Question

If $y=1-x+6x^2+3e^x$is a solution of a homogeneous fourth-order linear differential equation with constant coefficients, then the roots of the auxiliary equation are ______.

Short answer

The roots of the auxiliary equation is ${\text{y}}^{\text{(4)}}{\text{- y''' = 0}}_{}$.

Step-by-step solution

Define a solution of the differential equation.

A differential equation solution is a statement for the dependent variable in terms of one or more independent variables that fulfils the relation. All feasible solutions are included in the general solution, which mainly contains arbitrary constants or arbitrary functions.

A solution of the differential equation is y=-5e-x+10ex.

Let the general solution be${\text{y = 1 - x + 6x}}^{\text{2}}{\text{+ 3e}}^{\text{x}}$, which is in the form of${\text{y =c}}_{\text{1}}{\text{y}}_{\text{1}}{\text{+c}}_{\text{2}}{\text{y}}_{\text{2}}{\text{+c}}_{\text{3}}{\text{y}}_{\text{3}}{\text{+c}}_{\text{4}}{\text{y}}_{\text{4}}$.

Let the assumption of the solution of an DE be $y=x^m$. Then, the solution of the required DE are,

$$\begin{gathered} {\text{y}}_{\text{1}}{\text{= 1 =e}}^{\text{0}}{\text{=e}}^{{\text{m}}_{\text{1}}\text{x}} \\ {\text{y}}_{\text{2}}{\text{= x = xe}}^{\text{0}}{\text{= xe}}^{{\text{m}}_{\text{2}}\text{x}} \\ {\text{y}}_{\text{3}}{\text{=x}}^{\text{2}}{\text{=x}}^{\text{2}}{\text{e}}^{\text{0}}{\text{=x}}^{\text{2}}{\text{e}}^{{\text{m}}_{\text{3}}\text{x}} \\ {\text{y}}_{\text{4}}{\text{=e}}^{\text{x}}{\text{=e}}^{{\text{m}}_{\text{4}}\text{x}} \end{gathered}$$

Solve for roots:

$$\begin{gathered} {\text{m}}_{\text{1}}\text{= 0} \\ {\text{m}}_{\text{2}}\text{= 0} \\ {\text{m}}_{\text{3}}\text{= 0} \\ {\text{m}}_{\text{4}}\text{= 1} \end{gathered}$$

Hence, the auxiliary equation is,

$$\begin{gathered} m\times m\times m\left(m-1\right)=0 \\ {m}^3\left(m-1\right)=0 \\ {m}^4-m^3=0 \end{gathered}$$

Thus, the corresponding homogeneous equation is ${\text{y}}^{\text{(4)}}\text{- y''' = 0}$.