$\textsf{Verify the following that the given function is a solution of the differential equation}$
$y^{\prime\prime}+y=\sec t
\quad 0<t<\pi/2
\quad y=(\cos{t})\ln{t}+t\sin{t}$
ok presume this is a plug in thing
but the y'' is a little daunting
$\textsf{Verify the following that the given function is a solution of the differential equation}$
$y^{\prime\prime}+y=\sec t
\quad 0<t<\pi/2
\quad y=(\cos{t})\ln{t}+t\sin{t}$
ok presume this is a plug in thing
but the y'' is a little daunting
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 $\displaystyle \cos(t)$.
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:
$\displaystyle y''+y=\sec(t)$