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### Messages - Ka Hou Cheok

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

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

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##### 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.

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##### 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.

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##### 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 $$y=c_1e^t+c_2te^t+c_3e^{-t}$$

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##### 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 $$y=y_c+Y=C_1\exp(2t)+C_2\exp(-t)-2t^2+3t-\frac{7}{2}$$

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