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′=rxr−1
y″=r(r−1)xr−2
Next plug these into the original differential equation
x2r(r−1)xr−2+axrxr−1+bxr=0r(r−1)xr+arxr+bxr=0Multiplying the exponentsr(r−1)+ar+b=0Dividing by xrr2+(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.
Solve
x2y″+5xy′+3y=0.
Solution
Let
y=xr
We take two derivatives.
y′=rxr−1
y″=r(r−1)xr−2.
Next plug these into the original differential equation
x2r(r−1)xr−2+5x ;rxr−1+3xr=0r(r−1)xr+5rxr+3xr=0Multiplying the exponentsr(r−1)+5r+3=0Dividing by xrr2+4r+3=0(r+3)(r+1)=0
r=−3orr=−1.
The general solution is
y=c1x−3+c2x−1.
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 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(x−3lnx)=0.
Hence
y2=x−3lnx.
The general solution is
y=c1x−3+c2x−3lnx.
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+4iandy2=x2−4i
As with constant coefficients, we would like to express the solution without complex numbers. We have
x−2+2i=e(−2+2i)lnx=x−2e2lnxi=x−2[cos(2lnx)+isin(2lnx)].
Similarly
x−2−2i=e(−2−2i)lnx=x−2e−2lnxi=x−2[cos(2lnx)−isin(2lnx)].
By playing with constants we get the two solutions
y1=x−2cos(2lnx)andy2=x−2sin(2lnx)
The general solution is
y=c1x−2cos(2lnx)+c2x−2sin(2lnx)=x−2[c1cos(2lnx)+c2sin(2lnx)]
In summary, we have the following theorem.
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+c2xrlnx.
- Case 3: If r1=l+mi and r2=l−mi then the general solution is
y=xl(c1cos(mlnx)+c2sin(mlnx)).
Contributors and Attributions
- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.