Question

In Problems 23-26verify that the indicated function is an explicit solution of the given differential equation. Give an interval of definition Ifor each solution.

${\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathit{s}\mathit{e}\mathit{c}\mathbf{\left(}\mathit{l}\mathit{n}\mathit{x}\mathbf{\right)}\mathbf{;}\mathit{y}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\mathit{l}\mathit{n}\mathit{x}\mathbf{\right)}\mathit{l}\mathit{n}\mathbf{\left(}\mathit{c}\mathit{o}\mathit{s}\mathbf{\right(}\mathit{l}\mathit{n}\mathit{x}\mathbf{\left)}\mathbf{\right)}\mathbf{+}\mathbf{\left(}\mathit{l}\mathit{n}\mathit{x}\mathbf{\right)}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathit{l}\mathit{n}\mathit{x}\mathbf{\right)}$

Short answer

The indicated function is an explicit solution of the given differential equation and the interval is $e^{-\frac{\pi }{2}}<x<e^{\frac{x}{2}}$.

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,

$$\begin{gathered} \text{y' = cos(lnx)}\frac{\text{1}}{\text{cos(lnx)}}\left(\text{-}\frac{\text{sin(lnx)}}{\text{x}}\right)\text{+ ln(cos(lnx))}\left(\text{-}\frac{\text{sin(lnx)}}{\text{x}}\right)\text{+ lnx}\frac{\text{cos(lnx)}}{\text{x}}\text{+}\frac{\text{sin(lnx)}}{\text{x}} \\ \text{= -}\frac{\text{ln(cos(lnx))sin(lnx)}}{\text{x}}\text{+}\frac{\text{(lnx)cos(lnx)}}{\text{x}} \end{gathered}$$

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

$$\begin{gathered} \text{y'' = - x}\left[\text{ln(cos(lnx))}\frac{\text{cos(lnx)}}{\text{x}}\text{+ (lnx)}\frac{\text{1}}{\text{cos(lnx)}}\left(\text{-}\frac{\text{sin(lnx)}}{\text{x}}\right)\right]\frac{\text{1}}{{\text{x}}^{\text{2}}} \\ \text{+ ln(cos(lnx))sin(lnx)}\frac{\text{1}}{{\text{x}}^{\text{2}}}\text{+ x}\left[\text{(lnx)}\left(\text{-}\frac{\text{sin(lnx)}}{\text{x}}\right)\right]\text{+} \\ \left[\frac{\text{cos(lnx)}}{\text{x}}\right]\frac{\text{1}}{{\text{x}}^{\text{2}}}\text{- (lnx)cos(lnx)}\frac{\text{1}}{{\text{x}}^{\text{2}}} \\ \text{=}\frac{\text{1}}{{\text{x}}^{\text{2}}}\left[\text{- ln(cos(lnx))cos(lnx) +}\frac{{\text{sin}}^{\text{2}}\text{(lnx)}}{\text{cos(lnx)}}\text{+ ln(cos(lnx))sin(lnx)} \\ \text{- (lnx)sin(lnx) + cos(lnx) - (lnx)cos(lnx)} \\ \right] \end{gathered}$$

Determine the explicit solution and its interval.

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

$\begin{array}{rcl}& & {\text{x}}^{\text{2}}\text{y'' + xy' + y = - ln(cos(lnx))cos(lnx) +}\frac{{\text{sin}}^{\text{2}}\text{(lnx)}}{\text{cos(lnx)}}\\ & & \text{+ ln(cos(lnx))sin(lnx) - (lnx)sin(lnx)}\\ & & \text{+ cos(lnx) - (lnx)cos(lnx) - ln(cos(lnx))sin(lnx)}\\ & & \text{+ (lnx)cos(lnx) + cos(lnx)ln(cos(lnx))}\\ & & \text{=}\frac{{\text{sin}}^{\text{2}}\text{(lnx)}}{{\text{cos}}^{\text{2}}\text{(lnx)}}\text{+ cos(lnx)sin(lnx)}\\ & & \text{=}\frac{{\text{sin}}^{\text{2}}\text{(lnx) + cos(lnx)}}{\text{cos(lnx)}}\\ & & \text{=}\frac{\text{1}}{\text{cos(lnx)}}\\ & & \text{= sec(lnx)}\end{array}$

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.

The domain of Inx is $\text{(0,}\infty \text{)}$, so as $\mathrm{cosln}x>0$ when $-\frac{\pi }{2}<\ln x<\frac{\pi }{2}$, and also e^x is an increasing function. Then the interval is $e^{-\frac{\pi }{2}}<x<e^{\frac{x}{2}}$.