use the method of variation of parameters to find a particular solution of the given differential equation…

Explanation:

Step1: Find complementary function

The homogeneous equation is $y'' - 16y=0$. The characteristic equation is $r^{2}-16 = 0$, which gives $r=\pm4$. So, the complementary function $y_c = C_1e^{4x}+C_2e^{-4x}$, and $y_1 = e^{4x}$, $y_2=e^{-4x}$.

Step2: Calculate the Wronskian

The Wronskian $W=\begin{vmatrix}y_1&y_2\y_1'&y_2'\end{vmatrix}=\begin{vmatrix}e^{4x}&e^{-4x}\4e^{4x}&- 4e^{-4x}\end{vmatrix}=-4 - 4=-8$.

Step3: Find $u_1'$ and $u_2'$

We know that for $y''+p(x)y'+q(x)y = f(x)$ (here $p(x) = 0$, $q(x)=-16$, $f(x)=\cosh(4x)=\frac{e^{4x}+e^{-4x}}{2}$), $u_1'=-\frac{y_2f(x)}{W}$ and $u_2'=\frac{y_1f(x)}{W}$. $u_1'=-\frac{e^{-4x}\cdot\frac{e^{4x}+e^{-4x}}{2}}{-8}=\frac{1 + e^{-8x}}{16}$, $u_2'=\frac{e^{4x}\cdot\frac{e^{4x}+e^{-4x}}{2}}{-8}=-\frac{e^{8x}+1}{16}$.

Step4: Integrate $u_1'$ and $u_2'$

$u_1=\int\frac{1 + e^{-8x}}{16}dx=\frac{x}{16}-\frac{e^{-8x}}{128}$, $u_2=\int-\frac{e^{8x}+1}{16}dx=-\frac{e^{8x}}{128}-\frac{x}{16}$.

Step5: Find the particular solution

$y_p = u_1y_1+u_2y_2=\left(\frac{x}{16}-\frac{e^{-8x}}{128}\right)e^{4x}+\left(-\frac{e^{8x}}{128}-\frac{x}{16}\right)e^{-4x}=\frac{x}{16}(e^{4x}-e^{-4x})-\frac{1}{128}(e^{-4x}+e^{4x})=\frac{x}{8}\sinh(4x)-\frac{1}{64}\cosh(4x)$.

Answer:

$\frac{x}{8}\sinh(4x)-\frac{1}{64}\cosh(4x)$