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,\;\;}\kern-0.3pt

{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 \(t:\) \(Dy = {\large\frac{{dy}}{{dt}}\normalsize}.\) 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) + } \right.}\kern0pt

{\left. {{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}} }\right.}+{\left.{ \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

\[{{y^{\left( n \right)}} }={ {e^{ – nt}}\left[ {D\left( {D – 1} \right)\left( {D – 2} \right) \cdots }\right.}\kern0pt{\left.{ \left( {D – n + 1} \right)} \right]y.}\]

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:

\[y’ = k{x^{k – 1}},\]

\[y^{\prime\prime} = k\left( {k – 1} \right){x^{k – 2}},\]

\[{y^{\prime\prime\prime} = k\left( {k – 1} \right) \cdot}\kern0pt {\left( {k – 2} \right){x^{k – 3}},}\]

\[ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots\]

\[

{{y^{\left( n \right)}} }

= {\left[ {k\left( {k – 1} \right) \cdots }\right.}\kern0pt{\left.{ \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 }\kern0pt{ \left( {k – n + s + 1} \right)} \right]} }+{ {a_n} = 0,\;\;}\kern-0.3pt

{\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

\[ {{x^n}{y^{\left( n \right)}}\left( x \right) }+{ {a_1}{x^{n – 1}}{y^{\left( {n – 1} \right)}}\left( x \right) + \cdots } + {{a_{n – 1}}xy’\left( x \right) }+{ {a_n}y\left( x \right) = f\left( x \right),\;\;}\kern-0.3pt {x \gt 0.} \]

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.

## Solved Problems

Click or tap a problem to see the solution.