Toronto Math Forum
MAT244-2013S => MAT244 Math--Tests => MidTerm => Topic started by: Victor Ivrii on March 06, 2013, 09:07:57 PM
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Find a particular solution of equation
\begin{equation*}
(t^2-1) y''-2ty'+2 y=1.
\end{equation*}
Hint: use variation of parameters.
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By inspection, $y = 1/2$ is a solution.
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The homogeneous sol'ns were y1=t, y2=t2 right?
But why Wolfram alpha gives y(t) = (c_1 sqrt(t^2-1) (1-t)^(3/2))/sqrt(t+1)+(c_2 t sqrt(t^2-1) sqrt(1-t))/((t-1) sqrt(t+1))
...
the 2nd term of Wolfram is equivalent {if regardless of the signs (consider all things in roots being positive)} ; but I have no idea what the first term is ... hold on, it can be written as constant *(t-1)^2 , so it's still correct :P
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$$
y(t) =(c_1 \sqrt{t^2-1} (1-t)^{3/2})/\sqrt{t+1}+(c_2 t \sqrt{t^2-1} \sqrt{1-t})/((t-1) \sqrt{t+1})
$$
[Don't know how to get the codes working..]
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solution
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By inspection, $y = 1/2$ is a solution.
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
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By inspection, $y = 1/2$ is a solution.
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
I also did not see this during the test. Is there anyone who did?
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By inspection, $y = 1/2$ is a solution.
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
I also did not see this during the test. Is there anyone who did?
The hint led me to do variation right away...
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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.