What is the general solution of the differential equation # x^2 (d^2y)/(dx^2) - 3x dy/dx+y=sin(logx)/x #?

We have:

# x^2 (d^2y)/(dx^2) - 3x dy/dx+y=sin(logx)/x # ..... [A]

This is a Euler-Cauchy Equation (the power of #x# is the same as the degree of the differential in every occurrence of their product) 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:

# (e^t)^2 ((d^2y)/(dt^2)-dy/dt)e^(-2t) - 3(e^t) e^(-t)dy/dt+y=sin(loge^t)/((e^t)) #

# :. (d^2y)/(dt^2)-dy/dt - 3dy/dt+y = e^(-t)sint #

# :. (d^2y)/(dt^2) - 4dy/dt+y = e^(-t)sint # ..... [B]

This is now a second order linear non-Homogeneous Differentiation Equation. The standard approach is to find a solution, #y_c# of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, #y_p# of the non-homogeneous equation.

Complementary Function

The associated homogeneous equation is:

# (d^2y)/(dt^2) - 4dy/dt+y = 0 # ..... [C]

Which has the Auxiliary Equation:

# m^2-4m+1 = 0#

We can solve this quadratic equation, and we get two distinct real solutions::

# m=2+-sqrt(3) #

Thus the Homogeneous equation [C] has the solution:

# y_c = Ae^((2-sqrt(3))t) + Be^((2+sqrt(3))t)#

Particular Solution

With this particular equation [B], a probably solution is of the form:

# y = ae^(-t)sint + be^(-t)cost #
# \ \ = e^(-t)(asint + bcost) #

Where #a# and #b# are constants to be determined. Let us assume the above solution works, in which case be differentiating wrt #x# we have:

# y' \ \= e^(-t)(acost - bsint) - e^(-t)(asint + bcost) #
# \ \ \ \ \= e^(-t)( (a-b)cost - (a+b)sint ) #

# y'' = e^(-t)(-(a-b)sint - (a+b)cost ) - e^(-t)((a-b)cost - (a+b)sint ) #
# \ \ \ \ \ = e^(-t)(2bsint -2acost) #

Substituting into the Differential Equation #[B]# we get:

# e^(-t)(2bsint -2acost) - 4e^(-t)( (a-b)cost - (a+b)sint ) + e^(-t)(asint + bcost) = e^(-t)sint #

Equating coefficients of #cos(x)# and #sin(x)# we get:

#cos(x): -2a - 4(a-b) + b = 0 #
#sin(x): 2b +4 (a+b) + a = 1 #

Solving simultaneously we get:

# a=5/61 # and #b=6/61#

And so we form the Particular solution:

# y_p = 5/61e^(-t)sint + 6/61e^(-t)cost #

General Solution

Which then leads to the GS of [B}

# y(t) = y_c + y_p #
# \ \ \ \ \ \ \ = Ae^((2-sqrt(3))t) + Be^((2+sqrt(3))t) + 5/61e^(-t)sint + 6/61e^(-t)cost#

Now we initially used a change of variable:

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

So restoring this change of variable we get:

# y = Ae^((2-sqrt(3))lnx) + Be^((2+sqrt(3))lnx) + 5/61e^(-lnx)sinlnx + 6/61e^(-lnx)coslnx#

# :. y = Ax^(2-sqrt(3)) + Bx^(2+sqrt(3)) + 5/61e^(ln(1/x))sinlnx + 6/61e^(ln(1/x))coslnx#

# \ \ \ \ \ \ = Ax^(2-sqrt(3)) + Bx^(2+sqrt(3)) + 5/61 (1/x)sinlnx + 6/61 (1/x)coslnx #

# \ \ \ \ \ \ = Ax^(2-sqrt(3)) + Bx^(2+sqrt(3)) + (5sinlnx)/(61x) + (6coslnx)/(61x) #

Which is the General Solution of [A].