91影视

An explicit solution is one in which the dependent variable is expressed directly in terms of the independent variable and constants.

Let the first derivative of the function is,

\(\begin{array}{c}{\rm{y' = }}\frac{{\rm{d}}}{{{\rm{dx}}}}{\rm{(xsinx + cosxln(cosx))}}\\{\rm{ = xcosx + sinx + cosx}}\frac{{\rm{1}}}{{{\rm{cosx}}}}{\rm{( - sinx) - sinxln(cosx)}}\\{\rm{ = xcosx + sinx - sinx - sinxln(cosx)}}\\{\rm{ = xcosx - sinxln(cosx)}}\end{array}\)

Using the product rule, the second derivative of the function is,

\(\begin{array}{c}{\rm{y'' = }}\frac{{\rm{d}}}{{{\rm{dx}}}}{\rm{(xcosx - sinxln(cosx))}}\\{\rm{ = - xsinx + cosx - sinx}}\frac{{\rm{1}}}{{{\rm{cosx}}}}{\rm{( - sinx) - cosxln(cosx)}}\\{\rm{ = - xsinx + cosx + }}\frac{{{\rm{sin2}}}}{{{\rm{cosx}}}}{\rm{ - cosxln(cosx)}}\\{\rm{ = - (xsinx + cosxln(cosx)) + }}\frac{{{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{x + co}}{{\rm{s}}^2}x}}{{{\rm{cosx}}}}\end{array}\)

Substitute\(y\)and \(y''\) into the left-hand side of the differential equation.

\(\begin{array}{c}{\rm{y'' + y = - y + }}\frac{{\rm{1}}}{{{\rm{cosx}}}}\\{\rm{ = - y + secx}}\\{\rm{y'' + y = secx}}\end{array}\)

That is same as the right-hand side of the differential equation. Thus, the indicated function is an explicit solution of the given differential equation.

Since the given function is defined only when \({\rm{cosx > 0}}\), then the interval is \(I(2n\pi - \frac{\pi }{2},2n\pi + \frac{\pi }{2}),n = 1,2,3,...\).