calculuspowerup.com

Second Order Euler Equations

April 5th, 2009 |

Author: Merrin

We will now study a new class of differential equations similar to constant coefficient equations called Euler equations or Euler differential equations.

$ax^{2} \cfrac{d^{2} y}{dx^{2} } +bx\cfrac{dy}{dx} +cy=0$

This equation looks similar to the constant coefficient equation counterpart but there are some powers of x offsetting the derivatives. Recall the power rule.

${\cfrac{d}{dx} x^{n} =nx^{n-1}}$
${\cfrac{d^{2} }{dx^{2} } x^{n} =n(n-1)x^{n-2} }$

The factors of x in the variable coefficients build y back up to its original power in x if y is a power function. With this in mind we try a trial solution of

$y=x^{\lambda }$

${a\lambda (\lambda -1)x^{2} x^{\lambda -2} +b\lambda xx^{\lambda -1} +cx^{\lambda } =0}$

${a\lambda (\lambda -1)x^{\lambda } +b\lambda x^{\lambda } +cx^{\lambda } =0}$
$(a\lambda^2+(b-a)\lambda+c)x^\lambda=0$

This gives us an algebraic equation to solve instead of a differential equation like the characteristic equation for constant coefficient differential equations.

${a\lambda ^{2} +(b-a)\lambda +c=0}$
${\lambda =-\cfrac{b-a}{2a} \pm \cfrac{\sqrt{(b-a)^{2} -4ac} }{2a}}$
${\lambda =\lambda _{1} ,\lambda _{2} }$

The general solution is therefore, at least when the roots are different,

$y=c_{1} x^{\lambda _{1} } +c_{2} x^{\lambda _{2} }$

Example 1. Find the solution to the following Euler equation

$x^{2} \cfrac{d^{2} y}{dx^{2} } +x\cfrac{dy}{dx} -9y=0$

Solution 1. We try a trial substitution of the form

$y=cx^{\lambda }$

This gives us an equation

${\lambda (\lambda -1)+\lambda -9=0}$
${\lambda =\pm 3}$

The general solution is

$y=c_{1} x^{3} +c_{2} x^{-3}$

Curious as to what happens if the root repeats itself? Do we just multiply the solutions by another factor of x like we found for constant coefficients? It is likely not because we are not dealing with exponentials but power laws.

Example 2. Build a second order Euler equation with a repeated root and then solve it.
Solution 2. Well we need the discriminant to be zero so we have.

$\lambda =-\cfrac{b-a}{2a} \pm \cfrac{\sqrt{(b-a)^{2} -4ac} }{2a}$

We know that our repeated root will have a value of (a-b)/(2a), now we just have to pick the values such that

${(b-a)^{2} -4ac=0}$
${{\rm Let:}\, c=1,\, a=1} \\ {(b-1)^{2} =4}$
${b=1\pm 2=3,-1}$

It doesn’t matter which we pick. Let us take b=3 then

$x^{2} \cfrac{d^{2} y}{dx^{2} } +3x\cfrac{dy}{dx} +y=0$

This equation has the repeated root of value

$-\cfrac{b-a}{2a} =-\cfrac{3-1}{2} =-1$

This would imply the solution would be

$y=c_{1} x^{-1} +c_{2} x^{-1}$

Since the order of the equation is two we expect two different solutions. We will try our trick to find the second one.

$y=cu(x)x^{-1}$

Substituting this into the original equation we find.

${x^{2} \cfrac{d^{2} }{dx^{2} } (ux^{-1} )+3x\cfrac{d}{dx} (ux^{-1} )+(ux^{-1} )=0}$

${x^{2} \cfrac{d}{dx} (u'x^{-1} -ux^{-2} )+3x(u'x^{-1} -ux^{-2} )+ux^{-1} =0}$
${x^{2} (u''x^{-1} -u'x^{-2} -u'x^{-2} +2ux^{-3} )+3x(u'x^{-1} -ux^{-2} )+ux^{-1} =0}$

${u''x+u'(-1-1+3)+u(2x^{-1} -3x^{-1} +1x^{-1} )=0}$

As a result of the differentiation we have a new differential equation of u to solve.

${xu''+u'=0}$
${\cfrac{u''}{u'} =-\cfrac{1}{x} }$
${\ln \left|u'\right|=-\ln \left|x\right|+C}$
${\left|u'\right|=e^{-\ln \left|x\right|+C} =\cfrac{D}{x} }$
${u(x)=A\ln \left|x\right|+B}$

So our general solution turns out to be.

$y=c_{1} x^{-1} +c_{2} x^{-1} \ln \left|x\right|$

More advanced theory shows that there is a factor of

$(\ln \left|x\right|)^{n}$

for every root repeated n times in higher order equations.

Example 3. If the Euler characteristic equation has roots

$\lambda = a\pm ib$

what does the form of the general solution look like?

Solution 3. We can use Euler’s formula to exponentiate a complex power function in terms of e rather than x.

$x^{a+ib}=e^{(a+ib)\ln\left|x\right|}=x^a(\cos (b\ln\left|x\right|)+i\sin(b\ln\left|x\right|))$

So the general solution will look like

$y = x^a(A\cos (b\ln\left|x\right|)+B\sin(b\ln\left|x\right|))$

when you combine both solutions.

Euler equations are a cute example of a type of differential equation that share many of the properties of constant coefficient differential equations, yet with their own twist. Remember which trial solution to use. Use

$x^{\lambda }$