DE Undetermined Coefficient Method












0












$begingroup$


I would like to solve the following non-homogeneous second order differential equation with constant coefficients with the method of undetermined coefficient but I have some problem with the particular solution, can someone help me? Thanks in advance!

$2y^{''}+3y^{'}+y=t^{-2}$


My problem is that I don't know how to treat the negative power. For example considering $t^{2}$, I can write $Y=At^2+Bt+C$. But in this case what I should do?






ordinary-differential-equations







share|cite|improve this question









$endgroup$












  • $begingroup$
    @Moo. Did you try it ? I am almost sure that there is a typo since the particular solution involve very special functions.
    $endgroup$
    – Claude Leibovici
    Dec 2 '18 at 5:19










  • $begingroup$
    Thanks for the comments, so with undetermined coefficient is not possible to solve?
    $endgroup$
    – Fabio Taccaliti
    Dec 2 '18 at 5:41










  • $begingroup$
    What makes you think this is possible? Undetermined coefficients, at least the version in most books, applies only to functions $g(t)$ of very special form.
    $endgroup$
    – David C. Ullrich
    Dec 2 '18 at 14:32


















0












$begingroup$


I would like to solve the following non-homogeneous second order differential equation with constant coefficients with the method of undetermined coefficient but I have some problem with the particular solution, can someone help me? Thanks in advance!

$2y^{''}+3y^{'}+y=t^{-2}$


My problem is that I don't know how to treat the negative power. For example considering $t^{2}$, I can write $Y=At^2+Bt+C$. But in this case what I should do?






ordinary-differential-equations







share|cite|improve this question









$endgroup$












  • $begingroup$
    @Moo. Did you try it ? I am almost sure that there is a typo since the particular solution involve very special functions.
    $endgroup$
    – Claude Leibovici
    Dec 2 '18 at 5:19










  • $begingroup$
    Thanks for the comments, so with undetermined coefficient is not possible to solve?
    $endgroup$
    – Fabio Taccaliti
    Dec 2 '18 at 5:41










  • $begingroup$
    What makes you think this is possible? Undetermined coefficients, at least the version in most books, applies only to functions $g(t)$ of very special form.
    $endgroup$
    – David C. Ullrich
    Dec 2 '18 at 14:32
















0












0








0





$begingroup$


I would like to solve the following non-homogeneous second order differential equation with constant coefficients with the method of undetermined coefficient but I have some problem with the particular solution, can someone help me? Thanks in advance!

$2y^{''}+3y^{'}+y=t^{-2}$


My problem is that I don't know how to treat the negative power. For example considering $t^{2}$, I can write $Y=At^2+Bt+C$. But in this case what I should do?






ordinary-differential-equations







share|cite|improve this question









$endgroup$




I would like to solve the following non-homogeneous second order differential equation with constant coefficients with the method of undetermined coefficient but I have some problem with the particular solution, can someone help me? Thanks in advance!

$2y^{''}+3y^{'}+y=t^{-2}$


My problem is that I don't know how to treat the negative power. For example considering $t^{2}$, I can write $Y=At^2+Bt+C$. But in this case what I should do?






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 2 '18 at 4:47









Fabio TaccalitiFabio Taccaliti

968




968












  • $begingroup$
    @Moo. Did you try it ? I am almost sure that there is a typo since the particular solution involve very special functions.
    $endgroup$
    – Claude Leibovici
    Dec 2 '18 at 5:19










  • $begingroup$
    Thanks for the comments, so with undetermined coefficient is not possible to solve?
    $endgroup$
    – Fabio Taccaliti
    Dec 2 '18 at 5:41










  • $begingroup$
    What makes you think this is possible? Undetermined coefficients, at least the version in most books, applies only to functions $g(t)$ of very special form.
    $endgroup$
    – David C. Ullrich
    Dec 2 '18 at 14:32




















  • $begingroup$
    @Moo. Did you try it ? I am almost sure that there is a typo since the particular solution involve very special functions.
    $endgroup$
    – Claude Leibovici
    Dec 2 '18 at 5:19










  • $begingroup$
    Thanks for the comments, so with undetermined coefficient is not possible to solve?
    $endgroup$
    – Fabio Taccaliti
    Dec 2 '18 at 5:41










  • $begingroup$
    What makes you think this is possible? Undetermined coefficients, at least the version in most books, applies only to functions $g(t)$ of very special form.
    $endgroup$
    – David C. Ullrich
    Dec 2 '18 at 14:32


















$begingroup$
@Moo. Did you try it ? I am almost sure that there is a typo since the particular solution involve very special functions.
$endgroup$
– Claude Leibovici
Dec 2 '18 at 5:19




$begingroup$
@Moo. Did you try it ? I am almost sure that there is a typo since the particular solution involve very special functions.
$endgroup$
– Claude Leibovici
Dec 2 '18 at 5:19












$begingroup$
Thanks for the comments, so with undetermined coefficient is not possible to solve?
$endgroup$
– Fabio Taccaliti
Dec 2 '18 at 5:41




$begingroup$
Thanks for the comments, so with undetermined coefficient is not possible to solve?
$endgroup$
– Fabio Taccaliti
Dec 2 '18 at 5:41












$begingroup$
What makes you think this is possible? Undetermined coefficients, at least the version in most books, applies only to functions $g(t)$ of very special form.
$endgroup$
– David C. Ullrich
Dec 2 '18 at 14:32






$begingroup$
What makes you think this is possible? Undetermined coefficients, at least the version in most books, applies only to functions $g(t)$ of very special form.
$endgroup$
– David C. Ullrich
Dec 2 '18 at 14:32












1 Answer
1






active

oldest

votes


















2












$begingroup$

First, the characteristic polynomial



$$ 2r^2 + 3r + 1 = (2r+1)(r+1) = 0 implies r = -1, -1/2 $$



So the fundamental solution is
$$ y_h(t) = c_1 e^{-t} + c_2e^{-t/2} $$




Now, you probably won't be able to use undetermined coefficients to obtain the particular solution, but variation of parameters can be of use. Let



$$ y_p(t) = u(t)e^{-t} + v(t)e^{-t/2} $$



Then



begin{align}
e^{-t}u' + e^{-t/2}v' &= 0 \
-e^{-t}u' - frac12 e^{-t/2}v' &= frac12 t^{-2}
end{align}



$$ implies u' = -t^{-2}e^t , quad v' = t^{-2}e^{t/2} $$



The integrals are non-elementary, but can be expressed in terms of incomplete Gamma functions






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot, now it's clear!
    $endgroup$
    – Fabio Taccaliti
    Dec 2 '18 at 19:17











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

First, the characteristic polynomial



$$ 2r^2 + 3r + 1 = (2r+1)(r+1) = 0 implies r = -1, -1/2 $$



So the fundamental solution is
$$ y_h(t) = c_1 e^{-t} + c_2e^{-t/2} $$




Now, you probably won't be able to use undetermined coefficients to obtain the particular solution, but variation of parameters can be of use. Let



$$ y_p(t) = u(t)e^{-t} + v(t)e^{-t/2} $$



Then



begin{align}
e^{-t}u' + e^{-t/2}v' &= 0 \
-e^{-t}u' - frac12 e^{-t/2}v' &= frac12 t^{-2}
end{align}



$$ implies u' = -t^{-2}e^t , quad v' = t^{-2}e^{t/2} $$



The integrals are non-elementary, but can be expressed in terms of incomplete Gamma functions






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot, now it's clear!
    $endgroup$
    – Fabio Taccaliti
    Dec 2 '18 at 19:17
















2












$begingroup$

First, the characteristic polynomial



$$ 2r^2 + 3r + 1 = (2r+1)(r+1) = 0 implies r = -1, -1/2 $$



So the fundamental solution is
$$ y_h(t) = c_1 e^{-t} + c_2e^{-t/2} $$




Now, you probably won't be able to use undetermined coefficients to obtain the particular solution, but variation of parameters can be of use. Let



$$ y_p(t) = u(t)e^{-t} + v(t)e^{-t/2} $$



Then



begin{align}
e^{-t}u' + e^{-t/2}v' &= 0 \
-e^{-t}u' - frac12 e^{-t/2}v' &= frac12 t^{-2}
end{align}



$$ implies u' = -t^{-2}e^t , quad v' = t^{-2}e^{t/2} $$



The integrals are non-elementary, but can be expressed in terms of incomplete Gamma functions






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks a lot, now it's clear!
    $endgroup$
    – Fabio Taccaliti
    Dec 2 '18 at 19:17














2












2








2





$begingroup$

First, the characteristic polynomial



$$ 2r^2 + 3r + 1 = (2r+1)(r+1) = 0 implies r = -1, -1/2 $$



So the fundamental solution is
$$ y_h(t) = c_1 e^{-t} + c_2e^{-t/2} $$




Now, you probably won't be able to use undetermined coefficients to obtain the particular solution, but variation of parameters can be of use. Let



$$ y_p(t) = u(t)e^{-t} + v(t)e^{-t/2} $$



Then



begin{align}
e^{-t}u' + e^{-t/2}v' &= 0 \
-e^{-t}u' - frac12 e^{-t/2}v' &= frac12 t^{-2}
end{align}



$$ implies u' = -t^{-2}e^t , quad v' = t^{-2}e^{t/2} $$



The integrals are non-elementary, but can be expressed in terms of incomplete Gamma functions






share|cite|improve this answer











$endgroup$



First, the characteristic polynomial



$$ 2r^2 + 3r + 1 = (2r+1)(r+1) = 0 implies r = -1, -1/2 $$



So the fundamental solution is
$$ y_h(t) = c_1 e^{-t} + c_2e^{-t/2} $$




Now, you probably won't be able to use undetermined coefficients to obtain the particular solution, but variation of parameters can be of use. Let



$$ y_p(t) = u(t)e^{-t} + v(t)e^{-t/2} $$



Then



begin{align}
e^{-t}u' + e^{-t/2}v' &= 0 \
-e^{-t}u' - frac12 e^{-t/2}v' &= frac12 t^{-2}
end{align}



$$ implies u' = -t^{-2}e^t , quad v' = t^{-2}e^{t/2} $$



The integrals are non-elementary, but can be expressed in terms of incomplete Gamma functions







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 '18 at 6:27

























answered Dec 2 '18 at 15:51









DylanDylan

12.5k31026




12.5k31026












  • $begingroup$
    Thanks a lot, now it's clear!
    $endgroup$
    – Fabio Taccaliti
    Dec 2 '18 at 19:17


















  • $begingroup$
    Thanks a lot, now it's clear!
    $endgroup$
    – Fabio Taccaliti
    Dec 2 '18 at 19:17
















$begingroup$
Thanks a lot, now it's clear!
$endgroup$
– Fabio Taccaliti
Dec 2 '18 at 19:17




$begingroup$
Thanks a lot, now it's clear!
$endgroup$
– Fabio Taccaliti
Dec 2 '18 at 19:17


















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