Find the general solution to the given equation












0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    Try $x(t) = (c_1+c_2 t)e^t+(c_3+c_4 t)e^{-t}$
    $endgroup$
    – Cesareo
    Jan 26 at 23:40
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    Try $x(t) = (c_1+c_2 t)e^t+(c_3+c_4 t)e^{-t}$
    $endgroup$
    – Cesareo
    Jan 26 at 23:40














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















1












$begingroup$

Note that $e^t$ is already a homogeneous solution, that's why you can't use it as a particular guess. Try



$$ x_p(t) = Ate^t + Be^{-t} $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. Good to know this fact.
    $endgroup$
    – James Smith
    Jan 27 at 11:54











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Note that $e^t$ is already a homogeneous solution, that's why you can't use it as a particular guess. Try



$$ x_p(t) = Ate^t + Be^{-t} $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. Good to know this fact.
    $endgroup$
    – James Smith
    Jan 27 at 11:54
















1












$begingroup$

Note that $e^t$ is already a homogeneous solution, that's why you can't use it as a particular guess. Try



$$ x_p(t) = Ate^t + Be^{-t} $$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you. Good to know this fact.
    $endgroup$
    – James Smith
    Jan 27 at 11:54














1












1








1





$begingroup$

Note that $e^t$ is already a homogeneous solution, that's why you can't use it as a particular guess. Try



$$ x_p(t) = Ate^t + Be^{-t} $$






share|cite|improve this answer









$endgroup$



Note that $e^t$ is already a homogeneous solution, that's why you can't use it as a particular guess. Try



$$ x_p(t) = Ate^t + Be^{-t} $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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