Question

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

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{12}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{8}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

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

Step-by-step solution

Determine the general equation of the higher – order differential equation

The auxiliary equation is

$m^3-6m^2+12m-8=0$

By factoring, we find out the roots of the auxiliary equation as

$$\begin{gathered} m^3-6m^2+12m-8=0 \\ {m}^3-2m^2-4m^2+8m+4m-8=0 \\ {m}^2(m-2)-4m(m-2)+4(m-2)=0 \\ (m-2)\left(m^2-4m+4\right)=0 \\ (m-2)(m-2{)}^2=0 \\ (m-2{)}^3=0 \end{gathered}$$

Hence, roots of the auxiliary equation are $m_1=m_2=m_3=2$ and from Case II.

Repeated real roots, the general solution of the given differential equation is

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

Hence, the general solution is$\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{2}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathit{x}{\mathit{e}}^{\mathbf{2}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{3}}{\mathit{x}}^{\mathbf{2}}{\mathit{e}}^{\mathbf{2}\mathbf{x}}$