91Ó°ÊÓ

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.

The given differential equation is ${(D−1)}^2(D−6)(D+5)(D^2+1)(D^2+4)\left[y\right]=0$. To solve this equation we look at its auxillary equation which is .${(m+1)}^2{(m−6)}^3(m+5)(m^2+1)(m^2+4)=0$

The complete set of solution of auxillary equation is $\{1−,−1,6,66,−5,i,−i,2i,−2i\}$

To conclude that the general solution of the given differential equation is $y=C_1{e}^{−x}+C_{2}x{e}^{−x}+C_3{e}^{6x}+C_{4}x{e}^{6x}+C_5{x}^2{e}^{6x}+C_6{e}^{−5x}+C_7\cos x+C_8\sin x+C_9\cos 2x+C_{10}\sin 2xy=C_1{e}^{−x}+C_{2}x{e}^{−x}+C_3{e}^{6x}+C_{4}x{e}^{6x}+C_5{x}^2{e}^{6x}+C_6{e}^{−5x}+C_7\cos x+C_8\sin x+C_9\cos 2x+C_{10}\sin 2x$, where $C_i(1≤i≤10)$ are arbitrary constants.

The general solution of the given differential equation is$y=C_1{e}^{−x}+C_{2}x{e}^{−x}+C_3{e}^{6x}+C_{4}x{e}^{6x}+C_5{x}^2{e}^{6x}+C_6{e}^{−5x}+C_7\cos x+C_8\sin x+C_9\cos 2x+C_{10}\sin 2x$, where $C_i(1≤i≤7)$ are arbitrary constant.

Hence, the final answer is:

$y=C_1{e}^{−x}+C_{2}x{e}^{−x}+C_3{e}^{6x}+C_{4}x{e}^{6x}+C_5{x}^2{e}^{6x}+C_6{e}^{−5x}+C_7\cos x+C_8\sin x+C_9\cos 2x+C_{10}\sin 2x$