Question

Solve the Differential Equation `x^2y'' -xy'-8y=0`?

Answer 1

`y(x) = c_1/x^2+c_2x^4`

Explanation:

`x^2y''-xy'-8y=0`

This is a Cauchy-Euler differential equation of second order, so the resolution method is to substitute as possible solution:

`y(x) = x^n`

and its derivatives:

`y'(x) = nx^(n-1)`

`y''(x) = n(n-1)x^(n-2)`

Substituting in the original equation we have:

`x^2((n(n-1)x^(n-2))-nx x^(n-1)-8x^n =0`

`(n(n-1)x^n-nx^n-8x^n =0`

`x^n(n^2-n-n-8) =0`

`x^n(n^2-2n-8) =0`

So for `x!=0` we must have:

`n^2-2n-8 =0`

`n= (1+-sqrt(1+8))`

`n_1=-2`
`n_2 = 4`

The general solution is then:

`y(x) = c_1/x^2+c_2x^4`

Answer 2

`y = A/x^2+Bx^4`

Explanation:

We have:

`x^2y'' -xy'-8y=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) -xe^(-t)dy/dt-8y=0`

`:. ((d^2y)/(dt^2)-dy/dt) -dy/dt-8y=0`

`:. (d^2y)/(dt^2)-2dy/dt-8=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-8 = 0`

We can solve this quadratic equation, and we get two real and distint roots:

`m=-2,4`

Thus the Homogeneous equation [B] has the solution:

`y=Ae^(-2t)+Be^(4t)`

Now we initially used a change of variable:

`x = e^t => t=lnx`

So restoring this change of variable we get:

`y = Ae^(-2lnx)+Be^(4lnx)`

`:. y = Ae^(lnx^(-2))+Be^(lnx^4)`
`:. y = Ax^(-2)+Bx^4`
`:. y = A/x^2+Bx^4`

Which is the General Solution.