Question

Solve the differential equation:


`{x}^(2) \frac{ {d}^(2)y }{ {dx}^(2) } + x (dy)/(dx) + (x - 1)y = 0`
​

Answer 1

`y = C_1J_2(3√(x))+C_2Y_2(3√(x))`

Explanation:

`x^(2) (d^2 y)/(dx^2) + x(dx)/(dy) + (x-1)y =0`

Considering
`x=0` ordinary point we could use Frobenius method assuming

`$y = \sum_(n=0)^(\infty) c_nx^n$`

and consider the series representation given by the method to solve the equation. I think this way is harder.

In your case, It seems to have something to do with Bessel differential equation

`x^2 y'' + x y' + (x^2-n^2) y = 0`

that will have the same representation as the Frobenius method. To get the form
`(x^2-n^2)` as we have
`(x-1)` in the given equation we can substitute
`x` and take
`x = (t^2)/(9)`

Thus,

`(dy)/(dx) = (3)/(t)(dy)/(dt) \implies (d^2y)/(dx^2) = (9)/(t^2)(d^2y)/(dt^2)-(9)/(t^3)(dy)/(dt)`

`$\implies (1)/(3^2)\left(t^2\frac{{d}^2y}{{d}t^2}+t\frac{{d}y}{{d}t}+(t^2-3^2)y\right) = 0$`

The solution will have the format

`y( x) = {C_1}{J_n}( x) + {C_2}{Y_n}( x )`

such that

`${J_n}( x ) = \sum\limits_(p = 0)^\infty {\frac{{{{( { - 1} )}^p}}}{{\Gamma ( {p + 1} )\Gamma ( {p + n + 1} )}} {{\left( {(x)/(2)} \right)}^(2p + n)}}$`

`${Y_n}\left( x \right) = \frac{{{J_n}\left( x \right)\cos \pi n - {J_( - n)}\left( x \right)}}{{\sin \pi n}}$`

and
`C_1, C_2` are constants

And the order of the Bessel equation is
`n=2`

In our case, we can just write

`y(t) = C_1J_2(t)+C_2Y_2(t)`

`y = C_1J_2(3√(x))+C_2Y_2(3√(x))`