Solve the differential equation #x^2(d^2y)/(dx^2) + x dy/dx + y = x^m #?

We have:

# x^2(d^2y)/(dx^2) + x dy/dx + y = x^m # ..... [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:

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

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

# :. (d^2y)/(dt^2) + y = e^(tm) # ..... [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

# m^2+1 = 0#

We can solve this quadratic equation, and we get two complex conjugate solutions::

# m=+-i #

Thus the Homogeneous equation [B] has the solution:

# y_c = e^(0t)(Acos(1t)+Bsin(1t) ) #
# \ \ \ =Acost+Bsint #

Particular Solution

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

# y = ae^(tm) #

Where #a# is a constant to be determined by substitution

Let us assume the above solution works, in which case be differentiating wrt #x# we have:

# y' \ \= ame^(tm) #
# y'' = am^2e^(tm)#

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

# am^2e^(tm) + ae^(tm) = e^(tm) #
# :. am^2 + a = 1 #
# :. a(m^2 + 1) = 1 #
# :. a = 1/(m^2 + 1) #

And so we form the Particular solution:

# y_p = e^(tm)/(m^2 + 1) #

General Solution

Which then leads to the GS of [B}

# y(t) = y_c + y_p #
# \ \ \ \ \ \ \ = Acost+Bsint + e^(tm)/(m^2 + 1) #

Now we initially used a change of variable:

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

So restoring this change of variable we get:

# y = Acos(lnx)+Bsin(lnx) + ((e^t)^m)/(m^2 + 1) #
# \ \ = Acos(lnx)+Bsin(lnx) + ((x)^m)/(m^2 + 1) #

Which is the General Solution of [A].

# y = Acos(lnx)+Bsin(lnx) + (x^m)/(m^2 + 1) #