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### Messages - Meng Wu

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31
##### Term Test 2 / Re: TT2--P3
« 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 $$\textbf{x}(t)=\begin{pmatrix}e^t&e^{-2t}\\e^t&-2e^{-2t}\end{pmatrix}\begin{pmatrix}2\\1\end{pmatrix}+\ln(e^t+1)\begin{pmatrix}e^t\\e^t\end{pmatrix}+\begin{pmatrix}e^{-2t}\\-2e^{-2t}\end{pmatrix}$$

32
##### Term Test 2 / Re: TT2--P3
« 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}$$33 ##### Term Test 2 / Re: TT2--P1M « on: March 21, 2018, 08:14:49 PM » \underline{\text{Solution:}}$$\\$
Part(a) $\\$
First consider homogeneous equation: $$y''+4y=0$$
characteristic equation: $$r^2+4=0 \implies \cases{r_1=2i\\r_2=-2i}$$
Thus, the complementary solution $$y_c(t)=c_1\cos 2t+c_2\sin 2t$$
Now consider the nonhomogeneous equation $$y''+y=8\cos^{-1}(t)$$
Since $y_1(t)=\cos 2t$ and $y_2(t)=\sin 2t$, Wronskain $$W=[y_1,y_2](t)=\begin{array}{|c c|}\cos 2t&\sin 2t\\-2\sin 2t&2\cos 2t\end{array}=2\cos^22t+2\sin^22t=2 \neq0$$
Therefore, $y_1(t)$ and $y_2(t)$ form a fundamental set of solutions.$\\$
Use the Method of Variation of Parameters: $\\$
The particular solution \begin{align}Y(t)&=-y_1(t)\int_{t_0}^{t}{y_2(s)g(s)\over W[y_1,y_2](s)}ds+y_2(t)\int_{t_0}^{t}{y_1(s)g(s)\over W[y_1,y_2](s)}ds\\&=-\cos(2t)\int_{t_0}^{t}{\sin(2s)\cdot {8\cos^{-1}(s)}\over 2}ds+\sin(2t)\int_{t_0}^{t}{\cos(2s)\cdot {8\cos^{-1}(s)}\over 2}ds\\&=-\cos(2t)\int_{t_0}^{t}{2\sin(s)\cos(s)\cdot {8\over \cos(s)}\over 2}ds+\sin(2t)\int_{t_0}^{t}{(2\cos^2(s)-1)\cdot {8\cos^{-1}(s)}\over 2}ds\\&=-8\cos(2t)\int_{t_0}^{t}\sin(s)ds+4\sin(2t)\int_{t_0}^{t}[2\cos(s)-\sec(s)]\\&=8\cos(2t)\cos(t)+4\sin(2t)[2\sin(t)-\ln(\sec(t)+\tan(t))]\end{align}
Therefore, the general solution \begin{align}y(t)&=y_c(t)+Y(t)\\&=c_1\cos(2t)+c_2\sin(2t)+8\cos(2t)\cos(t)+4\sin(2t)[2\sin(t)-\ln(\sec(t)+\tan(t))]\end{align}
Part(b)$\\$
\begin{align}y(0)&=c_1\cos(0)+c_2\sin(0)+8\cos(0)\cos(0)+4\sin(0)[2\sin(0)-\ln(\sec(0)+\tan(0))]\\&=c_1+8=0 \implies c_1=-8\end{align}
$$y'(t)=-2c_1\sin(2t)+2c_2\cos(2t)-16\sin(2t)\cos(t)-8\cos(2t)\sin(t)+8\cos(2t)[2\sin(t)-\ln(\sec(t)+\tan(t)]+4\sin(2t)[2\cos(t)-{\sec(t)\tan(t)+\sec^2(t)\over \sec(t)+\tan(t)}]$$
$$y'(0)=2c_2=0 \implies c_2=0$$
Therefore, the general solution to the IVP is $$y(t)=-8\cos(2t)+8\cos(2t)\cos(t)+4\sin(2t)[2\sin(t)-\ln(\sec(t)+\tan(t))]$$

34
##### Term Test 2 / Re: TT2--P1D
« on: March 21, 2018, 05:14:19 PM »
\underline{\text{Solution:}}$$\\ Part(a) \\ First consider homogeneous equation:$$y''+y=0$$characteristic equation:$$r^2+1=0 \implies \cases{r_1=i\\r_2=-i}$$Thus, the complementary solution$$y_c(t)=c_1\cos t+c_2\sin t$$Now consider the nonhomogeneous equation$$y''+y=2\cos^{-2}(t)$$Since y_1(t)=\cos t and y_2(t)=\sin t, Wronskain$$W=[y_1,y_2](t)=\begin{array}{|c c|}\cos t&\sin t\\-\sin t&\cos t\end{array}=\cos^2t+\sin^2t=1 \neq0$$Therefore, y_1(t) and y_2(t) form a fundamental set of solutions.\\ Use the Method of Variation of Parameters: \\ The particular solution$$\begin{align}Y(t)&=-y_1(t)\int_{t_0}^{t}{y_2(s)g(s)\over W[y_1,y_2](s)}ds+y_2(t)\int_{t_0}^{t}{y_1(s)g(s)\over W[y_1,y_2](s)}ds\\&=-\cos(t)\int_{t_0}^{t}\sin(s)\cdot {2\cos^{-2}(s)}ds+\sin(t)\int_{t_0}^{t}{\cos(s) \cdot 2\cos^{-2}(s)}ds\\&=-2\cos(t)\int_{t_0}^{t}{\sin(s)\over\cos^2(s)}ds+2\sin(t)\int_{t_0}^{t}{\cos(s)\over \cos^2(s)}ds\\&=-2\cos(t)\int_{t_0}^{t}\sec(s)\tan(s)ds+2\sin(t)\int_{t_0}^{t}\sec(s)ds\\&=-2\cos(t)[\sec(t)]+2\sin(t)[\ln(\sec(t)+\tan(t))]\\&=-2+2\sin(t)[\ln(\sec(t)+\tan(t))]\end{align}$$Therefore, the general solution$$\begin{align}y(t)&=y_c(t)+Y(t)\\&=c_1\cos(t)+c_2\sin(t)+2\sin(t)[\ln(\sec(t)+\tan(t))]-2\end{align}$$Part(b)\\$$\begin{align}y(0)&=c_1\cos(0)+c_2\sin(0)+2\sin(0)[\ln(\sec(0)+\tan(0))]-2\\&=c_1-2=0 \implies c_1=2\end{align}y'(t)=-c_1\sin(t)+c_2\cos(t)+2\cos(t)[\ln(\sec(t)+\tan(t))]+2\sin(t){\sec(t)\tan(t)+\sec^2{(t)}\over \sec(t)+\tan(t)}y'(0)=-c_1\sin(0)+c_2\cos(0)+2\cos(0)[\ln(\sec(0)+\tan(0))]+2\sin(0){\sec(0)\tan(0)+\sec^2{(0)}\over \sec(0)+\tan(0)} \implies c_2=0$$Therefore, the general solution to the IVP is$$y(t)=2\cos(t)+2\sin(t)[\ln(\sec(t)+\tan(t))]-2$$35 ##### Term Test 2 / Re: TT2--P4M « on: March 21, 2018, 05:06:42 PM » The picture was too large to upload so I have a link where you can view it from: https://gyazo.com/7c7ba97b6a8bf65ea4c6434cf7eb04e7 You have to type it out. Prof. Victor will tell you the same 36 ##### Term Test 2 / Re: TT2--P1 « on: March 21, 2018, 03:18:38 PM » I think (15) should be$$\implies c_1-c_2-2=\pi$$Therefore,$$\cases{c_1=2+\pi\\c_2=0}$$(Maybe I'm wrong again ) 37 ##### Term Test 2 / Re: TT2--P1 « on: March 21, 2018, 03:08:05 PM » Don't know if my answer is correct or not. My brain was totally frozen for this question's integral during the test 38 ##### Term Test 2 / Re: TT2--P1 « on: March 21, 2018, 03:06:17 PM » \underline{\text{Solution:}}$$\\
Part(a) $\\$
First consider homogeneous equation: $$y''-y=0$$
characteristic equation: $$r^2-1=0 \implies \cases{r_1=1\\r_2=-1}$$
Thus, the complementary solution $$y_c(t)=c_1e^t+c_2e^{-t}$$
Now consider the nonhomogeneous equation $$y''-y={4\over e^{2t}+1}$$
Since $y_1(t)=e^t$ and $y_2(t)=e^{-t}$, Wronskain $$W=[y_1,y_2](t)=\begin{array}{|c c|}e^t &e^{-t}\\e^t&-e^{-t}\end{array}=e^t\cdot(-e^{-t})-e^{-t}\cdot e^t=-2 \neq0$$
Therefore, $y_1(t)$ and $y_2(t)$ form a fundamental set of solutions.$\\$
Use the Method of Variation of Parameters: $\\$
The particular solution \begin{align}Y(t)&=-y_1(t)\int_{t_0}^{t}{y_2(s)g(s)\over W[y_1,y_2](s)}ds+y_2(t)\int_{t_0}^{t}{y_1(s)g(s)\over W[y_1,y_2](s)}ds\\&=-e^t\int_{t_0}^{t}{e^{-s}\cdot {4\over e^{2s}+1}\over -2}ds+e^{-t}\int_{t_0}^{t}{e^s \cdot {4\over e^{2s}+1}\over -2}ds\\&=2e^t\int_{t_0}^{t}{e^{-s}\over e^{2s}+1}ds-2e^{-t}\int_{t_0}^{t}{e^s\over e^{2s}+1}ds\end{align}
For integral $\int_{t_0}^{t}{e^{-s}\over e^{2s}+1}ds$:$\\$
Let $$u=e^t \implies \cases{du=e^tdt \implies dt=e^{-t}du\\e^{-t}={1\over u}}$$
Thus \begin{align}\int_{t_0}^{t}{e^{-s}\over e^{2s}+1}ds&=\int_{t_0}^{t}{e^{-s}\over e^{2s}+1}e^{-s}du\\&=\int_{t_0}^{t}{1\over u^2(u^2+1)}du\\&=\int_{t_0}^{t}({1\over u^2}-{1\over u^2+1})du\\&=\int_{t_0}^{t}{1\over u^2}du-\int_{t_0}^{t}{1\over u^2+1}du\\&=-{1\over u}-\arctan(u)\\&=-\arctan(e^t)-e^{-t}\end{align}
For integral $\int_{t_0}^{t}{e^s\over e^{2s}+1}ds$:$\\$
$$\int_{t_0}^{t}{e^s\over e^{2s}+1}ds=\arctan(e^t)$$
Thus $$Y(t)=2e^t[-\arctan(e^t)-e^{-t}]-2e^{-t}[\arctan(e^t)]$$
Therefore, the general solution \begin{align}y(t)&=y_c(t)+Y(t)\\&=c_1e^t+c_2e^{-t}+2e^t[-\arctan(e^t)-e^{-t}]-2e^{-t}[\arctan(e^t)]\end{align}
Part(b)$\\$
\begin{align}y(0)&=c_1e^0+c_2e^0+2e^0[-\arctan(e^0)-e^0]-2e^0[\arctan(e^0)]=0\\&=c_1+c_2+2(-{\pi\over4}-1)-2{\pi \over4}=0\end{align}
$$\implies c_1+c_2=2+\pi$$
Note
$$y'(t)=c_1e^t-c_2e^{-t}-2e^t\arctan(e^t)-{2e^{2t}\over e^{2t}+1}+2e^{-t}\arctan(e^t)- {2\over e^{2t}+1}$$
\begin{align}y'(0)&=c_1e^0-c_2e^0-2e^0 \arctan(e^0)-{2e^0\over e^0+1}+2e^0\arctan(e^0)-{2\over e^0+1}=0\\&=c_1-c_2-2{\pi\over4}-1+2{\pi\over 4}-1\end{align}
$$\implies c_1-c_2=2$$
Thus, $$\cases{c_1=2+{\pi \over 2}\\c_2=-{\pi\over2}}$$
Therefore, general solution of the $IVP$ is $$y(t)=(2+{\pi \over 2})e^t-{\pi\over2}e^{-t}-2(e^t+e^{-t})\arctan(e^t)-2$$

39
##### Quiz-6 / Re: Q6--T0301
« on: March 16, 2018, 08:13:30 PM »
$$\textbf{x}'=\begin{pmatrix}3&-2\\4&-1\end{pmatrix}\textbf{x}$$
(a)$\\$
Find eigenvalues with $\det(\boldsymbol{A}-\lambda\boldsymbol{I_2})=0:$
\begin{aligned}\begin{array}{|c c|}3-\lambda&-2\\4&-1-\lambda\end{array}=0\\(3-\lambda)(-1-\lambda)-(-2)(4)=0\\\lambda^2- 2\lambda+5=0\end{aligned}
Since $\lambda={2\pm\sqrt{(-2)^2-4\cdot1\cdot 5}\over2},$ we have $\cases{\lambda_1=1+2i\\\lambda_2=1-2i}$ $\\$
Find egienvectors with $(\boldsymbol{A}-\lambda\boldsymbol{I_2})\textbf{x}=\boldsymbol{0}$, where $\textbf{x}$ represents the
eigenvectors.$\\$
When $\lambda=1+2i,$ $$\begin{pmatrix}2-2i&-2\\4&-2- 2i\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
Perform elementary row operation:
$$\begin{pmatrix}2-2i&-2\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
$$\implies \begin{pmatrix}x_1\\x_2\end{pmatrix}=x_2\begin{pmatrix}1\\1-i\end{pmatrix}\text{where the eigenvector is}\space \boldsymbol{\xi}^{(1)}=\begin{pmatrix}1\\1-i\end{pmatrix}.$$ Thus one solution of the given system is
$\textbf{x}^{(1)}(t)=\begin{pmatrix}1\\1-i\end{pmatrix}e^{(1+2i)t}.$ $\\$
When $\lambda=1+2i,$ $$\begin{pmatrix}2-2i&-2\\4&-2- 2i\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
Perform elementary row operation:
$$\begin{pmatrix}2-2i&-2\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
$$\implies \begin{pmatrix}x_1\\x_2\end{pmatrix}=x_2\begin{pmatrix}1\\1-i\end{pmatrix}\text{where the eigenvector is}\space \boldsymbol{\xi}^{(1)}=\begin{pmatrix}1\\1-i\end{pmatrix}.$$ Thus one solution of the given system is
$\textbf{x}^{(1)}(t)=\begin{pmatrix}1\\1-i\end{pmatrix}e^{(1+2i)t}.$ $\\$
When $\lambda=1-2i,$ $$\begin{pmatrix}2-2i&-2\\4&-2- 2i\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
Perform elementary row operation:
$$\begin{pmatrix}2-2i&-2\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
$$\implies \begin{pmatrix}x_1\\x_2\end{pmatrix}=x_2\begin{pmatrix}1\\1-i\end{pmatrix}\text{where the eigenvector is}\space \boldsymbol{\xi}^{(2)}=\begin{pmatrix}1\\1-i\end{pmatrix}.$$ Thus one solution of the given system is
$\textbf{x}^{(1)}(t)=\begin{pmatrix}1\\1-i\end{pmatrix}e^{(1+2i)t}.$ $\\$
To obtain a set of real-valued solution:
\begin{align}\textbf{x}^{(1)}(t)&=\begin{pmatrix}1\\1-i\end{pmatrix}e^t(\cos t+i\sin t)\\&=\begin{pmatrix}\cos 2t\\\cos 2t+\sin 2t\end{pmatrix}e^t+i\begin{pmatrix}\sin 2t\\\sin 2t-\cos 2t\end{pmatrix}e^t \\&=\boldsymbol{u}(t)+i\boldsymbol{v}(t)\end{align}
By the Theorem$\space 7.4.5$, we know that $\boldsymbol{u}(t)\space\text{and}\space\boldsymbol{v}(t)$ are also solutions of given
equation. $\\$
Therefore, the general solution of the given system equation expressed in terms of real-valued functions is
$$\textbf{x}(t)=c_1e^t\begin{pmatrix}\cos 2t\\\cos 2t+\sin 2t\end{pmatrix}+c_2e^t\begin{pmatrix}\sin 2t\\\sin 2t-\cos 2t\end{pmatrix}$$
(b)$\\$
All solutions approach to $\infty$ (unbounded) as $t\rightarrow \infty$.

40
##### Quiz-6 / Re: Q6--T0501
« on: March 16, 2018, 07:20:06 PM »
(a)$\\$
Find eigenvalues with $\det(\boldsymbol{A}-\lambda\boldsymbol{I_2})=0:$
\begin{aligned}\begin{array}{|c c|}2-\lambda&-5\\1&-2-\lambda\end{array}=0\\(2- \lambda)(-2-\lambda)-(-5)(1)=0\\\lambda^2+1=0\end{aligned}
We have $\cases{\lambda_1=i\\\lambda_2=-i}$ $\\$
Find egienvectors with $(\boldsymbol{A}-\lambda\boldsymbol{I_2})\textbf{x}=\boldsymbol{0}$, where $\textbf{x}$ represents the eigenvectors.$\\$
When $\lambda=i,$ $$\begin{pmatrix}2-i&-5\\1&-2-i\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
Perform elementary row operation:
$$\begin{pmatrix}2-i&-5\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
$$\implies \begin{pmatrix}x_1\\x_2\end{pmatrix}=x_2\begin{pmatrix}5\\2-i\end{pmatrix}\text{where the eigenvector is}\space \boldsymbol{\xi}^{(1)}=\begin{pmatrix}5\\2-i\end{pmatrix}.$$ Thus one solution of the given system is $\textbf{x}^{(1)}(t)=\begin{pmatrix}5\\2-i\end{pmatrix}e^{it}.$ $\\$
When $\lambda=-i,$ $$\begin{pmatrix}2+i&-5\\1&-2+i\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
Perform elementary row operation:
$$\begin{pmatrix}2+i&-5\\0&0\end{pmatrix}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$$
$$\implies \begin{pmatrix}x_1\\x_2\end{pmatrix}=x_2\begin{pmatrix}5\\2+i\end{pmatrix}\text{where the eigenvector is}\space \boldsymbol{\xi}^{(2)}=\begin{pmatrix}5\\2+i\end{pmatrix}.$$ Thus one solution of the given system is $\textbf{x}^{(2)}(t)=\begin{pmatrix}5\\2+i\end{pmatrix}e^{it}.$ $\\$
To obtain a set of real-valued solution:
\begin{align}\textbf{x}^{(1)}(t)&=\begin{pmatrix}5\\2-i\end{pmatrix}e^{0}(\cos t+i\sin t)\\&=\begin{pmatrix}5\cos t\\ 2\cos t+\sin t\end{pmatrix}+i\begin{pmatrix}5\sin t\\2\sin t-\cos t\end{pmatrix} \\&=\boldsymbol{u}(t)+i\boldsymbol{v}(t)\end{align}
By the Theorem$\space 7.4.5$, we know that $\boldsymbol{u}(t)\space\text{and}\space\boldsymbol{v}(t)$ are also solutions of given equation. $\\$
Therefore, the general solution of the given system equation expressed in terms of real-valued functions is $$\textbf{x}(t)=c_1\begin{pmatrix}5\cos t\\ 2\cos t+\sin t\end{pmatrix}+c_2 \begin{pmatrix}5\sin t\\2\sin t-\cos t\end{pmatrix}$$
(b)$\\$
All solutions are periodic as $t\rightarrow \infty$. The origin $\textbf{x}=\boldsymbol{0}$ is a center.

41
##### Quiz-6 / Q6--T0501
« on: March 16, 2018, 07:19:49 PM »
For the given equation:$\\$
a. Express the general solution of the given system of equations in terms of real-valued functions. $\\$
b. Also draw a direction field, sketch a few of the trajectories, and describe the behaviour of solutions as $t \rightarrow \infty$.
$$\textbf{x}'=\begin{pmatrix}2&-5\\1&-2\end{pmatrix}\textbf{x}$$

42
##### Quiz-4 / Re: Q4-T0101/T5101
« on: March 02, 2018, 05:38:02 PM »
$$y''+9y=9\sec^23t, 0<t<{\pi \over 6}$$
For homogeneous equation: $y''+9y=0$ $\\$
Characteristic equation: $$r^2+9=0 \implies \cases{r_1=3i\\r_2=-3i}$$
Complementary solution: \begin{align}y_c(t)&=c_1y_1(t)+c_2y_2(t)\\&=c_1\cos3t+c_2\sin3t\end{align}
For non-homogeneous equation: $y''+9y=9\sec^23t$ $\\$
$p(t)=0, q(t)=9,g(t)=9\sec^23t$ are continuous on $0< t<{\pi \over 6}$. $\\$
Now, $$W[y_1,y_2](t)=\begin{array}{|c c|} y_1(t) & y_2(t) \\ y_1'(t) & y_2'(t)\end{array}=\begin{array}{|c c|} \cos3t & \sin3t \\ -3\sin3t & 3\cos3t \end{array}=3$$
Thus, $y_1(t)$ and $y_2(t)$ form a fundamental set of solutions. $\\$
Therefore, \begin{align}u_1(t)=-\int{{y_2(t)g(t)\over W[y_1,y_2](t)}}&=-\int{{\sin3t\cdot 9\sec^23t\over 3}}dt\\&=-\int{3\sin3t{1\over \cos^23t}}dt \\&=-3\int{\sec3t\tan3t}dt\\&=-\sec3t\end{align}
\begin{align}u_2(t)=\int{{y_1(t)g(t)\over W[y_1,y_2](t)}}&=\int{{\cos3t\cdot 9\sec^23t\over 3}}dt\\&=\int{3\cos3t{1\over \cos^23t}}dt \\&=\int{3\sec3t}dt\\&=\ln|\sec3t+\tan3t|\end{align}
Hence, the particular solution is $y_p(t)=u_1(t)y_1(t)+u_2(t)y_2(t)$ $\\$
Thus, \begin{align}y_p(t)&=\cos3t\cdot (-\sec3t)+\sin3t\cdot(\ln|\sec3t+\tan3t|)\\&=\sin3t(\ln|\sec3t+\tan3t|)-1\end{align}
Therefore, the general solution is:
\begin{align}y(t)&=y_c(t)+y_p(t)\\&=c_1\cos3t+c_2\sin3t+\sin3t(\ln|\sec3t+\tan3t|)-1\end{align}

43
##### Quiz-4 / Re: Q4-T0201
« on: March 02, 2018, 05:36:54 PM »
$$y''+4y=3\csc2t, 0< t<{\pi \over 2}$$
For homogeneous equation: $y''+4y=0$ $\\$
Characteristic equation: $$r^2+4=0 \implies \cases{r_1=2i\\r_2=-2i}$$
Complementary solution: \begin{align}y_c(t)&=c_1y_1(t)+c_2y_2(t)\\&=c_1\cos2t+c_2\sin2t\end{align}
For non-homogeneous equation: $y''+4y=3\csc2t$ $\\$
$p(t)=0, q(t)=4,g(t)=3\csc2t$ are continuous on $0< t<{\pi \over 2}$. $\\$
Now, $$W[y_1,y_2](t)=\begin{array}{|c c|} y_1(t) & y_2(t) \\ y_1'(t) & y_2'(t)\end{array}=\begin{array}{|c c|} \cos2t & \sin2t \\ -2\sin2t & 2\cos2t \end{array}=2$$
Thus, $y_1(t)$ and $y_2(t)$ form a fundamental set of solutions. $\\$
Therefore, \begin{align}u_1(t)=-\int{{y_2(t)g(t)\over W[y_1,y_2](t)}}&=-\int{{\sin2t\cdot 3\csc2t\over 2}}dt\\&=-\int{{\sin2t\cdot {3\over \sin2t}\over 2}}dt \\&=-\int{{3\over 2}}dt\\&=-{3\over2}t\end{align}
\begin{align}u_2(t)=\int{{y_1(t)g(t)\over W[y_1,y_2](t)}}&=\int{{\cos2t\cdot 3\csc2t\over 2}}dt\\&={3\over 2}\int{{\cos2t \over \sin2t}}dt \\&={3 \over 2}\int{\cot2t}dt\\&={3 \over 4}\ln|\sin2t|\end{align}
Hence, the particular solution is $y_p(t)=u_1(t)y_1(t)+u_2(t)y_2(t)$ $\\$
Thus, \begin{align}y_p(t)&=\cos2t\cdot (-{3\over 2}t)+\sin2t\cdot({3 \over 4}\ln|\sin2t|)\\&={3 \over 4}\sin2t\ln|\sin2t|-{3\over 2}t\cos2t\end{align}
Therefore, the general solution is:
\begin{align}y(t)&=y_c(t)+y_p(t)\\&=c_1\cos2t+c_2\sin2t+{3 \over 4}\sin2t\ln|\sin2t|-{3\over 2}t\cos2t\end{align}

44
##### Quiz-4 / Re: Q4-T0301
« on: March 02, 2018, 05:35:25 PM »
$$y''-2y+y={e^t\over 1+t^2}$$
For homogeneous equation: $y''-2y+1=0$ $\\$
Characteristic equation: $$r^2-2r+1=0 \implies \cases{r_1=1\\r_2=1}$$
Complementary solution: \begin{align}y_c(t)&=c_1y_1(t)+c_2y_2(t)\\&=c_1e^t+c_2te^t\end{align}
For non-homogeneous equation: $y''-2y+y={e^t\over 1+t^2}$ $\\$
$p(t)=-2, q(t)=1,g(t)={e^t\over 1+t^2}$ are continuous everywhere.$\\$
Now, $$W[y_1,y_2](t)=\begin{array}{|c c|} y_1(t) & y_2(t) \\ y_1'(t) & y_2'(t)\end{array}=\begin{array}{|c c|} e^t & te^t \\ e^t & (1+t)e^t \end{array}=e^{2t}\neq 0$$
Thus, $y_1(t)$ and $y_2(t)$ form a fundamental set of solutions. $\\$
Therefore, \begin{align}u_1(t)=-\int{{y_2(t)g(t)\over W[y_1,y_2](t)}}&=-\int{te^t\cdot {e^t\over1+t^2}\over e^2t}dt\\&=-\int{t\over1+t^2}dt \\&=-{1\over2}\ln(1+t^2)\end{align}
\begin{align}u_2(t)=\int{{y_1(t)g(t)\over W[y_1,y_2](t)}}&=\int{{e^t\cdot {e^t\over1+t^2}\over e^{2t}}}dt\\&=\int{1\over 1+t^2}dt\\&=\arctan t\end{align}
Hence, the particular solution is $y_p(t)=u_1(t)y_1(t)+u_2(t)y_2(t)$ $\\$
Thus, \begin{align}y_p(t)&=e^t\cdot (-{1\over2}\ln(1+t^2))+te^t\cdot \arctan t\\&=-{1\over2}e^t\ln(1+t^2)+te^t\arctan t\end{align}
Therefore, the general solution is:
\begin{align}y(t)&=y_c(t)+y_p(t)\\&=c_1e^t+c_2te^t-{1\over2}e^t\ln(1+t^2)+te^t\arctan t\end{align}

45
##### Quiz-4 / Re: Q4-T0601
« on: March 02, 2018, 05:34:10 PM »
$$x^2y''+xy'+(x^2-0.25)y=3x^{3\over2}\sin(x); x>0; \\ y_1(x)=x^{-{1\over2}}\sin(x), y_2(x)=x^{-{1\over2}}\cos(x)$$
Hence,$$\cases{y_1(x)=x^{-{1\over2}}\sin x\\y_1'(x)=-{1\over2}x^{-{3\over 2}}\sin x+x^{-{1\over2}}\cos x\\y_1’’(x)={3\over4}x^{-{5\over2}}\sin x-x^{-{3\over2}}\cos x-x^{-{1\over2}}\sin x}$$
$$\cases{y_2(x)=x^{-{1\over2}}\cos(x)\\y_2’(x)=-{1\over2}x^{-{3\over2}}\cos x-x^{-{1\over2}}\sin x \\ y_2’’(x)={3\over4}x^{-{5\over2}}\cos x+x^{-{3\over2}}\sin x-x^{-{1\over2}}\cos x}$$
Substitute back into the homogeneous equation: $$x^2y''+xy'+(x^2-0.25)y=0$$
Verified that $y_1(x)$ and $y_2(x)$ both satisfy the corresponding homogeneous equation. $\\$
And the complementary solution $y_c(x)=c_1x^{-{1\over2}}\sin x+c_2x^{-{1\over2}}\cos x$ $\\$
Now divide both sides of the original equation by $x^2$:
$$y’’+{1\over x}y'+{x^2-0.25\over x^2}y=3x^{-{1\over2}}\sin x$$
Then $$p(t)={1\over x}, q(t)={x^2-0.25\over x^2}, g(t)=3x^{-{1\over2}}\sin x$$
$$W[y_1,y_2](x)=\begin{array}{|c c|} y_1(x)&y_2(x)\\y_1’(x)&y_2’(x)\end{array}=-x^{-1}\neq 0$$
Since the particular solution has the form: $$Y(x)=u_1(x)y_1(x)+u_2(x)y_2(x)$$
and \begin{align}u_1(x)&=-\int{{y_2(x)g(x)\over W[y_1,y_2](x)}}dx\\&=-\int{x^{-{1\over2}}\cos x\cdot 3x^{-{1\over2}}\sin x\over -x^{-1}}dx\\&={3\over2}\int{2\sin x\cos x}dx\\&={3\over2}\int{\sin 2x}dx\\&=-{3\over4}\cos 2x\end{align}
\begin{align}u_2(x)&=\int{y_1(x)g(x)\over W[y_1,y_2](x)}dx\\&=\int{x^{-{1\over2}}\sin x\cdot 3x^{-{1\over2}}\sin x\over -x^{-1}}dx\\&=-3\int{\sin^2 x}dx\\&=-3\int{1-\cos 2x\over 2}dx\\&=-{3\over2}x+{3\over4}\sin 2x\end{align}
Therefore, \begin{align}Y(x)&=-{3\over4}\cos 2x\cdot x^{-{1\over2}}\sin x+ (-{3\over2}x+{3\over4}\sin 2x)\cdot x^{-{1\over2}}\cos x\\&={3\over4}x^{-{1\over2}}(\sin 2x\cos x-\cos 2x\sin x)-{3\over 2}x^{-{1\over2}}\cos x\\&={3\over4}x^{-{1\over2}}(2\sin x\cos^2 x-(2\cos^2 x-\cos x)\sin x)-{3\over 2}x^{{1\over2}}\cos x\\&={3\over4}x^{-{1\over2}}\sin x-{3\over 2}x^{{1\over2}}\cos x\end{align}
Hence, the general solution:
\begin{align}y(x)&=y_c(x)+Y(x)\\&=c_1x^{-{1\over2}}\sin x +c_2x^{-{1\over2}}\cos x+ {3\over4}x^{-{1\over2}}\sin x-{3\over 2}x^{{1\over2}}\cos x\\&= (c_1+{3\over4})x^{-{1\over2}}\sin x+ c_2x^{-{1\over2}}\cos x-{3\over2}x^{1\over2}\cos x \\&=c_3x^{-{1\over2}}\sin x+ c_2x^{-{1\over2}}\cos x-{3\over2}x^{1\over2}\cos x\end{align}
where $c_3=c_1+{3\over4}$ $\\$
Therefore, the particular solution of the given nonhomogeneous equation is $$Y_{new}(t)=-{3\over2}x^{1\over2}\cos x$$

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