Author Topic: Problem 3 (main sitting)  (Read 1233 times)

Victor Ivrii

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Problem 3 (main sitting)
« on: November 19, 2019, 04:21:01 AM »
(a) Find the general solution of
$$
\mathbf{x}'=\begin{pmatrix} 1 &1\\
-2 &4\end{pmatrix}\mathbf{x}
$$
classify fixed point $(0,0)$ and sketch trajectories.

(b) Find the general solution
$$
\mathbf{x}'=\begin{pmatrix} 1 &1\\
-2 &4\end{pmatrix}\mathbf{x}+
\begin{pmatrix} \dfrac{e^{4t }}{e^{2t}+1} \\
0\end{pmatrix}.
$$

Yiheng Bian

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Re: Problem 3 (main sitting)
« Reply #1 on: November 19, 2019, 04:29:59 AM »
(a):
We solve homo firstly:
$$
det(A-{\lambda}I)=0\\

\begin{vmatrix}
1-\lambda & 1  \\
-2 & 4-\lambda
\end{vmatrix}=-5\lambda+{\lambda}^2+6=0\\
(\lambda-2)(\lambda-3)=0\\
\lambda_1=2,\lambda_2=3
$$
Then:
$$
(A-{\lambda}I)x=0
$$
$$
When \lambda=2
$$
$$
\begin{pmatrix}
-1 & 1 \\
-2 & 2
\end{pmatrix}
\quad= \begin{pmatrix}
0  \\
0
\end{pmatrix}
\quad
$$
RREF:
$$
\begin{pmatrix}
-1 & 1 \\
 0 & 0
\end{pmatrix}
\quad= \begin{pmatrix}
0  \\
0
\end{pmatrix}
\quad
$$
$$
\text{Let x_2=t, so }x_1=x_2=t\\
t*\begin{pmatrix}
1 \\
1
\end{pmatrix}
\quad
$$
$$
When \lambda=3
$$
$$
\begin{pmatrix}
-2 & 1 \\
-2 & 1
\end{pmatrix}
\quad= \begin{pmatrix}
0  \\
0
\end{pmatrix}
\quad
$$
RREF:
$$
\begin{pmatrix}
 2 & -1 \\
 0 & 0
\end{pmatrix}
\quad= \begin{pmatrix}
0  \\
0
\end{pmatrix}
\quad
$$
$$
\text{Let x_2=t, so }x_1=0.5t, x_2=t\\
t*\begin{pmatrix}
1 \\
2
\end{pmatrix}
\quad
$$
So, the general solution is :
$$
y= c_1e^{2t}\begin{pmatrix}
1 \\
1
\end{pmatrix}
\quad +c_2e^{3t}\begin{pmatrix}
1 \\
2
\end{pmatrix}
\quad
$$




(b):
$$
\phi = \begin{pmatrix}
e^{2t} & e^{3t} \\
e^{2t} & 2e^{3t}
\end{pmatrix}
\quad
$$
$$
\phi * u' = g(t)
$$
$$
\begin{pmatrix}
e^{2t} & e^{3t} \\
e^{2t} & 2e^{3t}
\end{pmatrix}
\quad *{\begin{pmatrix}
u_1'  \\
u_2'
\end{pmatrix}
\quad}=\begin{pmatrix}
\frac{e^{4t}}{e^{2t} + 1 }  \\
0
\end{pmatrix}
\quad
$$
Simplify we can get:
$$
u_1'=-\frac{-e^t}{e^{2t} + 1}\\
u_2'=\frac{2e^{2t}}{e^{2t}+1}
$$
Therefore:
$$
u_1=ln(e^{2t}+1)+c_1\\
u_2=-arctane^t+c_2
$$
Finally:
$$
x=\phi * u=(ln(e^{2t}+1)+c_1)*{\begin{pmatrix}
e^{2t}  \\
e^{2t}
\end{pmatrix}
\quad} +(-arctane^t+c_2)*\begin{pmatrix}
e^{3t}  \\
2e^{3t}
\end{pmatrix}
\quad
$$
OK, except LaTeX sucks:

1) * IS NOT a sign of multiplication
2)  "operators" should be escaped: \cos, \sin, \tan, \ln
« Last Edit: November 24, 2019, 09:32:25 AM by Victor Ivrii »

xuanzhong

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Re: Problem 3 (main sitting)
« Reply #2 on: November 19, 2019, 05:19:08 AM »
Here's the solution including sketching.

Aparna

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Re: Problem 3 (main sitting)
« Reply #3 on: November 19, 2019, 08:45:00 AM »
Computer-generated sketch:

Sifan Shao

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Re: Problem 3 (main sitting)
« Reply #4 on: November 19, 2019, 09:23:37 AM »
antiderivative of u1': let e^t = u, so e^2t+1 = u^2+1, du = e^t dt, so the formula becomes integral( - 1/(u^2+1)du), which is -arctan(u) -> -arctan(e^t)
antiderivative of u2': let e^2t = u, so the formula becomes integral(2u/(u+1)), which is by chain rule: ln(u+1) -> ln(e^2t+1)

Mingdi Xie

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Re: Problem 3 (main sitting)
« Reply #5 on: November 19, 2019, 09:41:06 PM »
This is my solution

Mingdi Xie

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Re: Problem 3 (main sitting)
« Reply #6 on: November 19, 2019, 09:44:19 PM »
There is a small typo in Yiheng Bian's solution, $u_1'$ and $u_2'$ should be alternate.

Victor Ivrii

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Re: Problem 3 (main sitting)
« Reply #7 on: November 24, 2019, 09:41:32 AM »
What everybody is missing:
Part of the problem "classify fixed point $(0,0)$".
It is unstable node,
« Last Edit: November 24, 2019, 09:43:04 AM by Victor Ivrii »