Question

In Problems23-26verify that the indicated function is an explicit solution of the given differential equation. Give an interval of definition Ifor each solution.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{secx}\mathbf{;}\mathbf{y}\mathbf{=}\mathbf{xsinx}\mathbf{+}\mathbf{\left(}\mathbf{cosx}\mathbf{\right)}\mathbf{ln}\mathbf{\left(}\mathbf{cosx}\mathbf{\right)}$

Short answer

The indicated function is an explicit solution of the given differential equation and the interval is $I(2n\pi -\frac{\pi }{2},2n\pi +\frac{\pi }{2}),n=1,2,3,...$.

Step-by-step solution

Define an explicit function

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

Determine the derivatives of the function

Let the first derivative of the function is,

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

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

$\begin{array}{c}\mathrm{y}\text{'}\text{'}=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{xcosx}-\mathrm{sinxln}(\mathrm{cosx}\left)\right)\\ =-\mathrm{xsinx}+\mathrm{cosx}-\mathrm{sinx}\frac{1}{\mathrm{cosx}}(-\mathrm{sinx})-\mathrm{cosxln}\left(\mathrm{cosx}\right)\\ =-\mathrm{xsinx}+\mathrm{cosx}+\frac{{\sin }^2\mathrm{x}}{\mathrm{cosx}}-\mathrm{cosxln}\left(\mathrm{cosx}\right)\\ =-(\mathrm{xsinx}+\mathrm{cosxln}(\mathrm{cosx}\left)\right)+\frac{{\sin }^2\mathrm{x}+{\cos }^{2}x}{\mathrm{cosx}}\end{array}$

Determine the explicit solution and its interval

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

$\begin{array}{c}\mathrm{y}\text{'}\text{'}+\mathrm{y}=-\mathrm{y}+\frac{1}{\mathrm{cosx}}\\ =-\mathrm{y}+\mathrm{secx}\\ \mathrm{y}\text{'}\text{'}+\mathrm{y}=\mathrm{secx}\end{array}$

That is the 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 $\mathrm{cosx}>0$,

Then the interval is $I(2n\pi -\frac{\pi }{2},2n\pi +\frac{\pi }{2}),n=1,2,3,...$.