Euler Equations

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)  =  a        and        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'  =  r xr-1        y''  =  r (r - 1) xr-2

Next plug these into the original differential equation

x2 r (r - 1)xr-2 + ax r xr-1 + b xr  =  0

r (r - 1)xr + ar xr + b xr  =  0        Multiplying the exponents

r (r - 1) + ar + b  =  0        Dividing by xr

r2 + (a - 1) r + b  =  0

We define

F(r)  =  r2 + (a - 1) 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    (Real Distinct Roots)

Solve

x2y'' + 5xy' + 3y  =  0

Solution

Let

y  =  xr

We take two derivatives.

y'  =  r xr-1        y''  =  r (r - 1) xr-2

Next plug these into the original differential equation

x2 r (r - 1)xr-2 + 5x r xr-1 + 3xr  =  0

r (r - 1)xr + 5r xr + 3xr  =  0        Multiplying the exponents

r (r - 1) + 5r + 3  =  0        Dividing by xr

r2 + 4 r + 3  =  0

(r + 3)(r + 1)  =  0

r  =  -3        or        r  =  -1

The general solution is

y  =  c1x -3 + c2x -1

Example    (Repeated Roots)

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  =  x -3

This gives us one solution and we could get another solution by reduction of order.  However, there is a more clever way.

We notice that both F(r) and F'(r) are zero at r  =  -3.

Notice also that the partial derivative

(xr)r  =  xr ln x

We have

Lr(xr)  =  [xrF(r)]r

or

L(xr ln r)  =  F(r) xr ln x  + xr Fr(r)  =  (r + 3)2 xr ln x  +2xr (r + 3)

Now plug in r  =  -3 to get

L(x -3 ln x)  =  0

Hence

y2  =  x -3 ln x

The general solution is

y  =  c1x -3 + c2x -3 ln x

Example  (Complex Roots)

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  =  2 - 4i

We get the solutions

y1  =  x2 + 4i     and        y2  =  x2 - 4i

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

x -2 + 2i  =  e(-2 + 2i)ln x  =  x-2e2lnx i  =  x-2[cos(2ln x) + i sin(2ln x)]

Similarly

x -2 - 2i  =  e(-2 - 2i)ln x  =  x -2e-2lnx i  =  x -2[cos(2ln x) - i sin(2ln x)]

By playing with constants we get the two solutions

y1  =  x -2 cos(2ln x)        and        y2  =  x -2 sin(2ln x)

The general solution is

y  =    c1x -2 cos(2ln x) + c2x -2 sin(2ln x)  =  x -2[c1cos(2ln x) + c2sin(2ln x)]

In summary, we have

 Theorem:      Solutions to Euler Equations Let           x2y'' + axy' + by  =  0 and let           F(r)  =  r2 + (a - 1)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 + c2xr ln x Case 3:  If  r1  =  l + mi  and r2  =  l - mi  then the general solution is             y  =  xl (c1 cos(m lnx) + c2 sin(m lnx))