# Differential Equations

## Higher Order Equations # Higher Order Euler Equation

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

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

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.

### Example 1

Find the general solution of the equation $${x^3}y^{\prime\prime\prime} – 2{x^2}y^{\prime\prime}$$ $$+\; 4xy’ – 4y$$ $$= 0$$ for $$x \gt 0.$$

### Example 2

Find the general solution of the equation $${x^4}{y^{IV}} + 6{x^3}y^{\prime\prime\prime}$$ $$+\; 9{x^2}y^{\prime\prime}$$ $$+\; 3xy’ + y$$ $$= 0$$ for $$x \gt 0.$$

### Example 3

Solve the differential equation $${x^3}y^{\prime\prime\prime} – 2{x^2}y^{\prime\prime}$$ $$+\; 6xy’$$ $$= x\left( {2\ln x + 1} \right)$$ for $$x \gt 0.$$
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Concept
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Problems 1-3