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

Question

Solve the differential equation using the method of variation of parameters. $$y ^ { tt } + y = \sec x , \frac { \pi } { 4 } < x < \frac { \pi } { 2 }$$
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 general solution of the associated homogeneous equation $y''+y=0$:
The characteristic equation is $r^2+1=0$, so the roots are $r=\pm i$. Thus, the general solution is $y_h(x)=c_1 \cos x + c_2 \sin x$.

Next, we find particular solutions $y_1(x)$ and $y_2(x)$ that satisfies the non-homogeneous equation for variation of parameters. We assume that $y_1(x)=u_1(x)\cos x + u_2(x)\sin x$ and $y_2(x)=v_1(x)\cos x + v_2(x)\sin x$. Then we have $$y_1''(x)+y_1(x)=\sec(x)$$ and $$y_2''(x)+y_2(x)=0.$$

For $y_2(x)$, since $y_h(x)$ is a solution, we can simply use $y_2(x)=c_1\cos(x)+c_2\sin(x)$.

For $y_1(x)$, we have
\begin{align*}
y_1''(x)+y_1(x)&=\sec x \
(-u_1(x)\sin x-u_2(x)\cos x)'+(u_1(x)\cos x+u_2(x)\sin x)\cos x+&(u_1(x)\sin x+u_2(x)\cos x)\sin x=\sec x\
(-u_1(x)\cos x+u_2(x)\sin x)+(u_1(x)\cos^2 x+u_2(x)\sin x\cos x+u_1(x)\sin x \cos x +u_2(x)\sin^2 x)&=\sec x\
(-u_1(x)\cos x+u_2(x)\sin x)+(u_1(x) \cos^2 x+u_2(x) \sin^2 x+u_1(x)\sin x\cos x)&=\sec x\
-u_1(x)\cos x+u_2(x)\sin x+u_1(x)\sin x \cos x&=\sec x\
\end{align*}

We solve this system of equations to find $u_1(x)$ and $u_2(x)$:
\begin{align*}
u_1(x)\cos x+u_2(x)\sin x&=0 \
u_1(x)\sin x-u_2(x)\cos x&=\sec x
\end{align*}

Multiplying the first equation by $\tan x$ and adding it to the second equation, we get
\begin{align*}
u_1(x)\sin x\tan x &=\sec x\tan x \
u_1(x)&=\cos x
\end{align*}

Plugging this result back into the first equation, we have
\begin{align*}
u_2(x)&=-\sin x
\end{align*}

Therefore, $y_1(x)=\cos x(c_1-x)+\sin x(c_2-\ln(\cos x))$.

Finally, the general solution of the non-homogeneous equation $y''+y=\sec x$ is the sum of the homogeneous and particular solutions: $$y(x)=y_h(x)+y_1(x)+y_2(x)=c_1\cos x+c_2\sin x+\cos x(c_1-x)+\sin x(c_2-\ln(\cos x))+c_3\cos x+c_4\sin x.$$

Simplifying, we get $$y(x)=(c_1+x)\sin x+(c_2+\ln(\cos x))\cos x.$$

Thus, the answer is E.

Solve the Differential Equation Using the Method of Variation of Parameters