Cauchy-Eulerequation

Cauchy-Euler equation
Definition1.An n th order Cauchy-Euler equation is
a n x n y(n)+a n−1x n−1y(n−1)+···+a1xy′+a0y=0.(1)
Please keep in mind the special structure of the coefficients.The i th coefficient of the
i th derivative of y is
a i x i for some constant a i.
The way wefind a solution to(1)is the same as the way we solved homogeneous linear differential equations with constant coefficients.The only difference is that we need tofind
a solution of the form
y=x m.
If we plug y=x m in(1),then we obtain
a n m(m−1)···(m−n+1)x m+a n−1m(m−1)···(m−n+2)x m+···a1mx m+a0x m=0.
Notice that x m is a common factor and if we pull that out,then we obtain
(a n m(m−1)···(m−n+1)+a n−1m(m−1)···(m−n+2)+···+a1m+a0)x m=0.
Therefore,
a n m(m−1)···(m−n+1)+a n−1m(m−1)···(m−n+2)+···+a1m+a0=0,
which is a polynomial of degree n in m,which means that there are at most n distinct real roots.
Case I.If there are exactly n distinct real roots m1,m2,...,m n,then as we have seen in the section4.3,
y1=x m1,y2=x m2,...,y n=x m n
constitute a fundamental set of solutions,that is,these are n linearly independent solutions.
Case II.If some of the roots are repeated,that is,if a root m i is repeated k times,which also means that the multiplicity of m i is k,then as we have already seen in the section
4.3,we obtain k linearly independent solutions corresponding to the root m i,
x m i,x m i ln x,x m i(ln x)2,...,x m i(ln x)k−1.
1
Case III.If there are complex roots,that is,if a+bi and a−bi are complex roots,then we know that
x a+bi,x a−bi
are linearly independent.However,these are not real functions.Hence,with the help
of the Euler formula
e iθ=cosθ+i sinθ,
we get
x a+bi=x a x bi=x a e ln x bi=x a e i(b ln x)=x a(cos(b ln x)+i sin(b ln x)), and
x a−bi=x a x−bi=x a e ln x−bi=x a e−i(b ln x)=x a(cos(b ln x)−i sin(b ln x)).
Therefore,we obtain two linearly independent real solutions
x a cos(b ln x),x a sin(b ln x).
I won’t derive the factor ln x in Case II,which can be done with Reduction
of order as was done in the section4.3.
Exercise1.Solve the following differential equation.
2x2y′′−3xy′+2y=0.(2)
Solution1.This is a2nd order Cauchy-Euler equation.So we try y=x m and determine m.Since
2x2(x m)′′−3x(x m)′+2x m=0⇔2m(m−1)−3m+2=0,
we need tofind m satisfying
2m(m−1)−3m+2=0⇔2m2−5m+2=(2m−1)(m−2)=0.
Hence,
m=1/2,2,
which implies that y1=x1/2,y2=x2are two linearly independent solutions and the general solution is
y=c1y1+c2y2=c1x1/2+c2x2.
2
Exercise2.Solve the following differential equation.
4x2y′′+8xy′+y=0.(3)
Solution2.This is a2nd order Cauchy-Euler equation.So we try y=x m and determine m.Since
4x2(x m)′′+8x(x m)′+x m=0⇔4m(m−1)+8m+1=0,
we need tofind m satisfying
4m(m−1)+8m+1=0⇔4m2+4m+1=(2m+1)2=0.
Hence,
m=−1/2
is the only repeated root.Since m=−1/2has multiplicity2,we obtain two linearly inde-pendent solutions
y1=x−1/2,y2=x−1/2ln x.
The general solution is
y=c1y1+c2y2=c1x−1/2+c2x−1/2ln x.
Exercise3.Solve the following differential equation.
x2y′′+xy′+y=0.(4)
Solution3.This is a2nd order Cauchy-Euler equation.So we try y=x m and determine m.Since
x2(x m)′′+x(x m)′+x m=0⇔m(m−1)+m+1=0,
we need tofind m satisfying
m(m−1)+m+1=0⇔m2+1=0.
Hence,
m=±i
are two complex roots.Since a=0and b=1,we obtain two linearly independent solutions y1=x a cos(b ln x)=cos(ln x),y2=x a sin(b ln x)=sin(ln x).
The general solution is
y=c1y1+c2y2=c1cos(ln x)+c2sin(ln x).
3。