Question

In Problems 19–30 proceed as in Example 5 to find a solution of the given initial-value problem.

Short answer

\(y = - \frac{3}{2}\cos x + \sin x(1 + \ln |\sec x + \tan x|) - 1\)

Step-by-step solution

The solve two initial-value problems

For this exercise, we solve two initial-value problems. First, we will be solving the associated homogeneous equation \({y^{\prime \prime }} + \)\(y = 0\) whose auxiliary equation is \({m^2} + 1 = 0\). Upon factoring and solving, the roots of this equation are \({m_1} = i\)and \({m_2} = - i\). Therefore, the general solution is

\(y(x) = {c_1}\cos x + {c_2}\sin x\)

where \({y_1}(x) = \cos x\) and \({y_2}(x) = \sin x\).

The derivative is \({y^\prime } = - {c_1}\sin x + {c_2}\cos x\). Applying the initial conditions \(y(\pi ) = \frac{1}{2}\)and \({y^\prime }(\pi ) = - 1\)gives

\(\begin{aligned}{l}{c_1}\cos \pi + {c_2}\sin \pi = \frac{1}{2} - {c_1}\\ = \frac{1}{2}{c_1} = - \frac{1}{2}\end{aligned}\)

\(\begin{aligned}{l} - {c_1}\sin \pi + {c_2}\cos \pi = - 1\\ - {c_2} = - 1\\{c_2} = 1\end{aligned}\)

Therefore, we have

\({y_h} = - \frac{1}{2}\cos x + \sin x\)

The Wronskian is

\(\begin{aligned}{l}W(t) = W(\cos t,\sin t)\\ = \left| {\begin{aligned}{*{20}{c}}{\cos t}&{\sin t}\\{ - \sin t}&{\cos t}\end{aligned}} \right|\\ = \cos t \cdot \cos t - \sin t \cdot ( - \sin t)\\ = {\cos ^2}t + {\sin ^2}t\\ = 1\end{aligned}\)

Using the Green’s function

The Green's function for the differential equation is

\(\begin{aligned}{l}G(x,t) = \frac{{{y_1}(t){y_2}(x) - {y_1}(x){y_2}(t)}}{{W(t)}}\\ = \frac{{\cos t\sin x - \cos x\sin t}}{1}\\ = \cos t\sin x - \cos x\sin t\end{aligned}\)

Since the given differential equation is already in the standard form (that is, the coefficient of \({y^{\prime \prime }}\)is 1 ), we consider \(f(x) = \) \({\sec ^2}x\). With the identification \(f(t) = {\sec ^2}t\) and \({x_0} = \pi \), the particular solution of \({y^{\prime \prime }} + y = {\sec ^2}x\)is

\(\begin{aligned}{l}{y_p} = \int_\pi ^x {(\cos t\sin x - \cos x\sin t)} {\sec ^2}tdt\\ = \int_\pi ^x {(\sin x\sec t)} dt - \int_\pi ^x {(\cos x\tan t\sec t)} dt\\ = \sin x\int_\pi ^x {\sec } tdt - \cos x\int_\pi ^x {\tan } t\sec tdt\\ = \sin x[\ln |\sec t + \tan t|]_\pi ^x - \cos x[\sec t]_\pi ^x\\ = \sin x[\ln |\sec x + \tan x| - \ln |\sec \pi + \tan \pi |] - \cos x[\sec x - \sec \pi ]\end{aligned}\)

\(\begin{aligned}{l} = \sin x[\ln |\sec x + \tan x| - \ln | - 1 + 0|] - \cos x[\sec x - ( - 1)]\\ = \sin x[\ln |\sec x + \tan x| - 0] - \cos x[\sec x + 1]\\ = \sin x(\ln |\sec x + \tan x|) - \cos x\sec x - \cos x\\ = \sin x(\ln |\sec x + \tan x|) - 1 - \cos x\end{aligned}\)

final proof

Thus, the general solution \(y = {y_h} + {y_p}\)of the given \({\rm{DE}}\) is

\(y = - \frac{1}{2}\cos x + \sin x + \sin x(\ln |\sec x + \tan x|) - 1 - \cos x\)

or upon simplification is,

\(y = - \frac{3}{2}\cos x + \sin x(1 + \ln |\sec x + \tan x|) - 1\)