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