1.3.13 Verify that the given function is a solution of the differential equation

Nov 2009
717
133
Wahiawa, Hawaii
$\textsf{Verify the following that the given function is a solution of the differential equation}$

$y^{\prime\prime}+y=\sec t
\quad 0<t<\pi/2
\quad y=(\cos{t})\ln{t}+t\sin{t}$

ok presume this is a plug in thing
but the y'' is a little daunting
 

romsek

MHF Helper
Nov 2013
6,665
3,002
California
it doesn't seem to be a solution.

Are there any typos?
 
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Dec 2011
2,314
915
St. Augustine, FL.
The given solution is:

\(\displaystyle y=\cos(t)\ln(\cos(t))+t\sin(t)\)

You initially stated the argument for the log function was \(\displaystyle t\) instead of \(\displaystyle \cos(t)\).
 
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Nov 2009
717
133
Wahiawa, Hawaii
ok so if
$$\displaystyle y=\cos(t)\ln(\cos(t))+t\sin(t)$$
then
$$\displaystyle y'= t \cos(t) - \sin(t) \log(\cos(t))$$
and
$$y''=(y')^2$$
 
Dec 2011
2,314
915
St. Augustine, FL.
I get:

\(\displaystyle y=\cos(t)\ln(\cos(t))+t\sin(t)\)

\(\displaystyle y'=-\sin(t)-\sin(t)\ln(\cos(t))+t\cos(t)+\sin(t)=-\sin(t)\ln(\cos(t))+t\cos(t)\)

\(\displaystyle y''=\sin^2(t)\sec(t)-\cos(t)\ln(\cos(t))-t\sin(t)+\cos(t)=\cos(t)+\sin^2(t)\sec(t)-y=(\cos^2(t)+\sin^2(t))\sec(t)-y=\sec(t)-y\)

And thus:

\(\displaystyle y''+y=\sec(t)\)
 
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Nov 2009
717
133
Wahiawa, Hawaii
whoa....

not sure how you got $y''$ is that from $W\vert A$
 
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