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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 $$\textbf{x}(t)=\begin{pmatrix}e^t&e^{-2t}\\e^t&-2e^{-2t}\end{pmatrix}\begin{pmatrix}2\\1\end{pmatrix}+\ln(e^t+1)\begin{pmatrix}e^t\\e^t\end{pmatrix}+\begin{pmatrix}e^{-2t}\\-2e^{-2t}\end{pmatrix}$$
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 $$\textbf{x}(t)=\begin{pmatrix}e^t&e^{-2t}\\e^t&-2e^{-2t}\end{pmatrix}\begin{pmatrix}2\\1\end{pmatrix}+\ln(e^t+1)\begin{pmatrix}e^t\\e^t\end{pmatrix}+\begin{pmatrix}e^{-2t}\\-2e^{-2t}\end{pmatrix}$$