Question

Solve the differential equation:
${x}^{2} \frac{ {d}^{2}y }{ {dx}^{2} } + x \frac{dy}{dx} + (x - 1)y = 0$
​

Answer 1

Answer:

$y = C_1J_2(3\sqrt{x})+C_2Y_2(3\sqrt{x})$

Step-by-step explanation:

$x^{2} \dfrac{d^2 y}{dx^2} + x\dfrac{dx}{dy} + (x-1)y =0$

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 $x = \dfrac{t^2}{9}$

Thus,

$\dfrac{dy}{dx} = \dfrac{3}{t}\dfrac{dy}{dt} \implies \dfrac{d^2y}{dx^2} = \dfrac{9}{t^2}\dfrac{d^2y}{dt^2}-\dfrac{9}{t^3}\dfrac{dy}{dt}$

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

The solution will have the format

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

such that

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

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

and $C_1, C_2$ are constants

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

In our case, we can just write

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

$y = C_1J_2(3\sqrt{x})+C_2Y_2(3\sqrt{x})$