Question

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

Answer 1

`y = (C_1 cos(4 log x)+C_2 sin(4 log x))/x`

Explanation:

Assuming that the differential equation reads

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

proposing a solution with the structure

`y = c x^alpha` and substituting

`x^2 alpha(alpha-1)cx^(alpha-2)+3alphacx^(alpha-1)+17cx^alpha =(alpha(alpha-1)+3alpha+17)c x^alpha = 0`

Solving now

`alpha(alpha-1)+3alpha+17=0` we obtain

`alpha = -1pm 4i` so the solutions are

`y = c_1 x^(-1-4i)+c_2x^(-1+4i)`

but

`x^(4i) = e^(i4 log x)` and

`e^(i4 log x) = cos(4logx)+isin(4logx)`

so we can reduce the solutions to the form

`y = (C_1 cos(4 log x)+C_2 sin(4 log x))/x`

Answer 2

`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.