Question

y"-16y'+64y=-2x^2-5e^(5x)

Answer 1

$y''-16y'+64y=-2x^2-5e^{5x}$

Let us solve a characteristic equation:

$k^2-16k+64=0$

$(k-8)^2=0$

$k_1= k_2=8$

The general solution is of the following form:

$y(x)=e^{8x}(C_1+C_2 x)+y_p(x),$ where $y_p(x)$ is a particular solution, $C_1, C_2$ are arbitrary constants.

The particular solution has the following form:

$y_p(x)=ax^2+bx+c+de^{5x}$, where $a,b,c$ and $d$ are constants.

$y_p'=2ax+b+5de^{5x}$

$y_p''=2a+25de^{5x}$

Put founded derivatives in the differential equation:

$2a+25de^{5x}-16(2ax+b+5de^{5x})+64(ax^2+bx+c+de^{5x})=-2x^2-5e^{5x}$

$2a+25de^{5x}-32ax-16b-80de^{5x}+64ax^2+64bx+64c+64de^{5x}=-2x^2-5e^{5x}$

$64ax^2+(-32a+64b)x+(2a-16b+64c)+9de^{5x}=-2x^2-5e^{5x}$

Equating the corresponding coefficients, we obtain a system of equations:

$\begin{cases} 64a=-2\\ -32a+64b=0\\2a-16b+64c=0\\9d=-5 \end{cases}$

$\begin{cases} a=-\frac{1}{32}\\ a=2b\\64c=-2a+16b\\9d=-5 \end{cases}$

$\begin{cases} a=-\frac{1}{32}\\ b=-\frac{1}{64}\\64c=\frac{1}{16}-\frac{1}{4}\\d=-\frac{5}{9} \end{cases}$

$\begin{cases} a=-\frac{1}{32}\\ b=-\frac{1}{64}\\c=-\frac{3}{1024}\\d=-\frac{5}{9} \end{cases}$

Finally, the solution of the differential equation is the following:

$y(x)=e^{8x}(C_1+C_2 x)-\frac{1}{32}x^2-\frac{1}{64}x-\frac{3}{1024}-\frac{5}{9}e^{5x}$