Answer
The given function is the solution of differential equation.
Work Step by Step
STEP 1
Differentiate the given function with respect to $t$.
$$\begin{aligned}y' &= \left( { - \sin t} \right)\ln \cos t + \left( {\cos t} \right)\left( {\frac{{ - \sin t}}{{\cos t}}} \right) + \sin t + t\cos t\\ &= - \sin t\ln \left( {\cos t} \right) - \sin t + \sin t + t\cos t\\ &= - \sin t\ln \left( {\cos t} \right) + t\cos t\end{aligned}$$
STEP 2
Differentiate again with respect to $t$.
$$\begin{aligned}y'' &= - \left[ {\cos t\ln \left( {\cos t} \right) + \sin t\left( {\frac{{ - \sin t}}{{\cos t}}} \right)} \right] + \cos t - t\sin t\\ &= - \cos t\ln \left( {\cos t} \right) + \frac{{{{\sin }^2}t}}{{\cos t}} + \cos t - t\sin t\\ &= - \cos t\ln \left( {\cos t} \right) + \frac{{1 - {{\cos }^2}t}}{{\cos t}} + \cos t - t\sin t\\ &= - \cos t\ln \left( {\cos t} \right) + \sec t - \cos t + \cos t - t\sin t\\ &= - \cos t\ln \left( {\cos t} \right) + \sec t - t\sin t\end{aligned}$$
STEP 3
Substitute the value of $y''$ into the given differential equation.
$$\begin{aligned} - \left( {\cos t} \right)\ln \cos t + \sec t - t\sin t + \left( {\cos t} \right)\ln \cos t + t\sin t &\stackrel{?}{=} \sec t\\\sec t &= \sec t\end{aligned}$$
Therefore, the given function is the solution of differential equation.
