Ordinary differential equation
In mathematics , an Euler–Cauchy equation , or Cauchy–Euler equation , or simply Euler's equation is a linear homogeneous ordinary differential equation with variable coefficients . It is sometimes referred to as an equidimensional equation. Because of its particularly simple equidimensional structure, the differential equation can be solved explicitly.
Let y (n ) (x ) be the n th derivative of the unknown function y (x ) . Then a Cauchy–Euler equation of order n has the form
a
n
x
n
y
(
n
)
(
x
)
+
a
n
−
1
x
n
−
1
y
(
n
−
1
)
(
x
)
+
⋯
+
a
0
y
(
x
)
=
0.
{\displaystyle a_{n}x^{n}y^{(n)}(x)+a_{n-1}x^{n-1}y^{(n-1)}(x)+\dots +a_{0}y(x)=0.}
The substitution
x
=
e
u
{\displaystyle x=e^{u}}
(that is,
u
=
ln
(
x
)
{\displaystyle u=\ln(x)}
; for
x
<
0
{\displaystyle x<0}
, one might replace all instances of
x
{\displaystyle x}
by
|
x
|
{\displaystyle |x|}
, which extends the solution's domain to
R
∖
{
0
}
{\displaystyle \mathbb {R} \setminus \{0\}}
) may be used to reduce this equation to a linear differential equation with constant coefficients. Alternatively, the trial solution
y
=
x
m
{\displaystyle y=x^{m}}
may be used to directly solve for the basic solutions.[1]
Second order – solving through trial solution[ edit ]
Typical solution curves for a second-order Euler–Cauchy equation for the case of two real roots
Typical solution curves for a second-order Euler–Cauchy equation for the case of a double root
Typical solution curves for a second-order Euler–Cauchy equation for the case of complex roots
The most common Cauchy–Euler equation is the second-order equation, appearing in a number of physics and engineering applications, such as when solving Laplace's equation in polar coordinates. The second order Cauchy–Euler equation is[1] [2]
x
2
d
2
y
d
x
2
+
a
x
d
y
d
x
+
b
y
=
0.
{\displaystyle x^{2}{\frac {d^{2}y}{dx^{2}}}+ax{\frac {dy}{dx}}+by=0.}
We assume a trial solution[1]
y
=
x
m
.
{\displaystyle y=x^{m}.}
Differentiating gives
d
y
d
x
=
m
x
m
−
1
{\displaystyle {\frac {dy}{dx}}=mx^{m-1}}
and
d
2
y
d
x
2
=
m
(
m
−
1
)
x
m
−
2
.
{\displaystyle {\frac {d^{2}y}{dx^{2}}}=m\left(m-1\right)x^{m-2}.}
Substituting into the original equation leads to requiring
x
2
(
m
(
m
−
1
)
x
m
−
2
)
+
a
x
(
m
x
m
−
1
)
+
b
(
x
m
)
=
0
{\displaystyle x^{2}\left(m\left(m-1\right)x^{m-2}\right)+ax\left(mx^{m-1}\right)+b\left(x^{m}\right)=0}
Rearranging and factoring gives the indicial equation
m
2
+
(
a
−
1
)
m
+
b
=
0.
{\displaystyle m^{2}+\left(a-1\right)m+b=0.}
We then solve for m . There are three particular cases of interest:
Case 1 of two distinct roots, m 1 and m 2 ;
Case 2 of one real repeated root, m ;
Case 3 of complex roots, α ± βi .
In case 1, the solution is
y
=
c
1
x
m
1
+
c
2
x
m
2
{\displaystyle y=c_{1}x^{m_{1}}+c_{2}x^{m_{2}}}
In case 2, the solution is
y
=
c
1
x
m
ln
(
x
)
+
c
2
x
m
{\displaystyle y=c_{1}x^{m}\ln(x)+c_{2}x^{m}}
To get to this solution, the method of reduction of order must be applied after having found one solution y = x m .
In case 3, the solution is
y
=
c
1
x
α
cos
(
β
ln
(
x
)
)
+
c
2
x
α
sin
(
β
ln
(
x
)
)
{\displaystyle y=c_{1}x^{\alpha }\cos(\beta \ln(x))+c_{2}x^{\alpha }\sin(\beta \ln(x))}
α
=
Re
(
m
)
{\displaystyle \alpha =\operatorname {Re} (m)}
β
=
Im
(
m
)
{\displaystyle \beta =\operatorname {Im} (m)}
For
c
1
,
c
2
∈
R
{\displaystyle c_{1},c_{2}\in \mathbb {R} }
.
This form of the solution is derived by setting x = e t and using Euler's formula
Second order – solution through change of variables[ edit ]
x
2
d
2
y
d
x
2
+
a
x
d
y
d
x
+
b
y
=
0
{\displaystyle x^{2}{\frac {d^{2}y}{dx^{2}}}+ax{\frac {dy}{dx}}+by=0}
We operate the variable substitution defined by
t
=
ln
(
x
)
.
{\displaystyle t=\ln(x).}
y
(
x
)
=
φ
(
ln
(
x
)
)
=
φ
(
t
)
.
{\displaystyle y(x)=\varphi (\ln(x))=\varphi (t).}
Differentiating gives
d
y
d
x
=
1
x
d
φ
d
t
{\displaystyle {\frac {dy}{dx}}={\frac {1}{x}}{\frac {d\varphi }{dt}}}
d
2
y
d
x
2
=
1
x
2
(
d
2
φ
d
t
2
−
d
φ
d
t
)
.
{\displaystyle {\frac {d^{2}y}{dx^{2}}}={\frac {1}{x^{2}}}\left({\frac {d^{2}\varphi }{dt^{2}}}-{\frac {d\varphi }{dt}}\right).}
Substituting
φ
(
t
)
{\displaystyle \varphi (t)}
the differential equation becomes
d
2
φ
d
t
2
+
(
a
−
1
)
d
φ
d
t
+
b
φ
=
0.
{\displaystyle {\frac {d^{2}\varphi }{dt^{2}}}+(a-1){\frac {d\varphi }{dt}}+b\varphi =0.}
This equation in
φ
(
t
)
{\displaystyle \varphi (t)}
is solved via its characteristic polynomial
λ
2
+
(
a
−
1
)
λ
+
b
=
0.
{\displaystyle \lambda ^{2}+(a-1)\lambda +b=0.}
Now let
λ
1
{\displaystyle \lambda _{1}}
and
λ
2
{\displaystyle \lambda _{2}}
denote the two roots of this polynomial. We analyze the case where there are distinct roots and the case where there is a repeated root:
If the roots are distinct, the general solution is
φ
(
t
)
=
c
1
e
λ
1
t
+
c
2
e
λ
2
t
,
{\displaystyle \varphi (t)=c_{1}e^{\lambda _{1}t}+c_{2}e^{\lambda _{2}t},}
where the exponentials may be complex.
If the roots are equal, the general solution is
φ
(
t
)
=
c
1
e
λ
1
t
+
c
2
t
e
λ
1
t
.
{\displaystyle \varphi (t)=c_{1}e^{\lambda _{1}t}+c_{2}te^{\lambda _{1}t}.}
In both cases, the solution
y
(
x
)
{\displaystyle y(x)}
may be found by setting
t
=
ln
(
x
)
{\displaystyle t=\ln(x)}
.
Hence, in the first case,
y
(
x
)
=
c
1
x
λ
1
+
c
2
x
λ
2
,
{\displaystyle y(x)=c_{1}x^{\lambda _{1}}+c_{2}x^{\lambda _{2}},}
and in the second case,
y
(
x
)
=
c
1
x
λ
1
+
c
2
ln
(
x
)
x
λ
1
.
{\displaystyle y(x)=c_{1}x^{\lambda _{1}}+c_{2}\ln(x)x^{\lambda _{1}}.}
Second order - solution using differential operators [ edit ]
Observe that we can write the second-order Cauchy-Euler equation in terms of a linear differential operator
L
{\displaystyle L}
as
L
y
=
(
x
2
D
2
+
a
x
D
+
b
I
)
y
=
0
,
{\displaystyle Ly=(x^{2}D^{2}+axD+bI)y=0,}
where
D
=
d
d
x
{\displaystyle D={\frac {d}{dx}}}
and
I
{\displaystyle I}
is the identity operator.
We express the above operator as a polynomial in
x
D
{\displaystyle xD}
rather than
D
{\displaystyle D}
. By the product rule,
(
x
D
)
2
=
x
D
(
x
D
)
=
x
(
D
+
x
D
2
)
=
x
2
D
2
+
x
D
.
{\displaystyle (xD)^{2}=xD(xD)=x(D+xD^{2})=x^{2}D^{2}+xD.}
So,
L
=
(
x
D
)
2
+
(
a
−
1
)
(
x
D
)
+
b
I
.
{\displaystyle L=(xD)^{2}+(a-1)(xD)+bI.}
We can then use the quadratic formula to factor this operator into linear terms. More specifically, let
λ
1
,
λ
2
{\displaystyle \lambda _{1},\lambda _{2}}
denote the (possibly equal) values of
−
a
−
1
2
±
1
2
(
a
−
1
)
2
−
4
b
.
{\displaystyle -{\frac {a-1}{2}}\pm {\frac {1}{2}}{\sqrt {(a-1)^{2}-4b}}.}
Then,
L
=
(
x
D
−
λ
1
I
)
(
x
D
−
λ
2
I
)
.
{\displaystyle L=(xD-\lambda _{1}I)(xD-\lambda _{2}I).}
It can be seen that these factors commute, that is,
(
x
D
−
λ
1
I
)
(
x
D
−
λ
2
I
)
=
(
x
D
−
λ
2
I
)
(
x
D
−
λ
1
I
)
{\displaystyle (xD-\lambda _{1}I)(xD-\lambda _{2}I)=(xD-\lambda _{2}I)(xD-\lambda _{1}I)}
. Hence, if
λ
1
≠
λ
2
{\displaystyle \lambda _{1}\neq \lambda _{2}}
, the solution to
L
y
=
0
{\displaystyle Ly=0}
is a linear combination of the solutions to each of
(
x
D
−
λ
1
I
)
y
=
0
{\displaystyle (xD-\lambda _{1}I)y=0}
and
(
x
D
−
λ
2
I
)
y
=
0
{\displaystyle (xD-\lambda _{2}I)y=0}
, which can be solved by separation of variables .
Indeed, with
i
∈
{
1
,
2
}
{\displaystyle i\in \{1,2\}}
, we have
(
x
D
−
λ
i
I
)
y
=
x
d
y
d
x
−
λ
i
y
=
0
{\displaystyle (xD-\lambda _{i}I)y=x{\frac {dy}{dx}}-\lambda _{i}y=0}
. So,
x
d
y
d
x
=
λ
i
y
∫
1
y
d
y
=
λ
i
∫
1
x
d
x
ln
y
=
λ
i
ln
x
+
C
y
=
c
i
e
λ
i
ln
x
=
c
i
x
λ
i
.
{\displaystyle {\begin{aligned}x{\frac {dy}{dx}}&=\lambda _{i}y\\\int {\frac {1}{y}}\,dy&=\lambda _{i}\int {\frac {1}{x}}\,dx\\\ln y&=\lambda _{i}\ln x+C\\y&=c_{i}e^{\lambda _{i}\ln x}=c_{i}x^{\lambda _{i}}.\end{aligned}}}
Thus the general solution is
y
=
c
1
x
λ
1
+
c
2
x
λ
2
{\displaystyle y=c_{1}x^{\lambda _{1}}+c_{2}x^{\lambda _{2}}}
.
If
λ
=
λ
1
=
λ
2
{\displaystyle \lambda =\lambda _{1}=\lambda _{2}}
, then we instead need to consider the solution of
(
x
D
−
λ
I
)
2
y
=
0
{\displaystyle (xD-\lambda I)^{2}y=0}
. Let
z
=
(
x
D
−
λ
I
)
y
{\displaystyle z=(xD-\lambda I)y}
, so that we can write
(
x
D
−
λ
I
)
2
y
=
(
x
D
−
λ
I
)
z
=
0.
{\displaystyle (xD-\lambda I)^{2}y=(xD-\lambda I)z=0.}
As before, the solution of
(
x
D
−
λ
I
)
z
=
0
{\displaystyle (xD-\lambda I)z=0}
is of the form
z
=
c
1
x
λ
{\displaystyle z=c_{1}x^{\lambda }}
. So, we are left to solve
(
x
D
−
λ
I
)
y
=
x
d
y
d
x
−
λ
y
=
c
1
x
λ
.
{\displaystyle (xD-\lambda I)y=x{\frac {dy}{dx}}-\lambda y=c_{1}x^{\lambda }.}
We then rewrite the equation as
d
y
d
x
−
λ
x
y
=
c
1
x
λ
−
1
,
{\displaystyle {\frac {dy}{dx}}-{\frac {\lambda }{x}}y=c_{1}x^{\lambda -1},}
which one can recognize as being amenable to solution via an integrating factor .
Choose
M
(
x
)
=
x
−
λ
{\displaystyle M(x)=x^{-\lambda }}
as our integrating factor. Multiplying our equation through by
M
(
x
)
{\displaystyle M(x)}
and recognizing the left-hand side as the derivative of a product, we then obtain
d
d
x
(
x
−
λ
y
)
=
c
1
x
−
1
x
−
λ
y
=
∫
c
1
x
−
1
d
x
y
=
x
λ
(
c
1
ln
(
x
)
+
c
2
)
=
c
1
ln
(
x
)
x
λ
+
c
2
x
λ
.
{\displaystyle {\begin{aligned}{\frac {d}{dx}}(x^{-\lambda }y)&=c_{1}x^{-1}\\x^{-\lambda }y&=\int c_{1}x^{-1}\,dx\\y&=x^{\lambda }(c_{1}\ln(x)+c_{2})\\&=c_{1}\ln(x)x^{\lambda }+c_{2}x^{\lambda }.\end{aligned}}}
Given
x
2
u
″
−
3
x
u
′
+
3
u
=
0
,
{\displaystyle x^{2}u''-3xu'+3u=0\,,}
we substitute the simple solution x m :
x
2
(
m
(
m
−
1
)
x
m
−
2
)
−
3
x
(
m
x
m
−
1
)
+
3
x
m
=
m
(
m
−
1
)
x
m
−
3
m
x
m
+
3
x
m
=
(
m
2
−
4
m
+
3
)
x
m
=
0
.
{\displaystyle x^{2}\left(m\left(m-1\right)x^{m-2}\right)-3x\left(mx^{m-1}\right)+3x^{m}=m\left(m-1\right)x^{m}-3mx^{m}+3x^{m}=\left(m^{2}-4m+3\right)x^{m}=0\,.}
For x m to be a solution, either x = 0 , which gives the trivial solution, or the coefficient of x m is zero. Solving the quadratic equation, we get m = 1, 3 . The general solution is therefore
u
=
c
1
x
+
c
2
x
3
.
{\displaystyle u=c_{1}x+c_{2}x^{3}\,.}
Difference equation analogue [ edit ]
There is a difference equation analogue to the Cauchy–Euler equation. For a fixed m > 0 , define the sequence f m (n ) as
f
m
(
n
)
:=
n
(
n
+
1
)
⋯
(
n
+
m
−
1
)
.
{\displaystyle f_{m}(n):=n(n+1)\cdots (n+m-1).}
Applying the difference operator to
f
m
{\displaystyle f_{m}}
, we find that
D
f
m
(
n
)
=
f
m
(
n
+
1
)
−
f
m
(
n
)
=
m
(
n
+
1
)
(
n
+
2
)
⋯
(
n
+
m
−
1
)
=
m
n
f
m
(
n
)
.
{\displaystyle {\begin{aligned}Df_{m}(n)&=f_{m}(n+1)-f_{m}(n)\\&=m(n+1)(n+2)\cdots (n+m-1)={\frac {m}{n}}f_{m}(n).\end{aligned}}}
If we do this k times, we find that
f
m
(
k
)
(
n
)
=
m
(
m
−
1
)
⋯
(
m
−
k
+
1
)
n
(
n
+
1
)
⋯
(
n
+
k
−
1
)
f
m
(
n
)
=
m
(
m
−
1
)
⋯
(
m
−
k
+
1
)
f
m
(
n
)
f
k
(
n
)
,
{\displaystyle {\begin{aligned}f_{m}^{(k)}(n)&={\frac {m(m-1)\cdots (m-k+1)}{n(n+1)\cdots (n+k-1)}}f_{m}(n)\\&=m(m-1)\cdots (m-k+1){\frac {f_{m}(n)}{f_{k}(n)}},\end{aligned}}}
where the superscript (k ) denotes applying the difference operator k times. Comparing this to the fact that the k -th derivative of x m equals
m
(
m
−
1
)
⋯
(
m
−
k
+
1
)
x
m
x
k
{\displaystyle m(m-1)\cdots (m-k+1){\frac {x^{m}}{x^{k}}}}
suggests that we can solve the N -th order difference equation
f
N
(
n
)
y
(
N
)
(
n
)
+
a
N
−
1
f
N
−
1
(
n
)
y
(
N
−
1
)
(
n
)
+
⋯
+
a
0
y
(
n
)
=
0
,
{\displaystyle f_{N}(n)y^{(N)}(n)+a_{N-1}f_{N-1}(n)y^{(N-1)}(n)+\cdots +a_{0}y(n)=0,}
in a similar manner to the differential equation case. Indeed, substituting the trial solution
y
(
n
)
=
f
m
(
n
)
{\displaystyle y(n)=f_{m}(n)}
brings us to the same situation as the differential equation case,
m
(
m
−
1
)
⋯
(
m
−
N
+
1
)
+
a
N
−
1
m
(
m
−
1
)
⋯
(
m
−
N
+
2
)
+
⋯
+
a
1
m
+
a
0
=
0.
{\displaystyle m(m-1)\cdots (m-N+1)+a_{N-1}m(m-1)\cdots (m-N+2)+\dots +a_{1}m+a_{0}=0.}
One may now proceed as in the differential equation case, since the general solution of an N -th order linear difference equation is also the linear combination of N linearly independent solutions. Applying reduction of order in case of a multiple root m 1 will yield expressions involving a discrete version of ln ,
φ
(
n
)
=
∑
k
=
1
n
1
k
−
m
1
.
{\displaystyle \varphi (n)=\sum _{k=1}^{n}{\frac {1}{k-m_{1}}}.}
(Compare with:
ln
(
x
−
m
1
)
=
∫
1
+
m
1
x
d
t
t
−
m
1
.
{\textstyle \ln(x-m_{1})=\int _{1+m_{1}}^{x}{\frac {dt}{t-m_{1}}}.}
)
In cases where fractions become involved, one may use
f
m
(
n
)
:=
Γ
(
n
+
m
)
Γ
(
n
)
{\displaystyle f_{m}(n):={\frac {\Gamma (n+m)}{\Gamma (n)}}}
instead (or simply use it in all cases), which coincides with the definition before for integer m .
^ a b c Kreyszig, Erwin (May 10, 2006). Advanced Engineering Mathematics . Wiley. ISBN 978-0-470-08484-7 .
^ Boyce, William E.; DiPrima, Richard C. (2012). Rosatone, Laurie (ed.). Elementary Differential Equations and Boundary Value Problems (10th ed.). pp. 272–273. ISBN 978-0-470-45831-0 .