# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Term Test 2 => Topic started by: Victor Ivrii on March 21, 2018, 02:57:33 PM

Title: TT2--P3
Post by: Victor Ivrii 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
$$\mathbf{x}'=\begin{pmatrix} 0 &1\\ 2 &-1\end{pmatrix}\mathbf{x}+ \begin{pmatrix} \frac{e^{2t }}{e^t+1} \\ \frac{e^{2t }}{e^t+1}\end{pmatrix},\qquad \mathbf{x}(0)=\begin{pmatrix} 3 \\ 0\end{pmatrix}.$$
Title: Re: TT2--P3
Post by: Meng Wu 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}$$Title: Re: TT2--P3 Post by: Meng Wu on March 21, 2018, 11:48:29 PM Part(b) \\ Now consider the non-homogeneous system: \\ First calculate the Wronskain$$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}$$Title: Re: TT2--P3 Post by: Meng Wu on March 21, 2018, 11:49:31 PM I honestly don't think I get part(b) right, correct me pelase :-\ Title: Re: TT2--P3 Post by: Jared Jubas-Malz on March 22, 2018, 01:18:14 AM I got a different answer for part (b). Aside from a missing constant, I got everything the same up until (6) so I will start from there. Rewriting (6) I got:$$\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}\$
Title: Re: TT2--P3
Post by: Victor Ivrii on March 24, 2018, 10:35:13 AM
One does not need to find fundamental matrix, it adds complexity to particular problem