# Differential Equations

## Second Order Equations # Second Order Euler Equation

A second order linear differential equation of the form

${x^2}y^{\prime\prime} + Axy' + By = 0,\;\;\; {x \gt 0}$

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:

$\cancel{e^{2t}}\cancel{e^{ - 2t}}\left( {\frac{{{d^2}y}}{{d{t^2}}} - \frac{{dy}}{{dt}}} \right) + A\cancel{e^t}\cancel{e^{ - t}}\frac{{dy}}{{dt}} + By = 0,\;\; \Rightarrow \frac{{{d^2}y}}{{d{t^2}}} - \frac{{dy}}{{dt}} + A\frac{{dy}}{{dt}} + By = 0,\;\; \Rightarrow \frac{{{d^2}y}}{{d{t^2}}} + \left( {A - 1} \right)\frac{{dy}}{{dt}} + By = 0.$

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

$\frac{{dy}}{{dx}} = k{x^{k - 1}},\;\; \frac{{{d^2}y}}{{d{x^2}}} = k\left( {k - 1} \right){x^{k - 2}}.$

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

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

The non-homogeneous Euler equation is written as

${x^2}\frac{{{d^2}y}}{{d{x^2}}} + Ax\frac{{dy}}{{dx}} + By = f\left( x \right),\;\;{x \gt 0}.$

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:

1. Find the general solution of the homogeneous Euler equation;
2. 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;
3. 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 $$2\text{).}$$

See solved problems on Page 2.