Question

In Problems 15 to 28 find the general solution of the given higher-order differential equation.

y''' + 3y'' + 3y' + y = 0

Short answer

$y=c_1{e}^{-x}+c_{2}x{e}^{-x}+c_3{x}^2{e}^{-x}$

Step-by-step solution

Differentiate the equation

Consider $y=e^{mx}$as the solution of the differential equation,

Substitute $y=e^{mx},y\text{'}=m{e}^{mx}$and $y\text{'}\text{'}=m^2{e}^{mx},y\text{'}\text{'}\text{'}=m^3{e}^{mx}$into $y\text{'}\text{'}\text{'}+3y\text{'}\text{'}+3y\text{'}+y=0$to obtain the auxiliary equation.

Substitute in the equation

$$\begin{gathered} e^{mx}\left(m^3+3m^2+3m+1\right)=0 \\ {m}^3{e}^{mx}+3m^2{e}^{mx}+3m{e}^{mx}+e^{mx}=0 \end{gathered}$$

For any real$x,e^{mx}\ne 0$, we have

$$\begin{gathered} m^3+3m^2+3m+1=0 \\ (m+1)\left(m^2+2m+1\right)=0 \\ (m+1)(m+1)(m+1)=0 \\ (m+1{)}^3=0 \end{gathered}$$

Then the roots are $m_{1,2,3}=-1$which are real and repeated.

Find the general solution of the Higher-order differential equation

Finally, since we have repeated roots $m_{1,2,3}=-1$, then we need to have additional linear independent solutions for this repeated root, then we obtain the general solution of our differential equation $y\text{'}\text{'}\text{'}+3y\text{'}\text{'}+3y\text{'}+y=0$as

$$\begin{gathered} y=n{e}^{mx}\left(c+lx+s{x}^2\right) \\ =n{e}^{-x}\left(c+lx+s{x}^2\right) \\ =c_1{e}^{-x}+c_{2}x{e}^{-x}+c_3{x}^2{e}^{-x} \end{gathered}$$

Where we have $c_1=n\times c,c_2=n\times land{c}_3=n\times s$are arbitrary constants.

The general solution is $\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{x}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{3}}{\mathit{x}}^{\mathbf{2}}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$.