# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Term Test 2 => Topic started by: Victor Ivrii on March 21, 2018, 03:01:41 PM

Title: TT2--P3D
Post by: Victor Ivrii on March 21, 2018, 03:01:41 PM
a. Find the general solution of
$$\mathbf{x}'=\begin{pmatrix} 0 &1\\ 3 &-2\end{pmatrix}\mathbf{x}$$
and sketch trajectories.

b. Solve
$$\mathbf{x}'=\begin{pmatrix} 0 &1\\ 3 &-2\end{pmatrix}\mathbf{x}+ \begin{pmatrix} \frac{e^{3t }}{e^t+1} \\ \frac{e^{3t }}{e^t+1}\end{pmatrix},\qquad \mathbf{x}(0)=\begin{pmatrix} 3-\ln2\\ -1-\ln2\end{pmatrix}.$$
Title: Re: TT2--P3D
Post by: Syed Hasnain on March 25, 2018, 07:59:52 PM
I have attached my solution.
Thanks
Title: Re: TT2--P3D
Post by: Victor Ivrii on March 27, 2018, 08:09:44 AM
Errors. Where arbitrary constant $K$ comes from? And readers should not be forced to download pdf files containing errors and clean drives after this.
Title: Re: TT2--P3D
Post by: Meng Wu on March 27, 2018, 11:37:27 PM
$\underline{\textbf{Solution:}}$ $\\$
$\textbf{(a)}$ $\\$
Find eigenvalues by $\det(A-\lambda I_2)=0$:
$$\begin{array}{|c c|}-\lambda&1\\3&-2-\lambda\end{array}=0 \implies \lambda^2+2\lambda-3=0\implies \cases{\lambda_1=1\\ \lambda_2=-3}$$
Find eigenvectors by $(A-\lambda I_2)\mathbf{x}=\boldsymbol 0$: $\\$
When $\lambda=1$, eigenvector $\boldsymbol{\xi}^{(1)}=\begin{pmatrix}1\\1\end{pmatrix}$. $\\$
When $\lambda=-3$, eigenvector $\boldsymbol{\xi}^{(2)}=\begin{pmatrix}1\\-3\end{pmatrix}$. $\\$
(Few steps omitted, since "Syed_Hasnain" got part(a) right already.)$\\$
Therefore, the general solution is $$\mathbf{x}(t)=c_1\begin{pmatrix}1\\1\end{pmatrix}e^t+c_2\begin{pmatrix}1\\-3\end{pmatrix}e^{-3t}$$
$\textbf{(b)}$ $\\$
$$W[\mathbf{x}^{(1)},\mathbf{x}^{(2)}](t)=\begin{array}{|c c|}e^t&e^{-3t}\\e^t&-3e^{-3t}\end{array}=-4e^{-2t}\neq 0$$
Thus, $\mathbf{x}^{(1)}(t)$ and $\mathbf{x}^{(2)}(t)$ form a fundamental set of solutions. $\\$
Hence the fundamental matrix $$\boldsymbol\Psi(t)=\begin{pmatrix}e^t&e^{-3t}\\e^t&-3e^{-3t}\end{pmatrix}$$
For the non-homogeneous system, we have the general solution$$\mathbf{x}=\boldsymbol\Psi(t)\mathbf{u}(t)$$
Since we know $$\boldsymbol\Psi(t)\mathbf{u}'(t)=\mathbf{g}(t)$$
$$\begin{pmatrix}e^t&e^{-3t}\\e^t&-3e^{-3t}\end{pmatrix}\begin{pmatrix}u_1'\\u_2'\end{pmatrix}=\begin{pmatrix}{e^{3t}\over e^t+1}\\{e^{3t}\over e^t+1}\end{pmatrix}$$
By row reduction: $$\begin{pmatrix}e^t&0\\0&-4e^{-3t}\end{pmatrix}\begin{pmatrix}u_1'\\u_2'\end{pmatrix}=\begin{pmatrix}{e^{3t}\over e^t+1}\\0\end{pmatrix}$$
Hence $$\cases{u_1'={e^{2t}\over e^t+1}\\u_2'=0} \implies \cases{u_1(t)=\int{{e^{2t}\over e^t+1}dt}=(e^t+1)-\ln(e^t+1)+c_1\\u_2(t)=\int{0dt}=c_2}$$
Thus \begin{align}\mathbf{x}&=\boldsymbol\Psi(t)\mathbf{u}(t)\\&=c_1\begin{pmatrix}1\\1\end{pmatrix}e^t+c_2\begin{pmatrix}1\\-3\end{pmatrix}e^{-3t}+\begin{pmatrix}1\\1\end{pmatrix}e^{2t}-\ln(e^t+1)\begin{pmatrix}1\\1\end{pmatrix}e^t\end{align}
Apply the given initial condition:$$\mathbf{x}(0)=\begin{pmatrix}c_1+c_2+1-\ln2\\c_1-3c_2+1-\ln2\end{pmatrix}=\begin{pmatrix}3-\ln2\\-1-\ln2\end{pmatrix}\implies \cases{c_1=1\\c_2=1}$$
Therefore, the general solution for the IVP:
$$\mathbf{x}=\begin{pmatrix}1\\1\end{pmatrix}e^t+\begin{pmatrix}1\\-3\end{pmatrix}e^{-3t}+\begin{pmatrix}1\\1\end{pmatrix}e^{2t}-\ln(e^t+1)\begin{pmatrix}1\\1\end{pmatrix}e^t$$