Mixing
Find the general solution to the given equation
$begingroup$
Find the general solution to the given equation
$$3x{"} - 4x{'} +x = e^t +e^{-t}$$
I know how to solve a given task using the Variation of Parameters method, however I would also like to know how to do that without it.
We have:
$$y = y_{h} + y_{p}$$
So finding the general solution to the homogenous equation is quite easy.
$$x_h = c_1e^{x/3} + c_2e^x$$
Now I have a problem with finding the particular solution to the non homogenous equation.
My first guess was to use $x = Ae^t + Be^{-t}$ and then using it in $3x{''}-4x'+x = e^t+e^{-t}$, but it does not seem to work (I don't really know why)
So my question is, what should be my $x$?
calculus ordinary-differential-equations
asked Jan 26 at 22:56
James SmithJames Smith
34817
$endgroup$
add a comment |
$begingroup$
Find the general solution to the given equation
$$3x{"} - 4x{'} +x = e^t +e^{-t}$$
I know how to solve a given task using the Variation of Parameters method, however I would also like to know how to do that without it.
We have:
$$y = y_{h} + y_{p}$$
So finding the general solution to the homogenous equation is quite easy.
$$x_h = c_1e^{x/3} + c_2e^x$$
Now I have a problem with finding the particular solution to the non homogenous equation.
My first guess was to use $x = Ae^t + Be^{-t}$ and then using it in $3x{''}-4x'+x = e^t+e^{-t}$, but it does not seem to work (I don't really know why)
So my question is, what should be my $x$?
calculus ordinary-differential-equations
asked Jan 26 at 22:56
James SmithJames Smith
34817
$endgroup$
$begingroup$
Try $x(t) = (c_1+c_2 t)e^t+(c_3+c_4 t)e^{-t}$
$endgroup$
– Cesareo
Jan 26 at 23:40
add a comment |
$begingroup$
Find the general solution to the given equation
$$3x{"} - 4x{'} +x = e^t +e^{-t}$$
I know how to solve a given task using the Variation of Parameters method, however I would also like to know how to do that without it.
We have:
$$y = y_{h} + y_{p}$$
So finding the general solution to the homogenous equation is quite easy.
$$x_h = c_1e^{x/3} + c_2e^x$$
Now I have a problem with finding the particular solution to the non homogenous equation.
My first guess was to use $x = Ae^t + Be^{-t}$ and then using it in $3x{''}-4x'+x = e^t+e^{-t}$, but it does not seem to work (I don't really know why)
So my question is, what should be my $x$?
calculus ordinary-differential-equations
asked Jan 26 at 22:56
James SmithJames Smith
34817
$endgroup$
Find the general solution to the given equation
$$3x{"} - 4x{'} +x = e^t +e^{-t}$$
I know how to solve a given task using the Variation of Parameters method, however I would also like to know how to do that without it.
We have:
$$y = y_{h} + y_{p}$$
So finding the general solution to the homogenous equation is quite easy.
$$x_h = c_1e^{x/3} + c_2e^x$$
Now I have a problem with finding the particular solution to the non homogenous equation.
My first guess was to use $x = Ae^t + Be^{-t}$ and then using it in $3x{''}-4x'+x = e^t+e^{-t}$, but it does not seem to work (I don't really know why)
So my question is, what should be my $x$?
calculus ordinary-differential-equations
calculus ordinary-differential-equations
asked Jan 26 at 22:56
James SmithJames Smith
34817
asked Jan 26 at 22:56
James SmithJames Smith
34817
asked Jan 26 at 22:56
James SmithJames Smith
34817
asked Jan 26 at 22:56
James SmithJames Smith
34817
asked Jan 26 at 22:56
James SmithJames Smith
34817
34817
$begingroup$
Try $x(t) = (c_1+c_2 t)e^t+(c_3+c_4 t)e^{-t}$
$endgroup$
– Cesareo
Jan 26 at 23:40
add a comment |
$begingroup$
Try $x(t) = (c_1+c_2 t)e^t+(c_3+c_4 t)e^{-t}$
$endgroup$
– Cesareo
Jan 26 at 23:40
$begingroup$
Try $x(t) = (c_1+c_2 t)e^t+(c_3+c_4 t)e^{-t}$
$endgroup$
– Cesareo
Jan 26 at 23:40
$begingroup$
Try $x(t) = (c_1+c_2 t)e^t+(c_3+c_4 t)e^{-t}$
$endgroup$
– Cesareo
Jan 26 at 23:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that $e^t$ is already a homogeneous solution, that's why you can't use it as a particular guess. Try
$$ x_p(t) = Ate^t + Be^{-t} $$
answered Jan 27 at 4:10
DylanDylan
14.1k31127
$endgroup$
$begingroup$
Thank you. Good to know this fact.
$endgroup$
– James Smith
Jan 27 at 11:54
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that $e^t$ is already a homogeneous solution, that's why you can't use it as a particular guess. Try
$$ x_p(t) = Ate^t + Be^{-t} $$
answered Jan 27 at 4:10
DylanDylan
14.1k31127
$endgroup$
$begingroup$
Thank you. Good to know this fact.
$endgroup$
– James Smith
Jan 27 at 11:54
add a comment |
$begingroup$
Note that $e^t$ is already a homogeneous solution, that's why you can't use it as a particular guess. Try
$$ x_p(t) = Ate^t + Be^{-t} $$
answered Jan 27 at 4:10
DylanDylan
14.1k31127
$endgroup$
$begingroup$
Thank you. Good to know this fact.
$endgroup$
– James Smith
Jan 27 at 11:54
add a comment |
$begingroup$
Note that $e^t$ is already a homogeneous solution, that's why you can't use it as a particular guess. Try
$$ x_p(t) = Ate^t + Be^{-t} $$
answered Jan 27 at 4:10
DylanDylan
14.1k31127
$endgroup$
Note that $e^t$ is already a homogeneous solution, that's why you can't use it as a particular guess. Try
$$ x_p(t) = Ate^t + Be^{-t} $$
answered Jan 27 at 4:10
DylanDylan
14.1k31127
answered Jan 27 at 4:10
DylanDylan
14.1k31127
answered Jan 27 at 4:10
DylanDylan
14.1k31127
answered Jan 27 at 4:10
DylanDylan
14.1k31127
14.1k31127
$begingroup$
Thank you. Good to know this fact.
$endgroup$
– James Smith
Jan 27 at 11:54
add a comment |
$begingroup$
Thank you. Good to know this fact.
$endgroup$
– James Smith
Jan 27 at 11:54
$begingroup$
Thank you. Good to know this fact.
$endgroup$
– James Smith
Jan 27 at 11:54
$begingroup$
Thank you. Good to know this fact.
$endgroup$
– James Smith
Jan 27 at 11:54
add a comment |
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$begingroup$
Try $x(t) = (c_1+c_2 t)e^t+(c_3+c_4 t)e^{-t}$
$endgroup$
– Cesareo
Jan 26 at 23:40