A second order linear differential equation of the form

\[{{x^2}y^{\prime\prime} + Axy’ + By = 0,\;\;\;}\kern-0.3pt{{x \gt 0}}\]

is called the Euler differential equation. It can be reduced to the linear homogeneous differential equation with constant coefficients. This conversion can be done in two ways.

### First Way of Solving an Euler Equation

We make the following substitution: \(x = {e^t}.\) Then the derivatives will be

\[

{y’ = \frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} }

= {\frac{{\frac{{dy}}{{dt}}}}{{{e^t}}} }

= {{e^{ – t}}\frac{{dy}}{{dt}},}

\]

\[

{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}}}}{{\frac{{dx}}{{dt}}}}\left( {{e^{ – t}}\frac{{dy}}{{dt}}} \right) }

= {\frac{{ – {e^{ – t}}\frac{{dy}}{{dt}} + {e^{ – t}}\frac{{{d^2}y}}{{d{t^2}}}}}{{{e^t}}} }

= {{e^{ – 2t}}\left( {\frac{{{d^2}y}}{{d{t^2}}} – \frac{{dy}}{{dt}}} \right).}

\]

Putting this into the original Euler equation gives:

\[\require{cancel}

{{\cancel{e^{2t}}\cancel{e^{ – 2t}}\left( {\frac{{{d^2}y}}{{d{t^2}}} – \frac{{dy}}{{dt}}} \right) }+{ A\cancel{e^t}\cancel{e^{ – t}}\frac{{dy}}{{dt}} + By = 0,\;\;}}\Rightarrow

{{\frac{{{d^2}y}}{{d{t^2}}} – \frac{{dy}}{{dt}} }+{ A\frac{{dy}}{{dt}} + By = 0,\;\;}}\Rightarrow

{{\frac{{{d^2}y}}{{d{t^2}}} + \left( {A – 1} \right)\frac{{dy}}{{dt}} }+{ By = 0.}}

\]

As it can be seen, we obtain the linear equation with constant coefficients. The corresponding characteristic equation has the form:

\[{k^2} + \left( {A – 1} \right)k + B = 0.\]

Now we can determine the roots of the characteristic equation and write the general solution for the function \(y\left( t \right).\) Then we can easily return to the function \(y\left( x \right)\) taking into account that

\[y\left( t \right) = y\left( {\ln x} \right).\]

### Second Way of Solving an Euler Equation

In the second method we look for a solution of the equation in the form of the power function \(y = {x^k},\) where \(k\) is an unknown number. It follows from here that

\[{\frac{{dy}}{{dx}} = k{x^{k – 1}},\;\;}\kern-0.3pt{\frac{{{d^2}y}}{{d{x^2}}} = k\left( {k – 1} \right){x^{k – 2}}.}\]

Substituting into the differential equation gives the following result:

\[

{{{x^2}k\left( {k – 1} \right){x^{k – 2}} }+{ Axk{x^{k – 1}} + B{x^k} = 0,\;\;}}\Rightarrow

{{k\left( {k – 1} \right){x^k} }+{ Ak{x^k} + B{x^k} = 0,\;\;}}\Rightarrow

{\left[ {k\left( {k – 1} \right) + Ak + B} \right]{x^k} }={ 0.}

\]

As \({x^k} \ne 0,\) then

\[

{k\left( {k – 1} \right) + Ak + B = 0,\;\;}\Rightarrow

{{k^2} + \left( {A – 1} \right)k + B = 0.}

\]

We get the same characteristic equation as in the first way. After finding the roots, one can write the general solution of the differential equation.

### Non-homogeneous Euler Equation

The non-homogeneous Euler equation is written as

\[

{{x^2}\frac{{{d^2}y}}{{d{x^2}}} + Ax\frac{{dy}}{{dx}} + By }={ f\left( x \right),\;\;}\kern-0.3pt

{{x \gt 0}.}

\]

If the right side has the form

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

we can easily construct the general solution similarly to the method of solving linear non-homogeneous differential equations with constant coefficients. The algorithm of the solution looks as follows:

- Find the general solution of the homogeneous Euler equation;
- Using the method of undetermined coefficients or the method of variation of constants, find a particular solution depending on the right side of the given non-homogeneous equation;
- The general solution of the non-homogeneous equation is the sum of the general solution of the homogeneous equation (step \(1\)) and a particular solution of the non-homogeneous equation (step \(2\text{).}\)

## Solved Problems

Click a problem to see the solution.