Question

Find a general solution to the givenhomogeneous equation.$(D+4)(D-3){(D+2)}^3{(D^2+4D+5)}^2{D}^5\left[y\right]=0$

Short answer

The answer to this problem is:

$y=c_1{e}^{-4x}+c_2{e}^{3x}+(c_3+c_{4}x+c_5{x}^2)e^{-2x}+(c_6+c_{7}x)e^{-2x}\cos x+(c_8+c_{9}x)\sin x+c_{10}+c_{11}x+c_{12}{x}^2+c_{13}{x}^3+c_{14}{x}^4$

Step-by-step solution

Homogenous Equation

A homogeneous system of linear equations is one in which all of the constant terms are zero. A homogeneous system always has at least one solution, namely the zero vector. When a row operation is applied to a homogeneous system, the new system is still homogeneous.

Solving of Homogenous Equation:

We will have to solve the following homogenous equation here;

$\begin{array}{l}D+4)\left(D-3\right){\left(D+2\right)}^2{\left(D^2+4D+5\right)}^2{D}^5\left[y\right]=0\\ (r+4)(r-3){\left(r+2\right)}^3{\left(r^2+4r+5\right)}^2{r}^5=0\\ r_1=-4\\ r_2=3\\ r_{3,4,5}=-2\\ r^2+4r+5=0\\ r_{6,7,8,9}=\frac{-4\pm \sqrt{4^2-4.5}}{2}\\ =-2\pm i\\ r_{10,11,12,13,14}=0\\ \\ y=c_1{e}^{-4x}+c_2{e}^{3x}+(c_3+c_{4}x+c_5{x}^2)e^{-2x}+(c_6+c_{7}x)e^{-2x}\cos x+(c_8+c_{9}x)\sin x+c_{10}+c_{11}x+c_{12}{x}^2+c_{13}{x}^3+c_{14}{x}^4\end{array}$

Hence, the final answer is:

$y=c_1{e}^{-4x}+c_2{e}^{3x}+(c_3+c_{4}x+c_5{x}^2)e^{-2x}+(c_6+c_{7}x)e^{-2x}\cos x+(c_8+c_{9}x)\sin x+c_{10}+c_{11}x+c_{12}{x}^2+c_{13}{x}^3+c_{14}{x}^4$