Question

use the method of variation of parameters to find a particular solution of the given differential equation y - 64y = cosh(8x)

Explanation:

Step1: Find the complementary function

The homogeneous equation is $y'' - 64y=0$. The characteristic equation is $r^{2}-64 = 0$, which factors as $(r - 8)(r + 8)=0$. So the roots are $r_1=8$ and $r_2=-8$. The complementary function $y_c = C_1e^{8x}+C_2e^{-8x}$.

Step2: Set up the form for the particular solution

We assume a particular solution of the form $y_p=u_1(x)e^{8x}+u_2(x)e^{-8x}$. We need to find $u_1(x)$ and $u_2(x)$.
We have the following system of equations based on the variation - of - parameters method:

$$\begin{cases}u_1'e^{8x}+u_2'e^{-8x}=0\\8u_1'e^{8x}-8u_2'e^{-8x}=\cosh(8x)\end{cases}$$

From the first equation, $u_2'=-u_1'e^{16x}$. Substitute into the second equation:
$8u_1'e^{8x}+8u_1'e^{8x}=\cosh(8x)$
$16u_1'e^{8x}=\cosh(8x)=\frac{e^{8x}+e^{-8x}}{2}$
$u_1'=\frac{1}{32}(1 + e^{-16x})$
Integrating $u_1'$ with respect to $x$ gives $u_1=\frac{1}{32}(x-\frac{1}{16}e^{-16x})$.
Since $u_2'=-u_1'e^{16x}$, then $u_2'=-\frac{1}{32}(e^{16x}+1)$. Integrating $u_2'$ with respect to $x$ gives $u_2=-\frac{1}{32}(\frac{1}{16}e^{16x}+x)$.

Step3: Calculate the particular solution

$y_p = u_1e^{8x}+u_2e^{-8x}$
$y_p=\frac{1}{32}(x-\frac{1}{16}e^{-16x})e^{8x}-\frac{1}{32}(\frac{1}{16}e^{16x}+x)e^{-8x}$
$y_p=\frac{1}{32}(xe^{8x}-\frac{1}{16}e^{-8x}-\frac{1}{16}e^{8x}-xe^{-8x})$
$y_p=\frac{1}{64}x\sinh(8x)-\frac{1}{512}\cosh(8x)$

Answer:

$y_p=\frac{1}{64}x\sinh(8x)-\frac{1}{512}\cosh(8x)$