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.

2 Answers
Jun 16, 2017

See below.

Explanation:

Assuming that the differential equation reads

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

The differential equation

x^2y''-3x y'+5y=0 is a linear homogeneous differential equation.

Proposing

y = c_0 x^alpha and substituting

(alpha(alpha-4)+5)c_0x^alpha = 0 and solving for alpha

alpha(alpha-4)+5=(alpha-2+i)(alpha-2-i)

then

y = x^2(c_1 x^i + c_2 x^-i)

but x = e^(log_e x) and x^(pm i)=e^(pm i log_e x)

now according to

e^(i z) = cos z + i sin z we have the general solution.

y = x^2(C_1 sin (log_e x) + C_2 cos(log_e x ))

Jun 21, 2017

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 A=0, but we cannot make any further assumptions about the constant B in the second solution