Question

Find a general solution to the givenhomogeneous equation.

${(D-1)}^3(D-2)(D^2+D+1){(D^2+6D+10)}^3\left[y\right]=0$

Short answer

The answer to this problem is:

$\begin{array}{l}y=c_1{e}^x+c_{2}x{e}^x+c_3{x}^2{e}^x+c_4{e}^{2x}+c_5{e}^{-\frac{x}{2}}\cos \frac{\sqrt{3}x}{2}+c_6{e}^{-\frac{x}{2}}\sin \frac{\sqrt{3}x}{2}+c_7{e}^{-3x}\cos x\\ +c_{8}x{e}^{-3x}\cos x+c_9{x}^2{e}^{-3x}\cos x+c_{10}{e}^{-3x}\sin x+c_{11}x{e}^{-3x}\sin x+c_{12}{x}^2{e}^{-3x}\sin x\end{array}$

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;

$\begin{array}{l}{\left(r-1\right)}^3(r-2)(r^2+r+1){\left(r^2+6r+10\right)}^3=0\\ r_1=1,r_2=2,r_3,r_4=-\frac{1}{2}\pm \frac{\sqrt{3}}{2}i,r_{5,6}=-3\pm i\end{array}$

Roots 1,-3+I and -3-I have multiplicity 3, while rest of all the roots are simple:

$\begin{array}{l}y=c_1{e}^x+c_{2}x{e}^x+c_3{x}^2{e}^x+c_4{e}^{2x}+c_5{e}^{-\frac{x}{2}}\cos \frac{\sqrt{3}x}{2}+c_6{e}^{-\frac{x}{2}}\sin \frac{\sqrt{3}x}{2}+c_7{e}^{-3x}\cos x\\ +c_{8}x{e}^{-3x}\cos x+c_9{x}^2{e}^{-3x}\cos x+c_{10}{e}^{-3x}\sin x+c_{11}x{e}^{-3x}\sin x+c_{12}{x}^2{e}^{-3x}\sin x\end{array}$

Hence, the final answer is:

$\begin{array}{l}y=c_1{e}^x+c_{2}x{e}^x+c_3{x}^2{e}^x+c_4{e}^{2x}+c_5{e}^{-\frac{x}{2}}\cos \frac{\sqrt{3}x}{2}+c_6{e}^{-\frac{x}{2}}\sin \frac{\sqrt{3}x}{2}+c_7{e}^{-3x}\cos x\\ +c_{8}x{e}^{-3x}\cos x+c_9{x}^2{e}^{-3x}\cos x+c_{10}{e}^{-3x}\sin x+c_{11}x{e}^{-3x}\sin x+c_{12}{x}^2{e}^{-3x}\sin x\end{array}$