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6.5: Euler Equations

( \newcommand{\kernel}{\mathrm{null}\,}\)

In this section, we will investigate the solutions of the most simple type of differential equations with regular singular points

x2y+axy+by=0

We can immediately see that 0 is a regular singular point of the differential equation since

xp(x)=aand x2q(x)=b

To solve the differential equation we assume that a solution is of the form

y=xr

Now we proceed in a similar way to how we tackled second order differential equations with constant coefficients. We take two derivatives.

y=rxr1

y=r(r1)xr2

Next plug these into the original differential equation

x2r(r1)xr2+axrxr1+bxr=0r(r1)xr+arxr+bxr=0Multiplying the exponentsr(r1)+ar+b=0Dividing by xrr2+(a1)r+b=0.

We define

F(r)=r2+(a1)r+b.

This is a quadratic in r. We will see as with second order homogeneous linear differential equations with constant coefficients, there will be three cases.

Example 6.5.1: Distinct Roots

Solve

x2y+5xy+3y=0.

Solution

Let

y=xr

We take two derivatives.

y=rxr1

y=r(r1)xr2.

Next plug these into the original differential equation

x2r(r1)xr2+5x ;rxr1+3xr=0r(r1)xr+5rxr+3xr=0Multiplying the exponentsr(r1)+5r+3=0Dividing by xrr2+4r+3=0(r+3)(r+1)=0

r=3orr=1.

The general solution is

y=c1x3+c2x1.

Example 6.5.2

Solve

x2y+7xy+9y=0.

Solution

Let

y=xr.

We have

F(r)=r2+6r+9=(r+3)2

which has the repeated root

r=3.

Hence a solution is

y1=x3.

This gives us one solution and we could get another solution by reduction of order. However, there is a more clever way by noting that both F(r) and F(r) are zero at r=3.

Notice also that the partial derivative

(xr)r=xrlnx.

We have

Lr(xr)=[xrF(r)]r

or

L(xrlnr)=F(r)xrlnx+xrFr(r)=(r+3)2xrlnx+2xr(r+3).

Now plug in r=3 to get

L(x3lnx)=0.

Hence

y2=x3lnx.

The general solution is

y=c1x3+c2x3lnx.

Example 6.5.3

Solve

x2y+5xy+8y=0.

Solution

Let

y=xr.

We have

F(r)=r2+4r+8.

which has complex roots

r=2+4i and r=24i

We get the solutions

y1=x2+4iandy2=x24i

As with constant coefficients, we would like to express the solution without complex numbers. We have

x2+2i=e(2+2i)lnx=x2e2lnxi=x2[cos(2lnx)+isin(2lnx)].

Similarly

x22i=e(22i)lnx=x2e2lnxi=x2[cos(2lnx)isin(2lnx)].

By playing with constants we get the two solutions

y1=x2cos(2lnx)andy2=x2sin(2lnx)

The general solution is

y=c1x2cos(2lnx)+c2x2sin(2lnx)=x2[c1cos(2lnx)+c2sin(2lnx)]

In summary, we have the following theorem.

Theorem: Solutions to Euler Equations

Let

x2y+axy+by=0

and let

F(r)=r2+(a1)r+b

have roots r1 and r2.

  • Case 1: If r1 and r2 are real and distinct, then the general solution is

y=c1xr1+c2xr2.

  • Case 2: If r1=r2=r then the general solution is

y=c1xr+c2xrlnx.

  • Case 3: If r1=l+mi and r2=lmi then the general solution is

y=xl(c1cos(mlnx)+c2sin(mlnx)).

Contributors and Attributions


This page titled 6.5: Euler Equations is shared under a not declared license and was authored, remixed, and/or curated by Larry Green.

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