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### Section 6.4 : Euler Equations

In this section we want to look for solutions to

\[\begin{equation}a{x^2}y'' + bxy' + cy = 0\label{eq:eq1}\end{equation}\]around \({x_0} = 0\). These types of differential equations are called **Euler Equations**.

Recall from the previous section that a point is an ordinary point if the quotients,

\[\frac{{bx}}{{a{x^2}}} = \frac{b}{{ax}}\hspace{0.25in}{\mbox{and }}\hspace{0.25in}\frac{c}{{a{x^2}}}\]have Taylor series around \({x_0} = 0\). However, because of the \(x\) in the denominator neither of these will have a Taylor series around \({x_0} = 0\) and so \({x_0} = 0\) is a singular point. So, the method from the previous section won’t work since it required an ordinary point.

However, it is possible to get solutions to this differential equation that aren’t series solutions. Let’s start off by assuming that \(x>0\) (the reason for this will be apparent after we work the first example) and that all solutions are of the form,

\[\begin{equation}y\left( x \right) = {x^r}\label{eq:eq2}\end{equation}\]Now plug this into the differential equation to get,

\[\begin{align*}a{x^2}\left( r \right)\left( {r - 1} \right){x^{r - 2}} + bx\left( r \right){x^{r - 1}} + c{x^r} & = 0\\ ar\left( {r - 1} \right){x^r} + b\left( r \right){x^r} + c{x^r} & = 0\\ \left( {ar\left( {r - 1} \right) + b\left( r \right) + c} \right){x^r} & = 0\end{align*}\]Now, we assumed that \(x>0\) and so this will only be zero if,

\[\begin{equation}ar\left( {r - 1} \right) + b\left( r \right) + c = 0\label{eq:eq3}\end{equation}\]So solutions will be of the form \(\eqref{eq:eq2}\) provided \(r\) is a solution to \(\eqref{eq:eq3}\). This equation is a quadratic in \(r\) and so we will have three cases to look at : Real, Distinct Roots, Double Roots, and Complex Roots.

#### Real, Distinct Roots

There really isn’t a whole lot to do in this case. We’ll get two solutions that will form a fundamental set of solutions (we’ll leave it to you to check this) and so our general solution will be,

\[y\left( x \right) = {c_1}{x^{{r_1}}} + {c_2}{x^{{r_2}}}\]We first need to find the roots to \(\eqref{eq:eq3}\).

\[\begin{align*}2r\left( {r - 1} \right) + 3r - 15 & = 0\\ 2{r^2} + r - 15 = \left( {2r - 5} \right)\left( {r + 3} \right) & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}{r_1} = \frac{5}{2},\,\,\,{r_2} = - 3\end{align*}\]The general solution is then,

\[y\left( x \right) = {c_1}{x^{\frac{5}{2}}} + {c_2}{x^{ - 3}}\]To find the constants we differentiate and plug in the initial conditions as we did back in the second order differential equations chapter.

\[y'\left( x \right) = \frac{5}{2}{c_1}{x^{\frac{3}{2}}} - 3{c_2}{x^{ - 4}}\] \[\left. \begin{align*}& 0 = y\left( 1 \right) = {c_1} + {c_2}\\ & 1 = y'\left( 1 \right) = \frac{5}{2}{c_1} - 3{c_2}\end{align*} \right\}\hspace{0.25in} \Rightarrow \hspace{0.25in}{c_1} = \frac{2}{{11}},\,\,{c_2} = - \frac{2}{{11}}\]The actual solution is then,

\[y\left( x \right) = \frac{2}{{11}}{x^{\frac{5}{2}}} - \frac{2}{{11}}{x^{ - 3}}\]With the solution to this example we can now see why we required \(x>0\). The second term would have division by zero if we allowed \(x=0\) and the first term would give us square roots of negative numbers if we allowed \(x<0\).

#### Double Roots

This case will lead to the same problem that we’ve had every other time we’ve run into double roots (or double eigenvalues). We only get a single solution and will need a second solution. In this case it can be shown that the second solution will be,

\[{y_2}\left( x \right) = {x^r}\ln x\]and so the general solution in this case is,

\[y\left( x \right) = {c_1}{x^r} + {c_2}{x^r}\ln x = {x^r}\left( {{c_1} + {c_2}\ln x} \right)\]We can again see a reason for requiring \(x>0\). If we didn’t we’d have all sorts of problems with that logarithm.

First the roots of \(\eqref{eq:eq3}\).

\[\begin{align*}r\left( {r - 1} \right) - 7r + 16 & = 0\\ {r^2} - 8r + 16 & = 0\\ {\left( {r - 4} \right)^2} & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}r = 4\end{align*}\]So, the general solution is then,

\[y\left( x \right) = {c_1}{x^4} + {c_2}{x^4}\ln x\]#### Complex Roots

In this case we’ll be assuming that our roots are of the form,

\[{r_{1,2}} = \lambda \pm \mu \,i\]If we take the first root we’ll get the following solution.

\[{x^{\lambda + \mu \,i}}\]This is a problem since we don’t want complex solutions, we only want real solutions. We can eliminate this by recalling that,

\[{x^r} = {{\bf{e}}^{\ln {x^r}}} = {{\bf{e}}^{r\ln x}}\]Plugging the root into this gives,

\[\begin{align*}{x^{\lambda + \mu \,i}} & = {{\bf{e}}^{\left( {\lambda + \mu \,i} \right)\ln x}}\\ & = {{\bf{e}}^{\lambda \ln x}}{{\bf{e}}^{\mu \,i\ln x}}\\ & = {{\bf{e}}^{\ln {x^\lambda }}}\left( {\cos \left( {\mu \ln x} \right) + i\sin \left( {\mu \ln x} \right)} \right)\\ & = {x^\lambda }\cos \left( {\mu \ln x} \right) + i{x^\lambda }\sin \left( {\mu \ln x} \right)\end{align*}\]Note that we had to use Euler formula as well to get to the final step. Now, as we’ve done every other time we’ve seen solutions like this we can take the real part and the imaginary part and use those for our two solutions.

So, in the case of complex roots the general solution will be,

\[y\left( x \right) = {c_1}{x^\lambda }\cos \left( {\mu \ln x} \right) + {c_2}{x^\lambda }\sin \left( {\mu \ln x} \right) = {x^\lambda }\left( {{c_1}\cos \left( {\mu \ln x} \right) + {c_2}\sin \left( {\mu \ln x} \right)} \right)\]Once again, we can see why we needed to require \(x > 0\).

Get the roots to \(\eqref{eq:eq3}\) first as always.

\[\begin{align*}r\left( {r - 1} \right) + 3r + 4 & = 0\\ {r^2} + 2r + 4 & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}{r_{1,2}} = - 1 \pm \sqrt 3 \,i\end{align*}\]The general solution is then,

\[y\left( x \right) = {c_1}{x^{ - 1}}\cos \left( {\sqrt 3 \ln x} \right) + {c_2}{x^{ - 1}}\sin \left( {\sqrt 3 \ln x} \right)\]We should now talk about how to deal with \(x < 0\) since that is a possibility on occasion. To deal with this we need to use the variable transformation,

\[\eta = - x\]In this case since \(x < 0\) we will get \(\eta > 0\). Now, define,

\[u\left( \eta \right) = y\left( x \right) = y\left( { - \eta } \right)\]Then using the chain rule we can see that,

\[u'\left( \eta \right) = - y'\left( x \right)\hspace{0.25in}{\mbox{and}}\hspace{0.25in}u''\left( \eta \right) = y''\left( x \right)\]With this transformation the differential equation becomes,

\[\begin{align*}a{\left( { - \eta } \right)^2}u'' + b\left( { - \eta } \right)\left( { - u'} \right) + cu & = 0\\ a{\eta ^2}u'' + b\eta u' + cu & = 0\end{align*}\]In other words, since \(\eta>0\) we can use the work above to get solutions to this differential equation. We’ll also go back to \(x\)’s by using the variable transformation in reverse.

\[\eta = - x\]Let’s just take the real, distinct case first to see what happens.

\[\begin{align*}u\left( \eta \right) & = {c_1}{\eta ^{{r_1}}} + {c_2}{\eta ^{{r_2}}}\\ y\left( x \right) & = {c_1}{\left( { - x} \right)^{{r_1}}} + {c_2}{\left( { - x} \right)^{{r_2}}}\end{align*}\]Now, we could do this for the rest of the cases if we wanted to, but before doing that let’s notice that if we recall the definition of absolute value,

\[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}x&{{\mbox{if }}x \ge 0}\\{ - x}&{{\mbox{if }}x < 0}\end{array}} \right.\]we can combine both of our solutions to this case into one and write the solution as,

\[y\left( x \right) = {c_1}{\left| x \right|^{{r_1}}} + {c_2}{\left| x \right|^{{r_2}}},\hspace{0.25in}x \ne 0\]Note that we still need to avoid \(x = 0\) since we could still get division by zero. However, this is now a solution for any interval that doesn’t contain \(x = 0\).

We can do likewise for the other two cases and the following solutions for any interval not containing \(x = 0\).

\[\begin{align*}y\left( x \right) & = {c_1}{\left| x \right|^r} + {c_2}{\left| x \right|^r}\ln \left| x \right|\\ y\left( x \right) & = {c_1}{\left| x \right|^\lambda }\cos \left( {\mu \ln \left| x \right|} \right) + {c_2}{\left| x \right|^\lambda }\sin \left( {\mu \ln \left| x \right|} \right)\end{align*}\]We can make one more generalization before working one more example. A more general form of an Euler Equation is,

\[a{\left( {x - {x_0}} \right)^2}y'' + b\left( {x - {x_0}} \right)y' + cy = 0\]and we can ask for solutions in any interval not containing \(x = {x_0}\). The work for generating the solutions in this case is identical to all the above work and so isn’t shown here.

The solutions in this general case for any interval not containing \(x = a\) are,

\[\begin{align*}y\left( x \right) & = {c_1}{\left| {x - a} \right|^{{r_1}}} + {c_2}{\left| {x - a} \right|^{{r_2}}}\\ y\left( x \right) & = {\left| {x - a} \right|^r}\left( {{c_1} + {c_2}\ln \left| {x - a} \right|} \right)\\ y\left( x \right) & = {\left| {x - a} \right|^\lambda }\left( {{c_1}\cos \left( {\mu \ln \left| {x - a} \right|} \right) + {c_2}\sin \left( {\mu \ln \left| {x - a} \right|} \right)} \right)\end{align*}\]Where the roots are solutions to

\[ar\left( {r - 1} \right) + b\left( r \right) + c = 0\]** Example 4 **Find the solution to the following differential equation on any interval not containing \(x = - 6\).

*Solution*

So, we get the roots from the identical quadratic in this case.

\[\begin{align*}3r\left( {r - 1} \right) + 25r - 16 & = 0\\ 3{r^2} + 22r - 16 & = 0\\ \left( {3r - 2} \right)\left( {r + 8} \right) & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}{r_1} = \frac{2}{3},\,\,{r_2} = - 8\end{align*}\]The general solution is then,

\[y\left( x \right) = {c_1}{\left| {x + 6} \right|^{\frac{2}{3}}} + {c_2}{\left| {x + 6} \right|^{ - 8}}\]