Solve the differential equation using the method of variation of parameters.
$$y ^ { tt } + y = \sec x , \quad \frac { \pi } { 4 }

Question

Solve the differential equation using the method of variation of parameters.
$$y ^ { tt } + y = \sec x , \quad \frac { \pi } { 4 } < x < \frac { \pi } { 2 }$$
A) $y ( x ) = c _ { 1 } + \left[ c _ { 2 } + \ln ( \sin x ) \right] \sin x$
B) $y ( x ) = \left( c _ { 1 } + x \right) \sin x + c _ { 2 } + \ln ( \cos x )$
C) $y ( x ) = \frac { \pi } { 2 } \sin x + \left[ \frac { \pi } { 2 } + \ln ( \cos x ) \right] \cos x$
D) $y ( x ) = \left[ \frac { \pi } { 4 } + \ln ( \cos x ) \right] \cos x$
E) $y ( x ) = \left( c _ { 1 } + x \right) \sin x + \left[ c _ { 2 } + \ln ( \cos x ) \right] \cos x$

Quizplus · Accepted Answer

First, we find the complementary solution of the differential equation $y_{cs}(x)=c_1\cos x+c_2\sin x$, where $c_1$ and $c_2$ are arbitrary constants.

Next, we assume a particular solution of the form $y_p(x)=u(x)\cos x+v(x)\sin x$, where $u(x)$ and $v(x)$ are functions to be determined. Taking derivatives, we have: 
$$y_p'(x)=-u(x)\sin x+v(x)\cos x+u'(x)\cos x+v'(x)\sin x$$
$$y_p''(x)=-u(x)\cos x-v(x)\sin x-2u'(x)\sin x+2v'(x)\cos x+u''(x)\cos x+v''(x)\sin x$$
Substituting these into the differential equation, we get: 
$$-u(x)\cos x-v(x)\sin x-2u'(x)\sin x+2v'(x)\cos x+u''(x)\cos x+v''(x)\sin x+\cos x=\sec x$$
Simplifying and grouping terms, we get: 
$$u''(x)\cos x+v''(x)\sin x-2u'(x)\sin x+2v'(x)\cos x=1$$
We now solve for $u'(x)$ and $v'(x)$ using Cramer's rule: 
$$u'(x)=\frac{\begin{vmatrix}\cos x&\sin x\1&0\end{vmatrix}}{\begin{vmatrix}\cos x&\sin x\-\sin x&\cos x\end{vmatrix}}=-\frac{1}{\cos x}$$
$$v'(x)=\frac{\begin{vmatrix}\cos x&\sin x\0&1\end{vmatrix}}{\begin{vmatrix}\cos x&\sin x\-\sin x&\cos x\end{vmatrix}}=\frac{\sin x}{\cos x^2}$$
Integrating with respect to $x$, we get: 
$$u(x)=\ln|\cos x|+c_1$$
$$v(x)=\int\frac{\sin x}{\cos x^2}\ dx=-\frac{1}{\cos x}+c_2$$
where $c_1$ and $c_2$ are arbitrary constants of integration.

Therefore, the general solution is given by: 
$$y(x)=y_{cs}(x)+y_p(x)=c_1\cos x+c_2\sin x+\left(\ln|\cos x|+c_1ight)\cos x+\left(-\frac{1}{\cos x}+c_2ight)\sin x$$
Simplifying, we get: 
$$y(x)=\left(c_1+\ln|\cos x|ight)\cos x+\left(c_2-\frac{1}{\cos x}ight)\sin x$$
Comparing with the answer choices, we see that the best choice is $\boxed{	extbf{(E)}}$.

Solve the Differential Equation Using the Method of Variation of Parameters