Cauchy-Euler Equations
The Cauchy-Euler equation has the form
a 0 x n y( n ) + a1x n!1y( n!1) + ..... + a n!1xy" + a n y = b(x)
where a0, a1, …. , an are constants.
Each term contains xky(k).
The transformation x = et reduces the equation to a linear O.D.E. with constant
coefficient in the variable t. Notice that we assume x > 0, and t = ln x.
Using chain rule, since y is function of t through x:
dy dy dx dy t dy
=
=
e =
x
dt dx dt dx
dx
then
dy dy
=
dx dt
Now,
2
d 2 y d ! dy $ d ! dy $ dx dy
d ! dy $
d 2 y dx
dy
t dy
2 d y
=
=
x
=
+
x
=
e
+
x
=
x
+
x
dt #" dx &%
dx
dx
dt 2 dt #" dt &% dt #" dx &% dt dx
dx 2 dt
dx 2
x
2
dy
2 d y
=
+x
dt
dx 2
then
x2
d 2 y d 2 y dy
=
'
dx 2 dt 2 dt
and
2
2
3
d 3y d ! d 2 y $ d ! dy
d 2 y dx
2 d y$
t dy
2t d y
2 d y t
= #
+x 2
+ 2e
+x
e =
& = #x + x
&=e
dx
dt 3 dt " dt 3 % dt " dx
dx 2 %
dx dt
dx 2
dx 3
2
2
3
2
3
3
dy
dy
dy
d2y
dy
2 d y
2 d y
3d y
2 d y
3d y
3d y
x +x
+ 2x
+x
=
+ 3x
+x
=
+3 2 '3 +x
dx
dt
dx 2
dx 2
dx 3 dt
dx 2
dx 3 dt
dt
dx 3
then
d 3y d 3y
d2y
dy
=
'
3
+2
3
3
2
dt
dx
dt
dt
Following the model, you can compute the higher order derivatives.
x3
Remark: For the second order equation, the transformation produces:
d2y
dy
ax 2 2 + bx + cy = g(x),!x > 0
dx
dx
taking!x = e t
a
d2y
dy
+ (b ! a) + cy = g(e t )
2
dt
dt
Example:
1) x2y’’ – 2xy’ + 2y = x3 , x > 0
Taking the transformation x = et the equation reduces to:
yt’’ + (-2 - 1) yt’ + 2y = e3t
or
yt’’ - 3yt’ + 2y = e3t
Corresponding homogeneous equation: yt’’ - 3yt’ + 2y = 0
Characteristic equation: r2 – 3r + 2 = (r - 2)(r - 1) = 0
Zeros are: r1 = 2, and r2 = 1
Fundamental set of solutions: F ={e2t, et}
Complementary solution: yc(t) = c1e2t + c2et.
Non-homogeneous term is: b(t) = e3t,
The UC set of e3t is S1 = {e3t}.
The candidate for particular solution is:
yp(t) = Ae3t
computing the derivatives,
yp’(x) = 3Ae3t
yp’’(x) = 9Ae3t
substituting into the equation,
9Ae3t – 9Ae3t + 2Ae3t = e3t
then
A=½
and
yp(t) = ½ e3t
y(t) = c1e2t + c2et + ½ e3t
t
since x = e , then the general solution of the original equation is:
y(x) = c1x2 + c2x + ½ x3
2) x2y’’ + 3xy’ - 8y = ln3x - ln x, x > 0
Taking the transformation x = et the equation reduces to:
yt’’ + (3 - 1) yt’ - 8y = t3 - t
or
yt’’ + 2yt’ - 8y = t3 - t
Corresponding homogeneous equation: yt’’ + 2yt’ - 8y = 0
Characteristic equation: r2 + 2r - 8 = (r + 4)(r - 2) = 0
Zeros are: r1 = 2, and r2 = -4
Fundamental set of solutions: F ={e2t, e-4t}
Complementary solution: yc(t) = c1e2t + c2e-4t.
Non-homogeneous term is: b(t) = t3 - t
The UC set of t3 – t is S1 = {t3, t2, t, 1}.
The candidate for particular solution is:
yp(t) = At3 + Bt2 + Ct + D
computing the derivatives,
yp’(x) = 3At2 + 2Bt + C
yp’’(x) = 6At + 2B
substituting into the equation,
6At + 2B + 6At2 + 4Bt + 2C - 8At3 - 8Bt2 - 8Ct - 8D = t3 - t
then
-8A = 1
A = -1/8
6A – 8B = 0
B = -6/64 = -3/32
6A + 4B – 8C = -1
C = -1/64
2B + 2C - 8D = 0
D = -7/256
and
yp(t) = -1/8 t3 – 3/32 t2 - 1/64 t – 7/256
y(t) = c1e2t + c2e-4t - 1/8 t3 – 3/32 t2 - 1/64 t – 7/256
since x = et and t = ln x, then the general solution of the original equation is:
y(x) = c1x2 + c2x-4 - 1/8 ln3x – 3/32 ln2x - 1/64 ln x – 7/256.
3) x3y’’’ - 3x2y’’ + 6xy’ - 6y = ln x, x > 0
Taking the transformation x = et the equation reduces to:
(yt’’’ -3yt’’ + 2yt’) – 3(yt’’ – yt’) + yt’’- 6y = t
or
yt’’’ - 6yt’’ + 11yt’ - 6y = t
Corresponding homogeneous equation: yt’’’ - 6yt’’ + 11yt’ - 6y = 0
Characteristic equation: r3 - 6r2 + 11r - 6 = (r -1)(r - 3)(r - 2) = 0
Zeros are: r1 = 1, r2 = 2, and r3 = 3
Fundamental set of solutions: F ={et, e2t, e3t}
Complementary solution: yc(t) = c1et + c2e2t + c3e3t.
Non-homogeneous term is: b(t) = t
The UC set of t is S1 = {t, 1}.
The candidate for particular solution is:
yp(t) = A + Bt
computing the derivatives,
yp’(x) = B
yp’’(x) = 0
yp’’’(x) = 0
substituting into the equation,
11B – 6A – 6Bt = t
then
-6B = 1
B = -1/6
11B – 6A = 0
11B = 6A = -1, B = -1/11
and
yp(t) = -1/6 – 1/11 t
y(t) = c1et + c2e2t + c3e3t – 1/6 - 1/11 t
since x = et and t = ln x, then the general solution of the original equation is:
y(x) = c1x + c2x2 + c3x3 – 1/6 - 1/11 ln x.
4) (x + 2)2 y’’ – (x + 2) y’ + y = 3x + 2
Take the transformation x + 2 = et, then t = ln(x + 2) and et – 2 = x
dt
1
=
dx x + 2
dy dy dt
1 dy
dy dy
=
=
!or ( x + 2 ) =
dx dt dx x + 2 dt
dx dt
2
2
d y d y dy
( x + 2 )2 2 = 2 !
dt
dx
dt
b(x) = 3x + 2, b(t) = 3(et – 2) + 2 = 3et – 4.
The transformed equation is:
yt’’ – 2yt’ + y = 3et - 4
Corresponding homogeneous equation: yt’’ – 2yt’ + y = 0
Characteristic equation: r2 - 2r + 1 = (r -1)2 = 0
Zeros are: r1 = 1 double zero
Fundamental set of solutions: F ={et, tet}
Complementary solution: yc(t) = c1et + c2tet .
Non-homogeneous term is: b(t) = 3et - 4
The UC set of et is S1 = {et}, and the UC set of 1 is S2 = {1}
Since S1 contains elements of the Fundamental Set of solutions, so replace S1 by S1* =
{t2et}
The candidate for particular solution is:
yp(t) = At2et + B
computing the derivatives,
yp’(x) = 2Atet + At2et
yp’’(x) = 2Aet + 4Atet + At2et
substituting into the equation,
[2A + 4At + At2 – 4At – 2At2 + At2]et + B = 2Aet + B = 3et - 4
then
2A = 3, A = 3/2 and B = -4
and
yp(t) = 3/2 t2et - 4
y(t) = c1et + c2tet + 3/2 t2et - 4
since x + 2 = et and t = ln (x + 2), then the general solution of the original equation is:
y(x) = c1 (x +2) + c2 (x + 2) ln(x +2) + 3/2 (x + 2)ln2(x + 2) – 4.