Let us consider the standard second order Cauchy-Euler's equation.

#ax^2(d^2y)/dx^2 + bx(dy)/dx + cy = 0#

For applying the standard method of Frobenious, let #y(x) = sum k_nx^(n + lamda)# be the trial solution.

This gives, #(dy)/dx = sum k_n(n + lamda)x^(n+lamda-1)#

and,

#(d^2y)/dx^2 = sum k_n (n + lamda)(n + lamda -1)x^(n + lamda -2)#

Substituting these expressions in the original equation,

#ax^2sum k_n (n + lamda)(n + lamda -1)x^(n + lamda -2) + bxsum k_n(n + lamda)x^(n+lamda-1) + csum k_nx^(n + lamda) = 0#

which gives,

#asum k_n (n + lamda)(n + lamda -1)x^(n + lamda) + bsum k_n(n + lamda)x^(n+lamda) + csum k_nx^(n + lamda) = 0#

further,

#sum k_n[a (n + lamda)(n + lamda -1) + b(n + lamda) + c]x^(n + lamda) = 0#

Now, equating to zero, the coefficient of #x^lamda# where, #n = 0#,

#k_0[a (lamda)(lamda -1) + b(lamda) + c] = 0#

Since #k_0 != 0# #implies a (lamda)(lamda -1) + b(lamda) + c = 0#

#implies alamda^2 - (a-b)lamda +c = 0#

This is the indical equation. Solving for #lamda#,

#lamda_1 = ((a-b)+sqrt((a-b)^2 - 4ac))/(2a)# ; #lamda_2 = ((a-b)-sqrt((a-b)^2 - 4ac))/(2a)#

Where #lamda_1# and #lamda_2# are the two roots of the indical equation.

considering now,

#sum k_n[a (n + lamda)(n + lamda -1) + b(n + lamda) + c]x^(n + lamda) = 0#

now, equating to zero the coefficient of #x^(n + lamda)#,

#k_n[a (n + lamda)(n + lamda -1) + b(n + lamda) + c] = 0#

Now for each value of #lamda# substituted in the above equation, we shall obtain that #k_0# and some other one coefficient is non zero while all others are zero.

Thus, for each of the two #lamda#, we shall obtain one solution.

If #y_1# and #y_2# are the two solutions (each containing one or two terms each), then the general solution is obtained by superposition of the two solutions (this is possible since, the equation is linear).

Thus, #y = Ay_1 + By_2# where A and B are arbitrary constants.

As an example, let us consider a very simple version of the equation,

#x^2(d^2y)/dx^2 - 6y = 0# where, #a = 1, b = 0# and #c = -6#

I've taken this equation, just for the purpose of illustrating the process. You can actually do it with any such equation with #b!=0# ofcourse.

So, let, #y(x) = sum k_nx^(n + lamda)# be the trial solution.

This gives, #(dy)/dx = sum k_n(n + lamda)x^(n+lamda-1)#

and,

#(d^2y)/dx^2 = sum k_n (n + lamda)(n + lamda -1)x^(n + lamda -2)#

Substituting these in the original equation, and doing simplifying,

#sum k_n[(n + lamda)(n + lamda -1) - 6]x^(n + lamda) = 0#

Now equating to zero, the coefficient of #x^(n + lamda)#

#k_0[(lamda)(lamda -1) - 6] = 0# and since #k_0 != 0#

implies, #[(lamda)(lamda -1) - 6] = 0#

which on simplifying gives, #lamda_1 = -2# and #lamda_2 = 3#

Consider now, #sum k_n[(n + lamda)(n + lamda -1) - 6]x^(n + lamda) = 0#

Equating to zero, the coefficient of #x^(n + lamda)#

#k_n[(n + lamda)(n + lamda -1) - 6] = 0#

Assuming #k_n !=0# , #[(n + lamda)(n + lamda -1) - 6] = 0#

For #lamda = -2#

#[(n - 2)(n - 3) - 6] = 0#

#implies n^2 - 5n = 0#

which gives, #n = 0,5# thus, for #lamda_1 = -2#, #k_0, k_5 !=0#

For #lamda = 3#

#[(n + 3)(n + 2) - 6] = 0#

#implies n^2 + 5n = 0#

Thus, #n = 0, -5#

Thus, only #k_0, k_-5 !=0#. Discarding #k_-5# for #lamda_2 = 3 , k_0 !=0#

Thus, if #y_1# is solution corresponding to #lamda_1 = -2#

#y_1 = sum k_nx^(n - 2) implies y_1 = k_0x^-2 + k_5x^(5-2)#

which gives, #y_1 = k_0x^-2 + k_5x^3# (All other terms are zero because the coefficients are zero other than #k_0 and k_5#)

Similarly, for #lamda_2 = 3#; #y_2 = sum k_nx^(n + 3)#

Implies, #y_2 = k_0x^3# (All other terms are zero other than #k_0#)

Thus, the general solution, #y = Ay_1 + By_2#

#implies y = Ak_0x^-2 + Ak_5x^3 + Bk_0x^3#

putting #C_1 = Ak_0# and #C_2 = Ak_5 + Bk_0#

Thus, we have #y = C_1x^-2 + C_2x^3# where #C_1# and #C_2# are completely arbitrary constants.