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  =  x^r 

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

        y'  =  r x^r-1        y''  =  r (r - 1) x^r-2 

Next plug these into the original differential equation

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

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

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

        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  =  x^r 

We take two derivatives.

        y'  =  r x^r-1        y''  =  r (r - 1) x^r-2 

Next plug these into the original differential equation

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

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

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

        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  =  x^r 

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

        (x^r)_r  =  x^r ln x

We have 

        L_r(x^r)  =  [x^rF(r)]_r 

or

        L(x^r ln r)  =  F(r) x^r ln x  + x^r F_r(r)  =  (r + 3)² x^r ln x  +2x^r (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  =  x^r 

We have 

        F(r) =  r² + 4r + 8  

which has complex roots

        r  =  2 + 4i    and     r  =  2 - 4i 

We get the solutions

        y₁  =  x^{2 + 4i}     and        y₂  =  x^{2 - 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^-2e^{2lnx 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

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