Differential Equations

2nd Order Equations

Second Order Euler Equation

Page 1
Theory
Page 2
Problems 1-4

A second order linear differental 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:

  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 \(2\text{).}\)

Solved Problems

Click on problem description to see solution.

 Example 1

Find the general solution of the differential equation \(4{x^2}y^{\prime\prime} + y = 0\) assuming that \(x \gt 0.\)

 Example 2

Find the general solution of the equation \({x^2}y^{\prime\prime} – xy’ – 8y \) \(= 0\) assuming that \(x \gt 0.\)

 Example 3

Find the general solution of the Euler equation \({x^2}y^{\prime\prime} + xy’ + y \) \(= 5{x^2}\) for \(x \gt 0.\)

 Example 4

Solve the non-homogeneous Euler equation \({x^2}y^{\prime\prime} – 2xy’ + 2y \) \(= 6{x^2} + 4\ln x\) assuming that \(x \gt 0.\)

Page 1
Theory
Page 2
Problems 1-4