I am required to show that the function:
$y(x)=\sqrt{x} \times J_{\frac{1}{4}}{(kx^2/2)}$ - which involves a Bessel function of order $\frac{1}{4}$
satisfies the differential equation of the form:
$y'' + k^2x^2y = 0$
I have used reduction of order - $y(x)= u(x)\sqrt{x}$ - and differentiating, then subbing into the differential equation I get the following equation, which is in the form:
$4x^2 u′′(x) + 4x u′(x)+ (4k^2x^4 − 1)u(x) = 0$
I then tried letting $t = (kx^2)/2$ but I can't get the Bessel Equation of order $\frac{1}{4}$.
Can someone please help me with getting the solution to the Bessel Equation. Thanks.
You got (which is correct) : $$4x^2 u′′(x) + 4x u′(x)+ (4k^2x^4 − 1)u(x) = 0$$ Then change of variable : $$t=\frac{kx^2}{2}\quad\to\quad \frac{dt}{dx}=kx$$ $\frac{du}{dx}=\frac{du}{dt}\frac{dt}{dx}=kx\frac{du}{dt}$
$\frac{d^2u}{dx^2}=\frac{d}{dx}\left(kx\frac{du}{dt}\right)= k\frac{du}{dt}+kx\frac{d^2u}{dt^2}\frac{dt}{dx}=k\frac{du}{dt}+k^2x^2\frac{d^2u}{dt^2}$ $$4x^2 \left(k\frac{du}{dt}+k^2x^2\frac{d^2u}{dt^2}\right) + 4x\left(kx\frac{du}{dt}\right)+ (4k^2x^4 − 1)u = 0$$
$$4k^2x^4\frac{d^2u}{dt^2}+8kx^2\frac{du}{dt} + (4k^2x^4 − 1)u = 0$$
$$16t^2\frac{d^2u}{dt^2}+16t\frac{du}{dt} + (16t^2 − 1)u = 0$$ $$\frac{d^2u}{dt^2}+\frac{1}{t}\frac{du}{dt} + (1 − \frac{1}{16t^2})u = 0$$ Compare to the Bessel's equation : $$\frac{d^2}{dt^2}J_\nu(t)+\frac{1}{t}\frac{d}{dt}J_\nu(t) + (1 − \frac{\nu^2}{t^2})J_\nu(t) = 0$$ $\nu^2=\frac{1}{16}\quad\to\quad \nu=\frac{1}{4}\quad\to\quad u=J_{1/4}(t)=J_{1/4}\left(\frac{kx^2}{2}\right)\quad\to\quad y=\sqrt(x)J_{1/4}\left(\frac{kx^2}{2}\right)$