What is solution for Bessel Equation of form $x^2y'' + xy' + (t^2x^2-1)y = 0$?

Question

$$x^2y'' + xy' + (t^2x^2-1)y = 0$$

Solution to SOLDE (second order linear differential equation) of form mentioned above which looks like Bessel Equation. For simplicity you can take x>0.

Answer 1

$$x^2y''(x)+xy'(x)+(x^2t^2-1)y(x)=0$$ Supposing that $t$ is a parameter not function of $x$, the change of variable $X=tx$ leads to : $$X^2y''(X)+Xy'(X)+(X^2-1)y(X)=0$$ This is a standard Bessel ODE which solution is : $$y(X)=c_1J_1(X)+c_2Y_1(X)$$ Bessel functions of the first and second kind respectively. $$y(x)=c_1J_1(tx)+c_2Y_1(tx)$$

NOTE after the Kartik Chhajed's comment :

If you are confused with the symbols $y'(x)$ and $y'(X)$, better use the less confusing $\frac{dy}{dx}$ and $\frac{dy}{dX}=\frac{dy}{d(tx)}=\frac{dy}{tdx}=\frac{1}{t}\frac{dy}{dx}$ , thus $y'(X)=\frac{1}{t}y'(x)$

and $\frac{d^2y}{dX^2}=\frac{1}{t^2}\frac{d^2y}{dx^2}$ , thus $y''(X)=\frac{1}{t^2}y''(x)$

Answer 2

Use the substitution $u=tx$. The idea is simplify like in a Cauchy-Euler ODE; then aply chain rule:

$$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=t\frac{dy}{du}\\\frac{d^2y}{dx^2}=\frac{d}{dx}\left(t\frac{dy}{du}\right)=t\frac{d\left(\frac{dy}{du}\right)}{dx}=t\frac{d\left(\frac{dy}{du}\right)}{du}\frac{du}{dx}$$

Then $$\frac{d^2y}{dx^2}=t^2\frac{d^2y}{du^2}$$

Substitute $y'$, $y''$ and $u=tx$ in the ODE

$$\frac{u^2}{t^2}[t^2y''(u)]+\frac ut[ty'(u)]+(u^2-1)y(u)=0\\u^2y''(u)+uy'(u)+(u^2-1)y(u)=0$$

Now it's a Bessel ODE which solution is:

$$y(u)=C_1J_1(u)+C_2J_{-1}(u)$$ or the same but using the Bessel of Second kind: $$y(u)=C_1J_1(u)+C_2Y_1(u)$$ Replacing U and that's all $$y(tx)=C_1J_1(tx)+C_2Y_1(tx)$$ *maybe the $t$ could get out of the bessel functions