Solve the Differential Equation # x^2y'' -xy'-8y=0 #?

#x^2y''-xy'-8y=0#

This is a Cauchy-Euler differential equation of second order, so the resolution method is to substitute as possible solution:

#y(x) = x^n#

and its derivatives:

#y'(x) = nx^(n-1)#

#y''(x) = n(n-1)x^(n-2)#

Substituting in the original equation we have:

#x^2((n(n-1)x^(n-2))-nx x^(n-1)-8x^n =0#

#(n(n-1)x^n-nx^n-8x^n =0#

#x^n(n^2-n-n-8) =0#

#x^n(n^2-2n-8) =0#

So for #x!=0# we must have:

#n^2-2n-8 =0#

#n= (1+-sqrt(1+8))#

#n_1=-2#
#n_2 = 4#

The general solution is then:

#y(x) = c_1/x^2+c_2x^4#

We have:

# x^2y'' -xy'-8y=0 # ..... [A]

This is a Euler-Cauchy Equation which is typically solved via a change of variable. Consider the substitution:

# x = e^t => xe^(-t)=1#

Then we have,

#dy/dx = e^(-t)dy/dt#, and, #(d^2y)/(dx^2)=((d^2y)/(dt^2)-dy/dt)e^(-2t)#

Substituting into the initial DE [A] we get:

# x^2((d^2y)/(dt^2)-dy/dt)e^(-2t) -xe^(-t)dy/dt-8y=0 #

# :. ((d^2y)/(dt^2)-dy/dt) -dy/dt-8y=0 #

# :. (d^2y)/(dt^2)-2dy/dt-8=0 # ..... [B]

This is now a second order linear homogeneous Differentiation Equation. The standard approach is to look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.

# m^2-2m-8 = 0#

We can solve this quadratic equation, and we get two real and distint roots:

# m=-2,4 #

Thus the Homogeneous equation [B] has the solution:

# y=Ae^(-2t)+Be^(4t)#

Now we initially used a change of variable:

# x = e^t => t=lnx #

So restoring this change of variable we get:

# y = Ae^(-2lnx)+Be^(4lnx)#

# :. y = Ae^(lnx^(-2))+Be^(lnx^4)#
# :. y = Ax^(-2)+Bx^4#
# :. y = A/x^2+Bx^4#

Which is the General Solution.