Author Topic: TT2--P3  (Read 3452 times)

Victor Ivrii

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TT2--P3
« on: March 21, 2018, 02:57:33 PM »
a. Find the general solution of
$$
\mathbf{x}'=\begin{pmatrix} 0 &1\\
2 &-1\end{pmatrix}\mathbf{x}
$$
and sketch trajectories.

b. Solve
$$
\mathbf{x}'=\begin{pmatrix} 0 &1\\
2 &-1\end{pmatrix}\mathbf{x}+
\begin{pmatrix} \frac{e^{2t }}{e^t+1} \\
\frac{e^{2t }}{e^t+1}\end{pmatrix},\qquad
\mathbf{x}(0)=\begin{pmatrix} 3 \\
0\end{pmatrix}.
$$

Meng Wu

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Re: TT2--P3
« Reply #1 on: March 21, 2018, 11:48:11 PM »
$\underline{\text{Solution:}}$$\\$
Part(a) $\\$
Find eigenvalues by $\det(A-\lambda I_2)=0$:
$$\begin{array}{|c c|}-\lambda&1\\2&-1-\lambda\end{array}=0 \implies \lambda^2+\lambda-2=0=(\lambda-1)(\lambda+2)=0 \implies \cases{\lambda_1=1\\ \lambda_2=-2}$$
Find eigenvectors by $(A-\lambda I_2)\textbf{x}=\boldsymbol 0$: $\\$
When $\lambda=1,$ $\\$
$$\begin{pmatrix}-1&1\\2&-2\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
$$\implies\begin{pmatrix}-1&1\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
$$\implies \begin{pmatrix}x_1\\x_2\end{pmatrix}=x_2\begin{pmatrix}1\\1\end{pmatrix}$$
Thus, the eigenvector $\boldsymbol{\xi}^{(1)}=\begin{pmatrix}1\\1\end{pmatrix}$, $\textbf{x}^{(1)}(t)=\begin{pmatrix}1\\1\end{pmatrix}e^t$. $\\$
When $\lambda=-2,$ $\\$
$$\begin{pmatrix}2&1\\2&1\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
$$\implies\begin{pmatrix}2&1\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
$$\implies \begin{pmatrix}x_1\\x_2\end{pmatrix}=x_2\begin{pmatrix}1\\-2\end{pmatrix}$$
Thus, the eigenvector $\boldsymbol{\xi}^{(2)}=\begin{pmatrix}1\\-2\end{pmatrix}$, $\textbf{x}^{(2)}(t)=\begin{pmatrix}1\\-2\end{pmatrix}e^{-2t}$.$\\$
Therefore, the general solution of given system is $$\textbf{x}(t)=c_1\begin{pmatrix}1\\1\end{pmatrix}e^t+c_2\begin{pmatrix}1\\-2\end{pmatrix}e^{-2t}$$

Meng Wu

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Re: TT2--P3
« Reply #2 on: March 21, 2018, 11:48:29 PM »
Part(b) $\\$
Now consider the non-homogeneous system: $\\$
First calculate the Wronskain $$W[\textbf{x}^{(1)},\textbf{x}^{(2)}](t)=\begin{array}{|c c|}e^t&e^{-2t}\\e^t&-2e^{-2t}\end{array}=-3e^{-t}\neq 0$$
Thus, $\textbf{x}^{(1)}(t)$ and $\textbf{x}^{(2)}(t)$ form a fundamental set of solutions. $\\$
Hence the fundamental matrix $$\boldsymbol\Psi(t)=\begin{pmatrix}e^t&e^{-2t}\\e^t&-2e^{-2t}\end{pmatrix}$$
Since the general solution for the non-homogeneous is:
$$\textbf{x}(t)=\boldsymbol{\Psi}(t)\boldsymbol{c}+\boldsymbol{\Psi}(t)\int_{t_0}^{t}\boldsymbol{\Psi}^{-1}(s)\boldsymbol{g}(s)ds$$
Using Quick Formula from linear algebra:
Let $$\boldsymbol{\Psi}(t)=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
$$\begin{align}\boldsymbol{\Psi}^{-1}(t)&={1\over \det(\boldsymbol{\Psi}(t))}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\\&=-{1\over 3}e^t\begin{pmatrix}-2e^{-2t}&-e^{-2t}\\-e^{t}&e^{t}\end{pmatrix}\\&=\begin{pmatrix}{2\over3}e^{-t}&{1\over3}e^{-t}\\{1\over3}e^{2t}&-{1\over3}e^{2t}\end{pmatrix}\end{align}$$
Hence, $$\begin{align}\boldsymbol{\Psi}^{-1}(t)\boldsymbol{g}(t)=\begin{pmatrix}{2\over3}e^{-t}&{1\over3}e^{-t}\\{1\over3}e^{2t}&-{1\over3}e^{2t}\end{pmatrix}\begin{pmatrix}{e^{2t}\over e^t+1}\\{e^{2t}\over e^t+1}\end{pmatrix}=\begin{pmatrix}{e^{t}\over e^t+1}\\0\end{pmatrix}\end{align}$$
Thus, $$\int_{t_0}^{t}\boldsymbol{\Psi}^{-1}(s)\boldsymbol{g}(s)ds=\int_{t_0}^{t}\begin{pmatrix}{e^{s}\over e^s+1}\\0\end{pmatrix}ds=\begin{pmatrix}\ln(e^t+1)\\k\end{pmatrix}$$
where $k$ is any arbitrary constant. $\\$Thus,
$$\begin{align}\boldsymbol{\Psi}(t)\int_{t_0}^{t}\boldsymbol{\Psi}^{-1}(s)\boldsymbol{g}(s)ds&=\begin{pmatrix}e^t&e^{-2t}\\e^t&-2e^{-2t}\end{pmatrix}\begin{pmatrix}{e^{s}\over e^s+1}\\0\end{pmatrix}\\&=\ln(e^t+1)\begin{pmatrix}e^t\\e^t\end{pmatrix}+k\begin{pmatrix}e^{-2t}\\-2e^{-2t}\end{pmatrix}\end{align}$$
For conveniently chosen $t_0=t$, we have $$\begin{align}\textbf{c}&=\boldsymbol{\Psi}^{-1}(t_0)\textbf{x}^{0}\\&=\begin{pmatrix}{2\over3}&{1\over3}\\{1\over3}&-{1\over3}\end{pmatrix}\begin{pmatrix}3\\0\end{pmatrix}\\&=\begin{pmatrix}2\\1\end{pmatrix}\implies \cases{c_1=2\\c_2=1}\end{align} $$
Therefore, the general solution for IVP is $$\textbf{x}(t)=\begin{pmatrix}e^t&e^{-2t}\\e^t&-2e^{-2t}\end{pmatrix}\begin{pmatrix}2\\1\end{pmatrix}+\ln(e^t+1)\begin{pmatrix}e^t\\e^t\end{pmatrix}+k\begin{pmatrix}e^{-2t}\\-2e^{-2t}\end{pmatrix}$$
where $k$ is any arbitrary constant.$\\$
If let $k=1$, we have $$\textbf{x}(t)=\begin{pmatrix}e^t&e^{-2t}\\e^t&-2e^{-2t}\end{pmatrix}\begin{pmatrix}2\\1\end{pmatrix}+\ln(e^t+1)\begin{pmatrix}e^t\\e^t\end{pmatrix}+\begin{pmatrix}e^{-2t}\\-2e^{-2t}\end{pmatrix}$$
« Last Edit: March 22, 2018, 08:58:05 AM by Meng Wu »

Meng Wu

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Re: TT2--P3
« Reply #3 on: March 21, 2018, 11:49:31 PM »
I honestly don't think I get part(b) right, correct me pelase  :-\

Jared Jubas-Malz

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Re: TT2--P3
« Reply #4 on: March 22, 2018, 01:18:14 AM »
I got a different answer for part (b). Aside from a missing constant, I got everything the same up until $(6)$ so I will start from there. Rewriting $(6)$ I got:
$$\textbf{x}(t)=(ln(e^t+1)+c_{1})e^t\begin{pmatrix}1\\1\end{pmatrix}+c_{2}e^{-2t}\begin{pmatrix}1\\-2\end{pmatrix}=ln(e^t+1)e^t\begin{pmatrix}1\\1\end{pmatrix}+c_{1}e^t\begin{pmatrix}1\\1\end{pmatrix}+c_{2}e^{-2t}\begin{pmatrix}1\\-2\end{pmatrix}$$
Plugging in the initial condition:
$$\textbf{x}(0)=\begin{pmatrix}3\\0\end{pmatrix}$$
This gives:
$$3=ln(2)+c_{1}+c_{2} \implies c_{1}=3-ln(2)-c_{2}\qquad(7)$$
and:
$$0=ln(2)+c_{1}-2c_{2}\qquad(8 )$$
Plugging $(7)$ into $(8 )$ gives $c_{2}=1$. Plugging this back into (7) gives $c_{1}=2-ln(2)$. Therefore the constants are:
$$c_{1}=2-ln(2)\quad and \quad c_{2}=1$$
Therefore the general solution would be:
$$\textbf{x}(t)=ln(e^t+1)e^t\begin{pmatrix}1\\1\end{pmatrix}+(2-ln(2))e^t\begin{pmatrix}1\\1\end{pmatrix}+e^{-2t}\begin{pmatrix}1\\-2\end{pmatrix}$$
« Last Edit: March 22, 2018, 01:27:35 AM by Jared Jubas-Malz »

Victor Ivrii

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Re: TT2--P3
« Reply #5 on: March 24, 2018, 10:35:13 AM »
One does not need to find fundamental matrix, it adds complexity to particular problem