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Messages - Siyuan Tao

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Term Test 2 / Re: TT2--P4
« on: March 21, 2018, 03:01:36 PM »
Find eigenvalues with det(A−λI)=0: det(X-λI)=

Since $\lambda={4\pm\sqrt{(-4)^2-4\cdot1\cdot13}\over2},$ we have $\cases{\lambda_1=2+3i\\\lambda_2=2-3i}$ $\\$
Find egienvectors with (A−λI)x=0, where x represents the eigenvectors:
when $\lambda=2+3i,$
Due to elementary row operation:
$$\implies\begin{pmatrix}x_1\\x_2\end{pmatrix}=x_2\begin{pmatrix}1\\i\end{pmatrix}\text{where the eigenvector is}\space
So, one of the solutions for the system is
$\textbf{x}^{(1)}(t)=\begin{pmatrix}1\\i\end{pmatrix}e^{(2+3i)t}.$ $\\$
to obtain a set of real-valued solution:
Therefore, the general solution of the given system equation expressed in terms of real-valued functions is

Quiz-1 / Re: Q1-T0301
« on: January 25, 2018, 08:54:03 AM »
It is from CH2.1 #40

According to the "Variation of Parameters", $g(t) = 3\cos(2t)$ is not everywhere zero, so we can assume the solution in the form
y = A(t) e^{-\int \frac{1}{t}\,dt} = A(t) e^{-\ln t} = A(t) t^{-1},
where $A$ is now a function of $t$.
By substituting for y in the given differential equation, $A(t)$ satistfies $A′(t) = 3t\cos(2t)$.
This implies to $A(t)$ by using"integration by parts" method.

Set  $u=3t$, $dv=\cos(2t)dt, du = 3 dt$, $v = 0.5\sin(2t)$,
change the $A(t)$ into $uv-\int vdu$ form and do the integration

Then we can get $$A(t) = \frac{3\cos(2t)}{4} + \frac{3t\sin(2t)}{2} + C$$.

So the solution is
$$y = \frac{3\cos(2t)}{4t} + \frac{3\sin(2t)}{2} + \frac{C}{t}.$$

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