MAT244-2013S > MidTerm

MT Problem 2b

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Jason Hamilton:

--- Quote from: Brian Bi on March 06, 2013, 10:01:33 PM ---By inspection, $y = 1/2$ is a solution.

--- End quote ---


Darn, why did I not see this...
Just goes to show that slowing down during a test and looking at the question with a calm mind can do wonders

Brian Bi:

--- Quote from: Jason Hamilton on March 06, 2013, 10:38:24 PM ---
--- Quote from: Brian Bi on March 06, 2013, 10:01:33 PM ---By inspection, $y = 1/2$ is a solution.

--- End quote ---


Darn, why did I not see this...
Just goes to show that slowing down during a test and looking at the question with a calm mind can do wonders

--- End quote ---
I also did not see this during the test. Is there anyone who did?

Jeong Yeon Yook:

--- Quote from: Brian Bi on March 06, 2013, 11:10:43 PM ---
--- Quote from: Jason Hamilton on March 06, 2013, 10:38:24 PM ---
--- Quote from: Brian Bi on March 06, 2013, 10:01:33 PM ---By inspection, $y = 1/2$ is a solution.

--- End quote ---


Darn, why did I not see this...
Just goes to show that slowing down during a test and looking at the question with a calm mind can do wonders

--- End quote ---
I also did not see this during the test. Is there anyone who did?

--- End quote ---

The hint led me to do variation right away...

Victor Ivrii:
I will not tell you if instructors knew about this "money for nothing" solutions (or may be some knew but kept for themselves) but this situation is not an uncommon in the real research as well. Sometimes after a year of hard work comes an idea which solves the problem on 1--2 pages and renders useless 50 page paper already submitted. And then years after one understands that these 50 pages were relevant to another problem ...

J. Y. Yook made few mistakes uncritically applying formula (forgetting that the senior coefficient is not $1$ so that the r.h.e. must be divided by it.
\begin{equation*}
\left\{\begin{aligned}
&u'_1 t + u'_2 (t^2+1)= 0,\\
&u'_1 + u'_2 \cdot 2t= \frac{1}{t^2-1}.
\end{aligned}
\right.
\end{equation*}
Then
\begin{equation*}
\left\{\begin{aligned}
&u'_1=\frac{1}{t^2-1}-\frac{2t^2}{(t^2-1)^2},\\
&u'_2=\frac{t}{(t^2-1)^2}
\end{aligned}
\right.
\end{equation*}
and
\begin{align*}
&u_2 = \int \frac{t}{(t^2-1)^2}\,dt =-\frac{1}{2  (t^2-1)} +c_2, \text{substitution $z=t^2-1$}\\
&u_1 = \int \frac{1}{t^2-1}dt -\int \frac{2t^2}{(t^2-1)^2}dt=\int \frac{1}{t^2-1}dt + \int t\cdot d\left(\frac{1}{t^2-1}\right)= \frac{t}{t^2-1}+c_1
\end{align*}
where we integrated by parts in $u_1$. So
\begin{equation*}
y = u_1 t+ u_2 (t^2+1)= \frac{1}{2} +c_1 t+c_2 (t^2+1)
\end{equation*}
which is the general solution containing the partial solution.

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