Problem 2 (night sections)

[Victor Ivrii]

Find the general solution of the given differential equation. Leave your answer in terms of one or more integrals.
\begin{equation*}
y'''-y'' + y'-y = \sec (t), \qquad -\frac{\pi}{2} < t < \frac{\pi}{2}.
\end{equation*}

[Ka Hou Cheok]

\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.

[Yangming Cai]

I am so impressed with your speed.

[Ka Hou Cheok]

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.

[Victor Ivrii]

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

[Ka Hou Cheok]

I replaced

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sin , cos , sec by

Code: [Select]

\sin, \cos, \sec and keyboard sign of integral by

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\int

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