Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.


Messages - Ka Hou Cheok

Pages: [1]
1
Quiz 3 / Re: Problem 2 (night sections)
« on: November 07, 2013, 09:42:05 AM »
I replaced
Code: [Select]
sin , cos , sec by
Code: [Select]
\sin, \cos, \sec and keyboard sign of integral by
Code: [Select]
\int

Thanks, Prof Victor. I'll use them well next time.

2
Quiz 3 / Re: Problem 2 (night sections)
« on: November 06, 2013, 09:51:57 PM »
I am so impressed with your speed.

I would take it as a compliment. Thanks.
I'm impressed and appreciate your results of the integrals. I was too lazy to integrate them.

3
Quiz 3 / Re: Problem 2 (night sections)
« on: November 06, 2013, 08:38:09 PM »
\end{equation*}

The responding characteristic equation is $$r^3-r^2+r-1=0$$ and we get $r_1=1, r_2=i, r_3=-i$. So $$y_c=c_1e^t+c_2\cos(t)+c_3\sin(t)$$

$$W=e^t((\sin^2(t)+\cos^2(t)-\sin(t)\cos(t))-(-\sin^2(t)-\cos^2(t)-\sin(t)\cos(t)))=2e^t$$
$$W_1=\cos^2(t)+\sin^2(t)=1\\
W_2=e^t(\sin(t)-\cos(t))\\
W_3=e^t(-\sin(t)-\cos(t))$$

$$u_1=\int \frac{(\sec(t))(1)}{2e^t}dt\\
u_2=\int \frac{(\sec(t))(e^t(\sin(t)-\cos(t))}{2e^t}dt\\
u_3=\int \frac{(\sec(t))(e^t(-\sin(t)-\cos(t))}{2e^t}dt$$

$$y=y_c+y_1u_1+y_2u_2+y_3u_3
=c_1e^t+c_2\cos(t)+c_3\sin(t)+\\
e^t\int \frac{(\sec(t))(1)}{2e^t}dt+
\cos(t)\int \frac{(\sec(t))(e^t(\sin(t)-\cos(t))}{2e^t}dt+
\sin(t)\int \frac{(\sec(t))(e^t(-\sin(t)-\cos(t))}{2e^t}dt$$

As the question stated my answer can be in terms of one or more integrals, hopefully I can stop here.

4
Quiz 3 / Re: Problem 1 (night sections)
« on: November 06, 2013, 08:32:38 PM »
\begin{equation*}
y'''-y''-y'+ y = 0
\end{equation*}
The responding characteristic equation is $$r^3-r^2-r+1=0$$
$$(r^3-r)-(r^2-1)=0$$
$$r(r^2-1)-(r^2-1)=0$$
$$(r-1)(r^2-1)=0$$
$$r_1=1, r_2=1, r_3=-1$$
So the general solution is $$y=c_1e^t+c_2te^t+c_3e^{-t}$$

5
Quiz 2 / Re: Problem 2, night sections
« on: October 30, 2013, 08:54:48 PM »
Let the solution $y=y_c+Y$,

The characteristic equation for the homogeneous equation $y''-y'-2y=0$ is
$$r^2-r-2=0$$
Solving the equation we have $r_1=2, r_2=-1$ and hence $$y_c=C_1\exp(2t)+C_2\exp(-t)$$

Let $Y=At^2+Bt+C$, $Y'=2tA+B$, $Y''=2A$.

$Y''-Y'-2Y=(2A)-(2tA+B)-2(At^2+Bt+C)=(-2A)t^2+(-2A-2B)t+(2A-B-2C)=-2t+4t^2$

By comparing the coefficients,
$$
\left\{\begin{aligned}
&-2A=4,\\
&-2A-2B=-2,\\
&2A-B-2C=0.
\end{aligned}\right.
$$
Then,
$$
\left\{\begin{aligned}
&A=-2,\\
&B=3,\\
&C=-7/2.
\end{aligned}\right.
$$
So, $Y=-2t^2+3t-\frac{7}{2}$ and hence $$y=y_c+Y=C_1\exp(2t)+C_2\exp(-t)-2t^2+3t-\frac{7}{2}$$

Pages: [1]