# What is the auxiliary equation for the differential equation x^2{y}'' +3x{y}'+y= 6?

## Question:

What is the auxiliary equation for the differential equation {eq}x^2{y}'' +3x{y}'+y= 6 {/eq}?

## Auxiliary Equation:

Consider the linear differential equation

{eq}\displaystyle ({x^2}{D^2} + axD + b)y = Q(x) {/eq}

Where a and b are the constants and Q(x) is a function of x and {eq}D = \frac{d}{{dx}}. {/eq}

For solving this equation we have to reduce it into linear differential equation with constant coefficients.

Let {eq}z = \log x, x = {e^z} {/eq}

{eq}{x^2}{D^2} = D'(D' - 1) {/eq}

{eq}xD = D' {/eq}

where {eq}D' = \frac{d}{{dz}} {/eq}

Putting above values in the Linear differential equation with variable coefficients.

Consider {eq}\displaystyle f(D') = 0 {/eq}

Replace D' by m we get auxiliary equation

{eq}\displaystyle f(m) = 0 {/eq}

Given differential equation is {eq}\displaystyle {x^2}y'' + 3xy' + y = 6 {/eq}.

It can be written in the form of

{eq}\displaystyle \left( {{x^2}{D^2} + 3xD + 1} \right)y = 6 {/eq}

Where {eq}\displaystyle D = \frac{d}{{dx}} {/eq}

Let {eq}\displaystyle z = \log x \Rightarrow x = {e^z} {/eq}

and {eq}\displaystyle {x^2}{D^2} = D'(D' - 1) {/eq}

{eq}\displaystyle xD = D' {/eq}

Where, {eq}\displaystyle D' = \frac{d}{{dz}} {/eq}

Plugging these substitutions in the given differential equation and we get

{eq}\displaystyle \left( {D'(D' - 1) + 3D' + 1} \right)y = 6 {/eq}

{eq}\displaystyle \left( {D{'^2} + 2D' + 1} \right)y = 6 {/eq}

{eq}\displaystyle f(D')y = 6 {/eq}

Where {eq}\displaystyle f(D') = \left( {D{'^2} + 2D' + 1} \right) {/eq}

Now we will find auxiliary equation.

Consider {eq}\displaystyle f(D') = 0 {/eq}

Replace D' by m we get

{eq}\displaystyle f(m) = 0 {/eq}

Then {eq}\displaystyle {m^2} + 2m + 1 = 0 {/eq}

{eq}\displaystyle {\left( {m + 1} \right)^2} = 0 {/eq}.

which is required auxiliary equation.