# What is the general solution of the differential equation? # x^2y'' +3xy'+17y=0 #

##### 2 Answers

#### Explanation:

Assuming that the differential equation reads

proposing a solution with the structure

Solving now

but

so we can reduce the solutions to the form

# y=(Acos(4lnx))/x+(Bsin(4lnx))/x#

#### Explanation:

We have:

# x^2y'' +3xy'+17y=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+17y=0 #

# :. ((d^2y)/(dt^2)-dy/dt) +3dy/dt+17y=0 #

# :. (d^2y)/(dt^2)+2dy/dt+17y=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+2m+17 = 0#

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

# m=-1+-4i#

Thus the Homogeneous equation [B] has the solution:

# y=e^(-t)(Acos4t+Bsin4t)#

Now we initially used a change of variable:

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

So restoring this change of variable we get:

# y=(x^(-1))(Acos(4lnx)+Bsin(4lnx))#

# :. y=(Acos(4lnx))/x+(Bsin(4lnx))/x#

Which is the General Solution.