 $$
I found the general solution of the D.E and I got the following correct solution:
$$ y = (x+C)left(frac{x+2}{x-2}right) $$
I know that this is the correct solution and the next part of the question says to determine whether there are any transient terms in the general solution. (Enter the transient terms as a comma-separated list; if there are none, enter NONE.)
The definition of a transient term from my understanding is a term that approaches zero as x goes to infinity.
Ive tried many answers and do not know how to find the transient terms. I thought there were none in this solution.
Can anyone guide me in understanding the problem?
Thanks.
ordinary-differential-equations
edited Dec 7 '18 at 12:19
Larry
2,1762828
asked Feb 9 '16 at 3:58
RichardRichard
6112
$endgroup$
Find the general solution of the given differential equation:
$$ (x^2-4)left(frac{dy}{dx}right) +4y = (x+2)^2 $$
I found the general solution of the D.E and I got the following correct solution:
$$ y = (x+C)left(frac{x+2}{x-2}right) $$
I know that this is the correct solution and the next part of the question says to determine whether there are any transient terms in the general solution. (Enter the transient terms as a comma-separated list; if there are none, enter NONE.)
The definition of a transient term from my understanding is a term that approaches zero as x goes to infinity.
Ive tried many answers and do not know how to find the transient terms. I thought there were none in this solution.
Can anyone guide me in understanding the problem?
Thanks.
ordinary-differential-equations
ordinary-differential-equations
edited Dec 7 '18 at 12:19
Larry
2,1762828
asked Feb 9 '16 at 3:58
RichardRichard
6112
edited Dec 7 '18 at 12:19
Larry
2,1762828
asked Feb 9 '16 at 3:58
RichardRichard
6112
edited Dec 7 '18 at 12:19
Larry
2,1762828
edited Dec 7 '18 at 12:19
Larry
2,1762828
edited Dec 7 '18 at 12:19
Larry
2,1762828
2,1762828
asked Feb 9 '16 at 3:58
RichardRichard
6112
asked Feb 9 '16 at 3:58
RichardRichard
6112
asked Feb 9 '16 at 3:58
RichardRichard
6112
6112
$begingroup$
$$lim_{x to infty} frac{1}{x-2} to 0$$ if that helps.
$endgroup$
– Mattos
Feb 9 '16 at 4:28
$begingroup$
Hello, I am from 2018
$endgroup$
– Aayush Paurana
Jul 5 '18 at 6:45
add a comment |
$begingroup$
$$lim_{x to infty} frac{1}{x-2} to 0$$ if that helps.
$endgroup$
– Mattos
Feb 9 '16 at 4:28
$begingroup$
Hello, I am from 2018
$endgroup$
– Aayush Paurana
Jul 5 '18 at 6:45
$begingroup$
$$lim_{x to infty} frac{1}{x-2} to 0$$ if that helps.
$endgroup$
– Mattos
Feb 9 '16 at 4:28
$begingroup$
$$lim_{x to infty} frac{1}{x-2} to 0$$ if that helps.
$endgroup$
– Mattos
Feb 9 '16 at 4:28
$begingroup$
Hello, I am from 2018
$endgroup$
– Aayush Paurana
Jul 5 '18 at 6:45
$begingroup$
Hello, I am from 2018
$endgroup$
– Aayush Paurana
Jul 5 '18 at 6:45
add a comment |
1 Answer
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The transient term of the ODE is a term such that upon taking the limit as t→infinity the term tends to 0. A non transient term that tends to a constant is called a steady state term.
answered Mar 12 '18 at 17:44
Hassan ShahzadHassan Shahzad
1
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add a comment |
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1 Answer
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$begingroup$
The transient term of the ODE is a term such that upon taking the limit as t→infinity the term tends to 0. A non transient term that tends to a constant is called a steady state term.
answered Mar 12 '18 at 17:44
Hassan ShahzadHassan Shahzad
1
$endgroup$
add a comment |
$begingroup$
The transient term of the ODE is a term such that upon taking the limit as t→infinity the term tends to 0. A non transient term that tends to a constant is called a steady state term.
answered Mar 12 '18 at 17:44
Hassan ShahzadHassan Shahzad
1
$endgroup$
add a comment |
$begingroup$
The transient term of the ODE is a term such that upon taking the limit as t→infinity the term tends to 0. A non transient term that tends to a constant is called a steady state term.
answered Mar 12 '18 at 17:44
Hassan ShahzadHassan Shahzad
1
$endgroup$
The transient term of the ODE is a term such that upon taking the limit as t→infinity the term tends to 0. A non transient term that tends to a constant is called a steady state term.
answered Mar 12 '18 at 17:44
Hassan ShahzadHassan Shahzad
1
answered Mar 12 '18 at 17:44
Hassan ShahzadHassan Shahzad
1
answered Mar 12 '18 at 17:44
Hassan ShahzadHassan Shahzad
1
answered Mar 12 '18 at 17:44
Hassan ShahzadHassan Shahzad
1
1
add a comment |
add a comment |
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$begingroup$
$$lim_{x to infty} frac{1}{x-2} to 0$$ if that helps.
$endgroup$
– Mattos
Feb 9 '16 at 4:28
$begingroup$
Hello, I am from 2018
$endgroup$
– Aayush Paurana
Jul 5 '18 at 6:45