Question

Find the general solution of x'=(2, 3, -1, -2)x+(e^t, t).

Answer 1

hey, girl, I really don't know how to solve it. @ybarrap

Answer 2

I don't really have a clue on this one, maybe it's the format. Looks like a vector, but then again not.

Answer 3

it's 2x2 matrix, 2 and 3 on the left, -1 and -2 on the right. e^t on top, t on bottom. but I know that the general solution for 2x2 matrix, I got the answer but not the other part.

Answer 4

\[x'=\begin{pmatrix}2&-1\\3&-2\end{pmatrix}x+\begin{pmatrix}e^t\\t\end{pmatrix}~~?\]
I suggest undetermined coefficients, but only because I don't remember how to use the other methods...

Do you need help with finding the eigenvalues, or the nonhomogeneous solution as well?

Answer 5

Anyway, here's the setup for the eigenvalues:
\[\begin{vmatrix}2-\lambda&-1\\3&-2-\lambda\end{vmatrix}=-(4-\lambda^2)+3=0~~\Rightarrow~~\lambda_1=1,\lambda_2=-1\]
Let \(\vec{\eta}_1\) be the associated eigenvector for \(\lambda_1\):
\[\begin{pmatrix}1&-1\\3&-3\end{pmatrix}\begin{pmatrix}\eta_1\\\eta_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\]
So you have \(\eta_1=\eta_2\). Simplest value to pick would be 1, so \(\vec{\eta}_1=\begin{pmatrix}1\\1\end{pmatrix}\).

Answer 6

Let \(\vec{\eta}_2\) be the eigenvector for \(\lambda_2\):
\[\begin{pmatrix}3&-1\\3&-1\end{pmatrix}\begin{pmatrix}\eta_1\\\eta_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\]
So you have \(3\eta_1=\eta_2\). Letting \(\eta_1=1\) gives you \(\eta_2=3\), so your other eigenvector is \(\vec{\eta}_2=\begin{pmatrix}1\\3\end{pmatrix}\).

Answer 7

That should take care of the homogeneous solution.

It's been a while since I've learned this stuff, but from what I remember, you use a guess of the form
\[\vec{x_p}=e^t\vec{A}+t\vec{B}+\vec{C}\]
or perhaps
\[\vec{x_p}=te^t\vec{A}+t\vec{B}+\vec{C}\]
or maybe even
\[\vec{x_p}=e^t\vec{A}+te^t\vec{B}+t\vec{C}+\vec{D}\]
As to which one is correct, I'm not sure.

Answer 8

As @SithsAndGiggles stated, the homogeneous part is \[x_h = C_1 e^t\left(\begin{matrix}1\\1\end {matrix}\right) + C_2e^{-t}\left(\begin{matrix}1\\3\end{matrix}\right)\]
your g(t) is \(g(t) = \left[\begin{matrix}e^t\\t\end{matrix}\right]\)
for non homogeneous part, you have to construct \(\psi (t)\) from the homogenous part , just take out the C and put e^t , e^-t inside.
\[\psi(t) = \left[\begin{matrix}e^t& e^{-t}\\e^t &3e^{-t}\end{matrix}\right]\]
\[\psi (t)^{-1}= \left[\begin{matrix}3e^{-t}&-e^{-t}\\-e^t&e^t\end{matrix}\right]\],
replace t by s, you have \(\psi(s)\) then apply formula to calculate the nonhomogeneous solution, say it is \(x_p\)
\[x_p= \psi(t)\int_0^{t }\psi (s)^{-1}g(s)ds\]

Answer 9

I calculate the integrand first, then put it under the integral, then time it with \(\psi (t)\)
integrand: \(\huge\psi (s)^{-1}*g(s) = \left[\begin{matrix}3e^{-s}&-e^{-s}\\-e^s&e^s\end{matrix}\right]\left[\begin{matrix}e^s\\s\end{matrix}\right]\)
\[=\huge\left[\begin{matrix}3- s e^{-s} \\e^s(s-e^s)\end{matrix}\right]\]

Answer 10

then, take integral of it
\[\huge \int_0^{t}\left[\begin{matrix}3-se^{-s}\\e^s(s-e^s)\end{matrix}\right]ds= \left[\begin{matrix}e^{-t}(t+1)\\te^t-\frac{e^{2t}}{2}-e^t+\frac{3}{2}\end{matrix}\right]\]

Answer 11

wow, i must love you so much to do a long thing like this , ha!!! to me, you are that special. ??? !!!

Answer 12

Now, time it with \(\psi (t)\) don't mess up the order
\[ \left[\begin{matrix}e^t& e^{-t}\\e^t &3e^{-t}\end{matrix}\right]\left[\begin{matrix}e^{-t}(t+1)\\te^t-\frac{e^{2t}}{2}-e^t+\frac{3}{2}\end{matrix}\right]\] you have the \(x_p\)
then you must add homogenous part and this part to get the final answer.

Answer 13

sorry girl, I have piano class at 12, must practice before class. cannot give you more. but I think you can finish it,right?

Answer 14

Nice job @SithsAndGiggles and @Loser66 . @Idealist, here is a nice explanation of the technique they used here: http://tutorial.math.lamar.edu/Classes/DE/NonhomogeneousSystems.aspx

Answer 15

Don't forget, once you have @SithsAndGiggles complementary solution \(x_h\) and @Loser66 particular solution \(x_p\), combine to get general solution \(x(t)=x_h+x_p\).

Answer 16

what is your question @Idealist

Answer 17

How did you get v(t)^-1?

Answer 18

it is inverse of \(\psi (t\) it's 2x2 matrix, quite easy to find it inverse, right?

Answer 19

What's the symbol v(t) called?

Answer 20

psi

Answer 21

Bye.

Answer 22

But I got a wrong answer for the inverse. How did you get yours?

Answer 23

oh yeah, you are right, I have to divide it by 1/2

Answer 24

I am sorry girl.

Answer 25

It's okay.

Answer 26

How did you multiply to g(s) and got the answer?

Answer 27

I don't get your question.

Answer 28

Like how did you multiply v(t)^-1 and g(s)

Answer 29

how to multiply a matrix with a matrix, right?

Answer 30

Yes.