# What is a second solution to the Differential Equation # x^2y'' -3xy'+5y=0 #?

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We are given that the answer does not contain cosine functions.

We are given that the answer does not contain cosine functions.

##### 2 Answers

See below.

#### Explanation:

Assuming that the differential equation reads

The differential equation

Proposing

then

but

now according to

Second solution is:

#y=Bx^2sin(lnx)#

#### Explanation:

If we assume the a corrected equation:

# x^2y'' -3xy'+5y=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) -3xe^(-t)dy/dt+5y=0 #

# :. ((d^2y)/(dt^2)-dy/dt) -3dy/dt-+5y=0 #

# :. (d^2y)/(dt^2)-4dy/dt+5y=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^2-4m+5 = 0#

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

# m=2+-i#

Thus the Homogeneous equation [B]:

# (d^2y)/(dt^2)-4dy/dt+5y=0 #

has the solution:

#y=e^(2t)(Acost+Bsint)#

Now we initially used a change of variable:

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

So restoring this change of variable we get:

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

# :. y=x^2(Acos(lnx)+Bsin(lnx)) #

# :. y=Ax^2cos(lnx)+Bx^2sin(lnx) #

Which is the General Solution.

From the given answer we can clearly infer that for the first part of the solution