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In Problems \(11 - 14\) verify that the indicated function is an explicit solution of the given differential equation. Assume an appropriate interval I of definition for each solution.

\(y'' + y = tanx; y = - (cosx) ln(secx + tanx)\).

Short Answer

Expert verified

The indicated function is an explicit solution of the given differential equation.

Step by step solution

01

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.

02

Determine the derivatives of the function.

Let the given function be \(y'' + y = tanx\).

Then, the first derivative of the function is,

\(y' = - (cos x)sec x + sin x ln(sec x + tan x)\)

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

\(\begin{aligned}{l}y'' &= ( - cos x(sec x tan x) + sin x sec x) + (sin x sec x + cos x ln(sec x + tan x))\\y'' &= - cos x(sec x tan x) + cos x ln( sec x + tan x) + 2sin x sec x\end{aligned}\)

03

Determine the explicit solution and its interval.

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

\(\begin{aligned}{c} - cos x(sec x tan x) + cos xln(sec x + tan x) + 2sin x sec x + ( - (cos x)ln(sec x + tan x)) &= tan x\\ - cos x(sec x tan x) + 2sin x sec x &= tan x\\ - \frac{1}{{sec x}}(sec x tan x) + 2sin x\frac{1}{{cos x}} &= tan x\\ - tan x + 2tan x &= tan x\\tan x &= tan x\end{aligned}\)

That is same as the right-hand side of the differential equation for every real number \(x\). Thus, the indicated function is an explicit solution of the given differential equation.

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