Question

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

Short answer

The general solution to the homogeneous equation is:

$y=C_1{e}^x+C_{2}x{e}^x+C_3{e}^{-3x}+C_4{e}^{-x}\sin 2x+C_{5}x{e}^{-x}\sin 2x+C_6{e}^{-x}\cos 2x+C_{7}x{e}^{-x}\cos 2x$

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 ${(D-1)}^2(D+3){(D^2+2D+5)}^2\left[y\right]=0$. To solve this equation we look at its auxillary equation which is ${(m-1)}^2(m+3){(m^2+2m+5)}^2=0$. To get all the solution of its equation we need to solve$m^2+2m+5=0$ .

We have

$m=\frac{-2\pm \sqrt{4-20}}{2}=-1\pm 2i$

Solving for general equation:

The complete set of solution of auxillary equation is .$\{1,1,-3,-1+2i,-1+2i,-1-2i,-1-2i\}$

To conclude that the general solution of the given differential equation is $y=C_1{e}^x+C_{2}x{e}^x+C_3{e}^{-3x}+C_4{e}^{-x}\sin 2x+C_{5}x{e}^{-x}\sin 2x+C_6{e}^{-x}\cos 2x+C_{7}x{e}^{-x}\cos 2x$, where $C_i(1\le i\le 7)$ are arbitrary constants.

The general solution of the given differential equation is$y=C_1{e}^x+C_{2}x{e}^x+C_3{e}^{-3x}+C_4{e}^{-x}\sin 2x+C_{5}x{e}^{-x}\sin 2x+C_6{e}^{-x}\cos 2x+C_{7}x{e}^{-x}\cos 2x$ , where $C_i(1\le i\le 7)$ are arbitrary constant.

Hence, the final answer is:$y=C_1{e}^x+C_{2}x{e}^x+C_3{e}^{-3x}+C_4{e}^{-x}\sin 2x+C_{5}x{e}^{-x}\sin 2x+C_6{e}^{-x}\cos 2x+C_{7}x{e}^{-x}\cos 2x$