Elementary ode non-homegeneous
$begingroup$
We would like to find a particular solution for $y''-y'-2y=2e^{-t}$.
The homogeneous equations associated has the solutions given by $psi_1(t)=e^{2t}$ and $psi_2(t)=e^{-t}$. Thus the Wronskian between $psi_{1}$, and $psi_{2}$ is given by $W=-3e^{t}$.
Now, by the variation of the parameters formula the particular solution by the ode is $x_{p}(t)=psi_{1}(t)alpha_{1}(t)+psi_{2}(t)alpha_{2}(t)$, where $alpha'=(-fpsi_{2})/W$ and $alpha'=(fpsi_{1})/W$, where $f=2e^{-t}$, then $alpha_{1}=(-2e^{-3t})/9$ and $alpha_{2}=(-2t)/3$
In short $x_p(t)=(-2e^{-t})/9 - (2te^{-t}/3)$, but the correct is $x_p(t)=(-2te^{-t})/3$. Is where my mistake?
ordinary-differential-equations
edited Dec 1 '18 at 16:51
user10354138
7,3772925
asked Dec 1 '18 at 16:45
C. JuniorC. Junior
670411
$endgroup$
add a comment |
$begingroup$
We would like to find a particular solution for $y''-y'-2y=2e^{-t}$.
The homogeneous equations associated has the solutions given by $psi_1(t)=e^{2t}$ and $psi_2(t)=e^{-t}$. Thus the Wronskian between $psi_{1}$, and $psi_{2}$ is given by $W=-3e^{t}$.
Now, by the variation of the parameters formula the particular solution by the ode is $x_{p}(t)=psi_{1}(t)alpha_{1}(t)+psi_{2}(t)alpha_{2}(t)$, where $alpha'=(-fpsi_{2})/W$ and $alpha'=(fpsi_{1})/W$, where $f=2e^{-t}$, then $alpha_{1}=(-2e^{-3t})/9$ and $alpha_{2}=(-2t)/3$
In short $x_p(t)=(-2e^{-t})/9 - (2te^{-t}/3)$, but the correct is $x_p(t)=(-2te^{-t})/3$. Is where my mistake?
ordinary-differential-equations
edited Dec 1 '18 at 16:51
user10354138
7,3772925
asked Dec 1 '18 at 16:45
C. JuniorC. Junior
670411
$endgroup$
3
$begingroup$
There are no mistakes. Particular solutions are only determined up to solution of the homogeneous equation, and in this case the difference $-2e^{-t}/9$ is a multiple of $psi_2$.
$endgroup$
– user10354138
Dec 1 '18 at 16:53
add a comment |
$begingroup$
We would like to find a particular solution for $y''-y'-2y=2e^{-t}$.
The homogeneous equations associated has the solutions given by $psi_1(t)=e^{2t}$ and $psi_2(t)=e^{-t}$. Thus the Wronskian between $psi_{1}$, and $psi_{2}$ is given by $W=-3e^{t}$.
Now, by the variation of the parameters formula the particular solution by the ode is $x_{p}(t)=psi_{1}(t)alpha_{1}(t)+psi_{2}(t)alpha_{2}(t)$, where $alpha'=(-fpsi_{2})/W$ and $alpha'=(fpsi_{1})/W$, where $f=2e^{-t}$, then $alpha_{1}=(-2e^{-3t})/9$ and $alpha_{2}=(-2t)/3$
In short $x_p(t)=(-2e^{-t})/9 - (2te^{-t}/3)$, but the correct is $x_p(t)=(-2te^{-t})/3$. Is where my mistake?
ordinary-differential-equations
edited Dec 1 '18 at 16:51
user10354138
7,3772925
asked Dec 1 '18 at 16:45
C. JuniorC. Junior
670411
$endgroup$
We would like to find a particular solution for $y''-y'-2y=2e^{-t}$.
The homogeneous equations associated has the solutions given by $psi_1(t)=e^{2t}$ and $psi_2(t)=e^{-t}$. Thus the Wronskian between $psi_{1}$, and $psi_{2}$ is given by $W=-3e^{t}$.
Now, by the variation of the parameters formula the particular solution by the ode is $x_{p}(t)=psi_{1}(t)alpha_{1}(t)+psi_{2}(t)alpha_{2}(t)$, where $alpha'=(-fpsi_{2})/W$ and $alpha'=(fpsi_{1})/W$, where $f=2e^{-t}$, then $alpha_{1}=(-2e^{-3t})/9$ and $alpha_{2}=(-2t)/3$
In short $x_p(t)=(-2e^{-t})/9 - (2te^{-t}/3)$, but the correct is $x_p(t)=(-2te^{-t})/3$. Is where my mistake?
ordinary-differential-equations
ordinary-differential-equations
edited Dec 1 '18 at 16:51
user10354138
7,3772925
asked Dec 1 '18 at 16:45
C. JuniorC. Junior
670411
edited Dec 1 '18 at 16:51
user10354138
7,3772925
asked Dec 1 '18 at 16:45
C. JuniorC. Junior
670411
edited Dec 1 '18 at 16:51
user10354138
7,3772925
edited Dec 1 '18 at 16:51
user10354138
7,3772925
edited Dec 1 '18 at 16:51
user10354138
7,3772925
7,3772925
asked Dec 1 '18 at 16:45
C. JuniorC. Junior
670411
asked Dec 1 '18 at 16:45
C. JuniorC. Junior
670411
asked Dec 1 '18 at 16:45
C. JuniorC. Junior
670411
670411
3
$begingroup$
There are no mistakes. Particular solutions are only determined up to solution of the homogeneous equation, and in this case the difference $-2e^{-t}/9$ is a multiple of $psi_2$.
$endgroup$
– user10354138
Dec 1 '18 at 16:53
add a comment |
3
$begingroup$
There are no mistakes. Particular solutions are only determined up to solution of the homogeneous equation, and in this case the difference $-2e^{-t}/9$ is a multiple of $psi_2$.
$endgroup$
– user10354138
Dec 1 '18 at 16:53
3
3
$begingroup$
There are no mistakes. Particular solutions are only determined up to solution of the homogeneous equation, and in this case the difference $-2e^{-t}/9$ is a multiple of $psi_2$.
$endgroup$
– user10354138
Dec 1 '18 at 16:53
$begingroup$
There are no mistakes. Particular solutions are only determined up to solution of the homogeneous equation, and in this case the difference $-2e^{-t}/9$ is a multiple of $psi_2$.
$endgroup$
– user10354138
Dec 1 '18 at 16:53
add a comment |
0
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$begingroup$
There are no mistakes. Particular solutions are only determined up to solution of the homogeneous equation, and in this case the difference $-2e^{-t}/9$ is a multiple of $psi_2$.
$endgroup$
– user10354138
Dec 1 '18 at 16:53