Solve the Differential Equation # x^2y'' -3x+5y=0 #?

We have:

# x^2y'' -3x+5y=0 #

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 we get:

# x^2((d^2y)/(dt^2)-dy/dt)e^(-2t) -3e^t+5y=0 #

# :. (d^2y)/(dt^2)-dy/dt +5y=3e^t # ..... [A]

This is now a second order linear 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-m+5 = 0#

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

# m=1/2(1+-sqrt(19)i) #

Thus the Homogeneous equation:

# (d^2y)/(dt^2)-dy/dt +5y=0 #

has the solution:

#y=e^(1/2t)(Acos(1/2sqrt(19)t) + Bcos(1/2sqrt(19)t))#

We now seek a Particular Solution of [A], which will be of the form:

# y = Ae^t #

Differentiating wrt #t# we get:

# y' = Ae^t #, and, ># y'' = Ae^t #

Substituting into the DE [A} we get:

# Ae^t-Ae^t+5Ae^t=3e^t #

Equating coefficients we have:

# A-A+5A=3 => A=3/5 #

Hence, the GS of [A] is the combination of the homogeneous solution and the particular solution, thus:

# y=e^(1/2t)(Acos(1/2sqrt(19)t) + Bcos(1/2sqrt(19)t)) +3/5e^t#

Now we initially used a change of variable:

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

So restoring this change of variable we get:

# y=e^(1/2lnx)(Acos(1/2sqrt(19)lnx) + Bcos(1/2sqrt(19)lnx)) +3/5e^lnx#

# :. y=e^(lnsqrt(x))(Acos(1/2sqrt(19)lnx) + Bcos(1/2sqrt(19)lnx)) +3/5x#

# :. y=sqrt(x)(Acos(1/2sqrt(19)lnx) + Bcos(1/2sqrt(19)lnx)) +3/5x#

Which is the General Solution