Author Topic: TT2--P3M  (Read 4108 times)

Victor Ivrii

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TT2--P3M
« on: March 21, 2018, 03:05:44 PM »
a. Find the general solution of
$$
\mathbf{x}'=
\begin{pmatrix} -1 &1\\
0 &-3\end{pmatrix}\mathbf{x}
$$
and sketch trajectories.
b. Solve
$$
\mathbf{x}'=\begin{pmatrix} -1 &1\\
0 &-3\end{pmatrix}\mathbf{x}+
\begin{pmatrix} \frac{12}{e^t+1} \\
\frac{12}{e^t+1}\end{pmatrix},\qquad
\mathbf{x}(0)=\begin{pmatrix} 1-3\ln 2 \\
-3-3\ln2\end{pmatrix}.
$$

Syed Hasnain

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Re: TT2--P3M
« Reply #1 on: March 26, 2018, 01:56:21 AM »
I have attached my solution,
Let me know about the mistakes...
thanks

Victor Ivrii

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Re: TT2--P3M
« Reply #2 on: March 27, 2018, 08:12:43 AM »
The answer is very long, not simplified and thus I have no time to check it.

Meng Wu

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Re: TT2--P3M
« Reply #3 on: March 28, 2018, 12:03:48 AM »
$\underline{\textbf{Solution:}}$ $\\$
$\textbf{(a)}$ $\\$
Find eigenvalues by $\det(A-\lambda I_2)=0$:
$$\begin{array}{|c c|}-1-\lambda&1\\0&-3-\lambda\end{array}=0 \implies \lambda^2+4\lambda+3=0\implies \cases{\lambda_1=-1\\ \lambda_2=-3}$$
Find eigenvectors by $(A-\lambda I_2)\mathbf{x}=\boldsymbol 0$: $\\$
When $\lambda=-1$, eigenvector $\boldsymbol{\xi}^{(1)}=\begin{pmatrix}1\\0\end{pmatrix}$. $\\$
When $\lambda=-3$, eigenvector $\boldsymbol{\xi}^{(2)}=\begin{pmatrix}1\\-2\end{pmatrix}$. $\\$
(Few steps omitted, since "Syed_Hasnain" got part(a) right already.)$\\$
Therefore, the general solution is $$\mathbf{x}(t)=c_1\begin{pmatrix}1\\0\end{pmatrix}e^{-t}+c_2\begin{pmatrix}1\\-2\end{pmatrix}e^{-3t}$$
$\textbf{(b)}$ $\\$
$$W[\mathbf{x}^{(1)},\mathbf{x}^{(2)}](t)=\begin{array}{|c c|}e^{-t}&e^{-3t}\\0&-2e^{-3t}\end{array}=-2e^{-4t}\neq 0$$
Thus, $\mathbf{x}^{(1)}(t)$ and $\mathbf{x}^{(2)}(t)$ form a fundamental set of solutions. $\\$
Hence the fundamental matrix $$\boldsymbol\Psi(t)=\begin{pmatrix}e^{-t}&e^{-3t}\\0&-2e^{-3t}\end{pmatrix}$$
For the non-homogeneous system, we have the general solution$$\mathbf{x}=\boldsymbol\Psi(t)\mathbf{u}(t)$$
Since we know $$\boldsymbol\Psi(t)\mathbf{u}'(t)=\mathbf{g}(t)$$
$$\begin{pmatrix}e^{-t}&e^{-3t}\\0&-2e^{-3t}\end{pmatrix}\begin{pmatrix}u_1'\\u_2'\end{pmatrix}=\begin{pmatrix}{12\over e^t+1}\\{12\over e^t+1}\end{pmatrix}$$
By row reduction: $$\begin{pmatrix}e^{-t}&0\\0&-2e^{-3t}\end{pmatrix}\begin{pmatrix}u_1'\\u_2'\end{pmatrix}=\begin{pmatrix}{18\over e^t+1}\\{12\over e^t+1}\end{pmatrix}$$
Hence $$\cases{u_1'={18e^{t}\over e^t+1}\\u_2'={-6e^{t}\over e^t+1}} \implies \cases{u_1(t)=\int{{18e^{t}\over e^t+1}dt}=18\ln(e^t+1)+c_1\\u_2(t)=\int{{-6e^{t}\over e^t+1}dt}=-6\ln(e^t+1)-3e^{-2t}+6e^t+c_2}$$
Thus $$\begin{align}\mathbf{x}&=\boldsymbol\Psi(t)\mathbf{u}(t)\\&=c_1\begin{pmatrix}1\\0\end{pmatrix}e^{-t}+c_2\begin{pmatrix}1\\-2\end{pmatrix}e^{-3t}+\begin{pmatrix}-3\\6\end{pmatrix}e^{-5t}+\begin{pmatrix}6\\-12\end{pmatrix}e^{-2t}+\ln(e^t+1)\begin{pmatrix}-6\\12\end{pmatrix}e^{-3t}+\ln(e^t+1)\begin{pmatrix}18\\0\end{pmatrix}e^{-t}\end{align}$$
Apply the given initial condition:$$\mathbf{x}(0)=\begin{pmatrix}12\ln2+c_1+c_2+3\\12\ln2-6-2c_2\end{pmatrix}=\begin{pmatrix}1-3\ln2\\-3-3\ln2\end{pmatrix}\implies \cases{c_1=-{1\over2}-{45\over2}\ln2\\c_2=-{3\over2}+{15\over2}\ln2}$$
Therefore, the general solution for the IVP:
$$\mathbf{x}=(-{1\over2}-{45\over2}\ln2)\begin{pmatrix}1\\0\end{pmatrix}e^{-t}+(-{3\over2}+{15\over2}\ln2)\begin{pmatrix}1\\-2\end{pmatrix}e^{-3t}+\begin{pmatrix}-3\\6\end{pmatrix}e^{-5t}+\begin{pmatrix}6\\-12\end{pmatrix}e^{-2t}+\ln(e^t+1)\begin{pmatrix}-6\\12\end{pmatrix}e^{-3t}+\ln(e^t+1)\begin{pmatrix}18\\0\end{pmatrix}e^{-t}$$

Jared Jubas-Malz

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Re: TT2--P3M
« Reply #4 on: March 28, 2018, 01:26:54 AM »
I got something different for part (b). I got that $u_{2}'$ should be:
$$\begin{align}u_{2}'=\frac{-6e^{3t}}{e^t+1}\end{align}$$
Therefore $u_{2}$ would be:
$$\begin{align}u_{2}(t) = -3e^{2t}+6e^t-6\ln(e^t+1)+c_{2}\end{align}$$
Then the general solution would be:
$$\begin{align}\textbf{x}(t)=(18\ln(e^t+1)+c_{1})e^{-t}\begin{pmatrix}1\\0\end{pmatrix}+(-3e^{2t}+6e^t-6\ln(e^t+1)+c_{2})e^{-3t}\begin{pmatrix}1\\-2\end{pmatrix}\end{align}$$
Subbing in the initial condition $\textbf{x}(0)=\begin{pmatrix}1-3\ln2\\-3-3\ln2\end{pmatrix}$:
$$\begin{align}1-3\ln2=c_{1}+3+c_{2}\end{align}$$
$$\begin{align}-3-3\ln2=-6-2c_{2}\end{align}$$
Solving these equations for the constants gives:
$$\begin{align}c_{1}=\frac{-19-9\ln2}{2}\end{align}$$
$$\begin{align}c_{2}=\frac{-3+3\ln2}{2}\end{align}$$
Therefore the general solution for the IVP is:
$$\begin{align}\textbf{x}(t)=\frac{-19-9\ln2}{2}e^{-t}\begin{pmatrix}1\\0\end{pmatrix}+\frac{-3+3\ln2}{2}e^{-3t}\begin{pmatrix}1\\-2\end{pmatrix}+18\ln(e^t+1)e^{-t}\begin{pmatrix}1\\0\end{pmatrix}-3e^{-t}\begin{pmatrix}1\\-2\end{pmatrix}+6e^{-2t}\begin{pmatrix}1\\-2\end{pmatrix}-6\ln(e^t+1)e^{-3t}\begin{pmatrix}1\\-2\end{pmatrix}
\end{align}$$
« Last Edit: March 28, 2018, 01:29:45 AM by Jared Jubas-Malz »

Meng Wu

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Re: TT2--P3M
« Reply #5 on: March 28, 2018, 09:11:15 AM »
Thanks for the correction. I made a mistake copying the result integral.
I got the same result.
I think there is a calculation error when computing the initial condition, hence the $c_1,c_2$ values. (if I am not mistaken :-\)
$$c_1+c_2=-2-15\ln2$$ and $$-2c_2=3-15\ln2$$
Therefore, $$c_1=-{1\over2}-{45\over2}\ln2$$
$$c_2=-{3\over2}+{15\over2}\ln2$$
« Last Edit: March 28, 2018, 09:15:25 AM by Meng Wu »

Jared Jubas-Malz

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Re: TT2--P3M
« Reply #6 on: March 28, 2018, 10:43:24 AM »
I found my mistake. I think you're correct.