Stuck while solving systems of linear differential equations
I want to solve the following differential equation:
$frac{dx}{dt}=-3x+y-8 \ frac{dy}{dt}=x-3y+8. $
I first rewrote the equations using the differential operator $D=frac{d}{dt}$ and eliminated the variable $y$.
begin{equation}
(D+3)x-y=-8, \x+(D+3)y=8.
end{equation}
therefore
$ (D^2+6D+8)x=-8D(1)-16$
which is equivalent to
begin{equation}
frac{d^2x}{dt^2}+6frac{dx}{dt}+8x=-8frac{d(1)}{dt}-16.
end{equation}
But the equation is not homogeneous! How would you solve it? Any help regardless of its form or length will be highly appreciated.
differential-equations
asked Dec 2 at 2:13
Ko Byeongmin
1326
add a comment |
I want to solve the following differential equation:
$frac{dx}{dt}=-3x+y-8 \ frac{dy}{dt}=x-3y+8. $
I first rewrote the equations using the differential operator $D=frac{d}{dt}$ and eliminated the variable $y$.
begin{equation}
(D+3)x-y=-8, \x+(D+3)y=8.
end{equation}
therefore
$ (D^2+6D+8)x=-8D(1)-16$
which is equivalent to
begin{equation}
frac{d^2x}{dt^2}+6frac{dx}{dt}+8x=-8frac{d(1)}{dt}-16.
end{equation}
But the equation is not homogeneous! How would you solve it? Any help regardless of its form or length will be highly appreciated.
differential-equations
asked Dec 2 at 2:13
Ko Byeongmin
1326
add a comment |
I want to solve the following differential equation:
$frac{dx}{dt}=-3x+y-8 \ frac{dy}{dt}=x-3y+8. $
I first rewrote the equations using the differential operator $D=frac{d}{dt}$ and eliminated the variable $y$.
begin{equation}
(D+3)x-y=-8, \x+(D+3)y=8.
end{equation}
therefore
$ (D^2+6D+8)x=-8D(1)-16$
which is equivalent to
begin{equation}
frac{d^2x}{dt^2}+6frac{dx}{dt}+8x=-8frac{d(1)}{dt}-16.
end{equation}
But the equation is not homogeneous! How would you solve it? Any help regardless of its form or length will be highly appreciated.
differential-equations
asked Dec 2 at 2:13
Ko Byeongmin
1326
I want to solve the following differential equation:
$frac{dx}{dt}=-3x+y-8 \ frac{dy}{dt}=x-3y+8. $
I first rewrote the equations using the differential operator $D=frac{d}{dt}$ and eliminated the variable $y$.
begin{equation}
(D+3)x-y=-8, \x+(D+3)y=8.
end{equation}
therefore
$ (D^2+6D+8)x=-8D(1)-16$
which is equivalent to
begin{equation}
frac{d^2x}{dt^2}+6frac{dx}{dt}+8x=-8frac{d(1)}{dt}-16.
end{equation}
But the equation is not homogeneous! How would you solve it? Any help regardless of its form or length will be highly appreciated.
differential-equations
differential-equations
asked Dec 2 at 2:13
Ko Byeongmin
1326
asked Dec 2 at 2:13
Ko Byeongmin
1326
asked Dec 2 at 2:13
Ko Byeongmin
1326
asked Dec 2 at 2:13
Ko Byeongmin
1326
asked Dec 2 at 2:13
Ko Byeongmin
1326
1326
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Given
$$dfrac{dx}{dt}=-3x+y-8 \ dfrac{dy}{dt}=x-3y+8$$
From the first equation we have
$$y = x' + 3x + 8 implies y' = x'' + 3 x'$$
Substituting into the second equation, we have
$$x'' + 3 x = x - 3(x' + 3x + 8) + 8$$
Simplifying
$$x'' + 6 x' + 8 x + 16 = 0$$
This results in
$$x(t) = c_1 e^{-4 t}+c_2 e^{-2 t}-2$$
You can use this result to find $y(t)$.
There are many other ways to solve these types of problems, but I used your approach.
answered Dec 2 at 2:33
Moo
5,53131020
Thanks a lot!! I wish I had more time to study the theory of ODEs...
– Ko Byeongmin
Dec 2 at 2:36
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Given
$$dfrac{dx}{dt}=-3x+y-8 \ dfrac{dy}{dt}=x-3y+8$$
From the first equation we have
$$y = x' + 3x + 8 implies y' = x'' + 3 x'$$
Substituting into the second equation, we have
$$x'' + 3 x = x - 3(x' + 3x + 8) + 8$$
Simplifying
$$x'' + 6 x' + 8 x + 16 = 0$$
This results in
$$x(t) = c_1 e^{-4 t}+c_2 e^{-2 t}-2$$
You can use this result to find $y(t)$.
There are many other ways to solve these types of problems, but I used your approach.
answered Dec 2 at 2:33
Moo
5,53131020
Thanks a lot!! I wish I had more time to study the theory of ODEs...
– Ko Byeongmin
Dec 2 at 2:36
add a comment |
Given
$$dfrac{dx}{dt}=-3x+y-8 \ dfrac{dy}{dt}=x-3y+8$$
From the first equation we have
$$y = x' + 3x + 8 implies y' = x'' + 3 x'$$
Substituting into the second equation, we have
$$x'' + 3 x = x - 3(x' + 3x + 8) + 8$$
Simplifying
$$x'' + 6 x' + 8 x + 16 = 0$$
This results in
$$x(t) = c_1 e^{-4 t}+c_2 e^{-2 t}-2$$
You can use this result to find $y(t)$.
There are many other ways to solve these types of problems, but I used your approach.
answered Dec 2 at 2:33
Moo
5,53131020
Thanks a lot!! I wish I had more time to study the theory of ODEs...
– Ko Byeongmin
Dec 2 at 2:36
add a comment |
Given
$$dfrac{dx}{dt}=-3x+y-8 \ dfrac{dy}{dt}=x-3y+8$$
From the first equation we have
$$y = x' + 3x + 8 implies y' = x'' + 3 x'$$
Substituting into the second equation, we have
$$x'' + 3 x = x - 3(x' + 3x + 8) + 8$$
Simplifying
$$x'' + 6 x' + 8 x + 16 = 0$$
This results in
$$x(t) = c_1 e^{-4 t}+c_2 e^{-2 t}-2$$
You can use this result to find $y(t)$.
There are many other ways to solve these types of problems, but I used your approach.
answered Dec 2 at 2:33
Moo
5,53131020
Given
$$dfrac{dx}{dt}=-3x+y-8 \ dfrac{dy}{dt}=x-3y+8$$
From the first equation we have
$$y = x' + 3x + 8 implies y' = x'' + 3 x'$$
Substituting into the second equation, we have
$$x'' + 3 x = x - 3(x' + 3x + 8) + 8$$
Simplifying
$$x'' + 6 x' + 8 x + 16 = 0$$
This results in
$$x(t) = c_1 e^{-4 t}+c_2 e^{-2 t}-2$$
You can use this result to find $y(t)$.
There are many other ways to solve these types of problems, but I used your approach.
answered Dec 2 at 2:33
Moo
5,53131020
edited Dec 2 at 2:37
answered Dec 2 at 2:33
Moo
5,53131020
answered Dec 2 at 2:33
Moo
5,53131020
answered Dec 2 at 2:33
Moo
5,53131020
5,53131020
Thanks a lot!! I wish I had more time to study the theory of ODEs...
– Ko Byeongmin
Dec 2 at 2:36
add a comment |
Thanks a lot!! I wish I had more time to study the theory of ODEs...
– Ko Byeongmin
Dec 2 at 2:36
Thanks a lot!! I wish I had more time to study the theory of ODEs...
– Ko Byeongmin
Dec 2 at 2:36
Thanks a lot!! I wish I had more time to study the theory of ODEs...
– Ko Byeongmin
Dec 2 at 2:36
add a comment |
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