# Higher Order Euler Equation

## Trigonometry # Higher Order Euler Equation

The $$n$$th order Euler equation can be written as

${x^n}{y^{\left( n \right)}} + {a_1}{x^{n - 1}}{y^{\left( {n - 1} \right)}} + \cdots + {a_{n - 1}}xy' + {a_n}y = 0,\;\; x \gt 0,$

where $${a_1}, \ldots ,{a_n}$$ are constants.

We have previously considered the second-order Euler equation. With some substitutions, this equation reduces to a homogeneous linear differential equation with constant coefficients. Such transformations are also used in the case of the $$n$$th order equation. Let us consider two methods for solving equations of this type.

## $$1.$$ Solving the $$N$$th Order Euler Equation Using the Substitution $$x = {e^t}$$

With the substitution $$x = {e^t},$$ the $$n$$th order Euler equation can be reduced to an equation with constant coefficients. We express the derivative of $$y$$ in terms of the new variable $$t.$$ This is conveniently done using the differential operator $$D.$$ In the formulas below, the operator $$D$$ denotes the first derivative with respect to $$t:$$ $$Dy = {\frac{{dy}}{{dt}}}.$$ Thus, we obtain:

$y' = \frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{\frac{{dy}}{{dt}}}}{{{e^t}}} = {e^{ - t}}\frac{{dy}}{{dt}} = {e^{ - t}}Dy,$
$y^{\prime\prime} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{d}{{dx}}\left( {{e^{ - t}}\frac{{dy}}{{dt}}} \right) = \frac{{\frac{d}{{dt}}}}{{{e^t}}}\left( {{e^{ - t}}\frac{{dy}}{{dt}}} \right) = {e^{ - t}}\frac{d}{{dt}}\left( {{e^{ - t}}\frac{{dy}}{{dt}}} \right) = {e^{ - t}}\left( { - {e^{ - t}}\frac{{dy}}{{dt}} + {e^{ - t}}\frac{{{d^2}y}}{{d{t^2}}}} \right) = {e^{ - 2t}}\left( {{D^2} - D} \right)y = {e^{ - 2t}}\left[ {D\left( {D - 1} \right)} \right]y,$
$y^{\prime\prime\prime} = \frac{d}{{dx}}\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right) = \frac{d}{{dx}}\left[ {{e^{ - 2t}}\left( {\frac{{{d^2}y}}{{d{t^2}}} - \frac{{dy}}{{dt}}} \right)} \right] = \frac{{\frac{d}{{dt}}}}{{{e^t}}}\left[ {{e^{ - 2t}}\left( {\frac{{{d^2}y}}{{d{t^2}}} - \frac{{dy}}{{dt}}} \right)} \right] = {e^{ - t}}\left[ { - 2{e^{ - 2t}}\left( {\frac{{{d^2}y}}{{d{t^2}}} - \frac{{dy}}{{dt}}} \right) + {e^{ - 2t}}\left( {\frac{{{d^3}y}}{{d{t^3}}} - \frac{{{d^2}y}}{{d{t^2}}}} \right)} \right] = {e^{ - 3t}}\left( { - 2\frac{{{d^2}y}}{{d{t^2}}} + 2\frac{{dy}}{{dt}} + \frac{{{d^3}y}}{{d{t^3}}} - \frac{{{d^2}y}}{{d{t^2}}}} \right) = {e^{ - 3t}}\left( {\frac{{{d^3}y}}{{d{t^3}}} - 3\frac{{{d^2}y}}{{d{t^2}}} + 2\frac{{dy}}{{dt}}} \right) = {e^{ - 3t}}\left( {{D^3} - 3{D^2} + 2D} \right)y = {e^{ - 3t}}\left[ {D\left( {D - 1} \right)\left( {D - 2} \right)} \right]y.$

The derivative of an arbitrary $$n$$th order with respect to $$t$$ is described by the expression

${y^{\left( n \right)}} = {e^{ - nt}}\left[ {D\left( {D - 1} \right)\left( {D - 2} \right) \cdots \left( {D - n + 1} \right)} \right]y.$

It is seen that after the substitution of the derivatives in the original Euler equation all the exponential factors are eliminated because

${x^n} = {e^{nt}}.$

As a result, the left side will consist of the derivatives of the function $$y$$ with respect to $$t$$ with constant coefficients. The general solution of this equation can be found by standard methods. At the end of the solution it is necessary to go back from $$t$$ to the original independent variable $$x$$ substituting $$t = \ln x.$$

## (2.\) Solving the $$N$$th Order Euler Equation Using the Power Function $$y = {x^k}$$

Consider another way of solving the Euler equations. Assume that the solution has the form of the power function $$y = {x^k},$$ where the index $$k$$ is defined in the course of the solution. The derivatives of the function $$y$$ can easily be expressed as follows:

$y' = k{x^{k - 1}},$
$y^{\prime\prime} = k\left( {k - 1} \right){x^{k - 2}},$
$y^{\prime\prime\prime} = k\left( {k - 1} \right) \left( {k - 2} \right){x^{k - 3}},$
$\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots$
${y^{\left( n \right)}} = \left[ {k\left( {k - 1} \right) \cdots \left( {k - n + 1} \right)} \right]{x^{k - n}}.$

Substituting this into the initial homogeneous Euler equation and cancelling by $$y = {x^k} \ne 0,$$ we immediately obtain the auxiliary equation:

$\left[ {k\left( {k - 1} \right) \cdots \left( {k - n + 1} \right)} \right] + {a_1}\left[ {k\left( {k - 1} \right) \cdots \left( {k - n + 2} \right)} \right] + \cdots + {a_{n - 1}}k + {a_n} = 0,$

which can be written in a more compact form as

$\sum\limits_{s = 0}^{n - 1} {{a_s}\left[ {k\left( {k - 1} \right) \cdots \left( {k - n + s + 1} \right)} \right]} + {a_n} = 0,\;\; \text{where}\;\;{a_0} = 1.$

Solving the auxiliary equation, we find its roots and then construct the general solution of differential equation. In the final expression we must return to the original variable $$x$$ using the substitution $$t = \ln x.$$

## Higher Order Nonhomogeneous Euler Equation

In the general case, the nonhomogeneous Euler equation can be represented as

${x^n}{y^{\left( n \right)}}\left( x \right) + {a_1}{x^{n - 1}}{y^{\left( {n - 1} \right)}}\left( x \right) + \cdots + {a_{n - 1}}xy'\left( x \right) + {a_n}y\left( x \right) = f\left( x \right),\;\; x \gt 0.$

Using the substitution $$y = {e^t},$$ any nonhomogeneous Euler equation can be transformed into a nonhomogeneous linear equation with constant coefficients. Moreover, if the right-hand side of the original equation has the form

$f\left( x \right) = {x^\alpha }{P_m}\left( {\ln x} \right),$

where $${P_m}$$ is a polynomial of degree $$m,$$ then the resulting particular solution of the nonhomogeneous equation can be found by the method of undetermined coefficients.