Cauchy-Euler Equations The Cauchy-Euler equation has the form a0x

nyn( )+ a1x

n!1yn!1( )

+ .....+ an!1x "y + any = 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

dt=dy

dx

dx

dt=dy

dxet =

dy

dxx

then

xdy

dx=dy

dt

Now, d2y

dt2=d

dt

dy

dt

!

"#$

%&=d

dtxdy

dx

!

"#$

%&=dx

dt

dy

dx+ x

d

dt

dy

dx

!

"#$

%&= et

dy

dx+ x

d2y

dx2dx

dt= x

dy

dx+ x2

d2y

dx2

=dy

dt+ x2

d2y

dx2

then

x2d2y

dx2=d2y

dt2'dy

dt

and d3y

dt3=d

dt

d2y

dt3

!

"#

$

%& =

d

dtxdy

dx+ x2

d2y

dx2

!

"#

$

%& = e

t dy

dx+ x

d2y

dx2dx

dt+ 2e2t

d2y

dx2+ x2

d3y

dx3et =

xdy

dx+ x2

d2y

dx2+ 2x2

d2y

dx2+ x3

d3y

dx3=dy

dt+ 3x2

d2y

dx2+ x3

d3y

dx3=dy

dt+ 3d2y

dt2' 3dy

dt+ x3

d3y

dx3

then

x3d3y

dx3=d3y

dt3' 3d2y

dt2+ 2

dy

dt

Following the model, you can compute the higher order derivatives. Remark: For the second order equation, the transformation produces:

ax2d2y

dx2+ bx

dy

dx+ cy = g(x),!x > 0

taking!x = et

ad2y

dt2+ (b ! a)

dy

dt+ cy = g(et )

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 since x = et, 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

dx=

1

x + 2

dy

dx=dy

dt

dt

dx=

1

x + 2

dy

dt!or x + 2( )

dy

dx=dy

dt

x + 2( )2 d

2y

dx2=d2y

dt2!dy

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.