The Cauchy Euler Equation:
The second order Cauchy Euler Equation is:
dy
d2 y
+ bx
+ cy = 0
dx2
dx
This equation can be transformed into a second order liner differential equaax2
tion with constant coefficients with use of the substitution:
x = et
Differentiating with respect to t gives:
dx
= et
dt
So
dx
=x
dt
dy
In the Cauchy Euler equation we see the term x dx
which we need to substi-
tute for, so I will multiply both sides of
dy
dx dy
=x
dt dx
dx
Differentiating
dy
dt
dx
dt
= x by
dy
dx :
This simplifies to
dy
dy
=x
dt
dx
dy
= x dx
with respect to x gives:
dy
d2 y
d dy
=
+x 2
dx dt
dx
dx
2
d y
In the Cauchy Euler equation we see the term x2 dx
2 which we need to
substitute for, so I will multiply the left side of the above equation by
dx
dt
and
the right hand side by x (Remember they are equal):
d dy dx
=
dx dt dt
dy
d2 y
+x 2 x
dx
dx
dy
Replacing x dx
with
dy
dt
This simplifies to
dy
d2 y
d2 y
=x
+ x2 2
2
dt
dx
dx
2
d y
and solving for x2 dx
2 gives:
d2 y
d2 y dy
= 2 −
2
dx
dt
dt
So under this substitution the Cauchy Euler equation becomes:
x2
a
d2 y dy
−
dt2
dt
+b
dy
+ cy = 0
dt
This simplifies to the second order linear equation with constant coefficients:
a
d2 y
dy
+ (b − a)
+ cy = 0
dt2
dt
Which we can solve by finding the roots of the characteristic polynomial:
ar2 + (b − a)r + c = 0
Again there are three cases: the roots are real and distinct, the roots are
real and repeated or the roots are complex.
Case 1: we have two real and distinct roots r1 and r2 . Then the solutions
to the differential equation are:
y1 = er1 t
and
y 2 = e r2 t
Since x = et , t = ln(x) Making the solution:
y1 = er1 ln(x) = eln(x
r1
r2
)
and
y2 = er2 ln(x) = eln(x
y 1 = x r1
and
y 2 = x r2
)
So
Case 2: The roots are real and repeated r1 = r2 . Then the solutions to the
differential equation are:
y 1 = e r1 t
and
y2 = ter1
Since x = et , t = ln(x) Making the solution:
y1 = er1 ln(x) = eln(x
r1
)
r1
y2 = ln(x)er1 ln(x) = ln(x)eln(x
and
)
So
y1 = xr1
and
y2 = ln(x)xr1
Case 3: The roots are complex r1 = α + βı and r2 = α − βı. Then the
solutions to the differential equation are:
y1 = eαt cos(βt)
and
y2 = eαt sin(βt)
Since x = et , t = ln(x) Making the solution:
y1 = xα cos(β ln(x))
An Example: Solve:
and
y2 = xα sin(β ln(x))
x2 y 00 − 5xy 0 + 13y = 0
Forming the characteristic polynomial
r2 + (−5 − 1)r + 13 = 0
r2 − 6r + 13 = 0
(r − 3)2 = −4
So we have complex roots so the solution is:
y = C1 e3x cos(2x) + C2 e3x sin(2x)
1.
Solve:
x2
dy
d2 y
+ 7x
+ 8y = 0
dx2
dx
2.
Solve:
x2
d2 y
dy
+ 9x
+ 12y = 0
dx2
dx
3.
Solve:
x2
d2 y
dy
− 11x
+ 36y = 0
2
dx
dx
4.
Solve:
x2
d2 y
dy
− 4x
+ 6y = x3 ln(x)
dx2
dx
5.
Solve:
x2
d2 y
+ y = x2
dx2
r = 3 ± 2ı