Question

Find a general solution to the givenhomogeneous equation.${(D+1)}^2{(D-6)}^3(D+5)(D^2+1){(D^2+4)}^2\left[y\right]=0$

Short answer

The general solution to the homogeneous equation is:

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

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:

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 .${(m+1)}^2{(m-6)}^3(m+5)(m^2+1)(m^2+4)=0$

 Step 3: Solving for general equation:

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 $C_i(1\le i\le 10)$ 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 $C_i(1\le i\le 7)$ are arbitrary constant.

Hence, the final answer is:

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