Question

Need Help solve
\(\large x'=\begin{bmatrix}
-2 & 1\\
1 &-2
\end{bmatrix}x+\begin{bmatrix}
1\\ 3

\end{bmatrix}e^{-2t}\)

Answer 1

I call your function x(t) as below:
\[\left(\begin{matrix}a(t) \\ b(t)\end{matrix}\right)\]
so I can write these equations:
\[a'=-2a+b+2e ^{-2t},b'=a-2b+3e ^{-2t}\]

Answer 2

aha k i dont mind :)

Answer 3

adding to the doubled first equation the second equation, we can write:\[2a'+b'=-3a+5e ^{-2t}\]
deriving one time that equation we have:
\[2a''+b''=-3a'-10e ^{-2t}\]

Answer 4

so:
\[b''(t)=-3a'-10e ^{-2t}-2a''\]

Answer 5

now I derive the firs equation two times with respect to t, and I can get this:
\[a'''=-2a''+b''+4e _{^{-2t}}\]
inserting the expression for b'' above, we can write:
\[a'''+4a''+3a'=-10e ^{-2t}\]
which can be easily soved

Answer 6

i see ok so by solving characteristic equation
r^3+4r^2+3r=0
r(r^2+4r+3)=0
r(r+1)(r+3)=0
r=0,-1,-3

i would have homogeneous solution
\(a_H=c_1+c_2e^{-t}+c_2e^{-3t}\)

Answer 7

now i have to look for particular solution ?

Answer 8

please try with this function:
\[a _{p}(t)=A*e ^{-2t}\]
where A is a real constant

Answer 9

you should get this:
\[a _{p}=-5e ^{-2t}\]

Answer 10

@Michele_Laino shouldn't that equation in \(a\) be
\[a'''+4a''+3a'=-\color{red}6e ^{-2t}~~?\]

Answer 11

so, a(t) is:
\[a(t)=c _{1}+c _{2}e ^{-t}+c _{3}e ^{-3t}-5e ^{-2t}\]

Answer 12

I will check, please wait... @SithsAndGiggles

Answer 13

you are right! @SithsAndGiggles it is -6e^(-2t)

Answer 14

@Marki please note the post of @SithsAndGiggles on my error of computation

Answer 15

Fortunately the homogeneous solution doesn't change :)

Answer 16

that's right! @SithsAndGiggles

Answer 17

so our solution for a(t) should be this:
\[a(t)=c _{1}+c _{2}e ^{-t}+c _{3}e ^{-3t}-3e ^{-2t}\]

Answer 18

@Marki note that elimination is not the only way to obtain a solution. You can also try using undetermined coefficients and the usual eigenvector approach for the homogeneous solution. (I prefer this method because you get \(x_1=a\) and \(x_2=b\) right away.)

Answer 19

hmm
i would assume
\(a=Ae^{-2t}\)
\(a'=-2Ae^{-2t}\)
\(a''=4Ae^{-2t}\)
\(a'''=-8Ae^{-2t}\)

so
-8A+16A-6A=-6
A=-1/3

Answer 20

please, I got: -14+16=+2, so A=-3 I think

Answer 21

whereas our solution for b(t) is:
\[b(t)=-3c _{1}t+c _{2}e ^{-t}-c _{3}e ^{-3t}-e ^{-2t}\]

Answer 22

this my way to solve , but i was looking for other comfortable methods

A is Diagonalizable
eigen values and corresponding vectors

-1,(1,1)
-3,(1,-1)
and i get some direct solution with some x(t)=Ty(t)=|1 1 ;1 -1||y1;y2|

at the end i got some long solution xD
i'll post if u interested

Answer 23

ok!

Answer 24

i got this (in other method)

\(\large a(t)=-2e^{-t} (e^{-t}-1) +c_1e^{-t}-e^{-3t}(e^t-1)+c_2e^{-3t}\)


\(\large b(t)=-2e^{-t} (e^{-t}-1) +c_1e^{-t}+e^{-3t}(e^t-1)-c_2e^{-3t}\)

Answer 25

wew!
but i completely got ur method hmm
but in exam mostly i'll use this since they might give 3*3

Answer 26

thanks both @Michele_Laino @SithsAndGiggles

Answer 27

thanks! for your matrix method! @Marki

Answer 28

thanks! for your correction @SithsAndGiggles

Answer 29

Perhaps I made a mistake myself, but I'm getting a different solution than what elimination gives.
\[{\bf x}'=\begin{pmatrix}-2&1\\1&-2\end{pmatrix}{\bf x}+\begin{pmatrix}1\\3\end{pmatrix}e^{-2t}\]
The coefficient matrix has the same eigenvalues/vectors, \(\lambda_1=-3\) with \(\vec{\eta}_1=\begin{pmatrix}-1\\1\end{pmatrix}\) and \(\lambda_2=-1\) with \(\vec{\eta}_2=\begin{pmatrix}1\\1\end{pmatrix}\) (which agrees with Marki's work).

The homogeneous solution is then
\[{\bf x}_h=C_1\begin{pmatrix}-1\\1\end{pmatrix}e^{-3t}+C_2\begin{pmatrix}1\\1\end{pmatrix}e^{-t}\]
For the non-homogeneous part, we set \({\bf x}=Ae^{-2t}\), so that \({\bf x}'=-2Ae^{-2t}\), then we substitute into the original equation:
\[\begin{align*}-2Ae^{-2t}&=\begin{pmatrix}-2&1\\1&-2\end{pmatrix}Ae^{-2t}+\begin{pmatrix}1\\3\end{pmatrix}e^{-2t}\\\\
\vec{0}&=\left(\left[\begin{pmatrix}-2&1\\1&-2\end{pmatrix}-2I\right]A+\begin{pmatrix}1\\3\end{pmatrix}\right)e^{-2t}\\\\
\begin{pmatrix}-4&1\\1&-4\end{pmatrix}A&=\begin{pmatrix}-1\\-3\end{pmatrix}\\\\
A&=\frac{7}{15}\begin{pmatrix}1\\1\end{pmatrix}
\end{align*}\]
So the final solution would be (according to this work)
\[{\bf x}=C_1\begin{pmatrix}-1\\1\end{pmatrix}e^{-3t}+C_2\begin{pmatrix}1\\1\end{pmatrix}e^{-t}+\frac{7}{15}\begin{pmatrix}1\\1\end{pmatrix}e^{-2t}\]

Answer 30

Oh I think I see the mistake, I did \(\color{red}{-2IA}\) instead of \(\color{red}{+2IA}\)...

Answer 31

It should be
\[A=-\begin{pmatrix}3\\1\end{pmatrix}\]

Answer 32

i thinkits ok
method are much imp than solution

Answer 33

It helps when WA agrees :)
http://www.wolframalpha.com/input/?i=x%27%3D-2x%2By%2Be%5E%28-2t%29%2C+y%27%3Dx-2y%2B3e%5E%28-2t%29

Answer 34

to me or u ?

Answer 35

Both! It's nice knowing you have the right solution