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Show that the general solution of \[ \dfrac {d^2 y }{d x^2} + 4x^2 y=0 \ is \ y=\sqrt{z} [A J_{1/4} (x^2) + B \ J_{-1/4} (x^2)] \] where A and B are constants.
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Let  $u=x^2 \ or \ x=\sqrt{u} \ \ \to (1)$

Differentiate w.r.t.x,

$\therefore \dfrac {du}{dx} = 2x =2\sqrt{u} \ \to (2) \ (from \ 1)$

$\therefore \dfrac{dy}{dx} = \dfrac {dy}{du} \times \dfrac {du}{dx} = 2 \sqrt{u} \dfrac {dy}{du} \ (from \ 2)$

Again, differentiate w.r.t.x,

$\therefore \dfrac {d}{dx} \left ( \dfrac {dy}{dx} \right )= \dfrac {d}{dx} \left ( 2 \sqrt{u} \dfrac {dy}{du} \right )$

$\therefore \dfrac {d^2 y}{dx^2} = \dfrac {d}{du} \left ( 2 \sqrt{u} \dfrac {dy}{du} \right ) \times \dfrac {du}{dx}$

$\therefore \dfrac {d^2y}{dx^2} = 2 \left (\sqrt{u}\cdot \dfrac {d^2 y}{du^2} + \dfrac {dy}{du}\cdot \dfrac {1}{2 \sqrt{u}} \right )\times 2 \sqrt{u} \ \ (from \ 2)$

$\therefore \dfrac {d^2 y}{dx^2}= 4 u\dfrac {d^2 y}{du^2} + 2 \dfrac {dy}{du} \ \ \to (3)$

Given D.E. is

$\dfrac {d^2y}{dx^2} + 4x^2 y = 0$

$\therefore 4u \dfrac {d^2 y}{du^2}+ 2 \dfrac {dy}{du}+ 4 uy=0 \ \to (4) \ \ (from \ 1 \ \& \ 3)$

$Now, put y= u^{1/4} t \to (5)$

Differentiate w.r.t. u, 

$\therefore \dfrac {dy}{du}= 4^{1/4}\cdot \dfrac {dt}{du}+ t\cdot \dfrac {1}{4}u^{-3/4} \ \ \to (6) \ and,$

Again differentiate w.r.t. u,

$\therefore \dfrac {d^2 y}{du^2} = \left [u^{1/4}\cdot \dfrac {d}{du} \left ( \dfrac {dt}{du} \right ) + \dfrac {dt}{du}\cdot \dfrac {1}{4} u^{-3/4} \right ] + \dfrac {1}{4} \left [t\cdot \dfrac {-3}{4}u^{-7/4} + u^{-3/4} \cdot \dfrac {dt}{du}\right ] $

$= u^{1/4} \dfrac {d^2t}{du^2} + \dfrac {1}{4}u^{-3/4} \dfrac {dt}{du}- \dfrac {3}{16}tu^{-7/4} + \dfrac {1}{4} u^{-3/4}\dfrac {dt}{du}$

$\therefore 4u^{-5/4} \dfrac {d^2 t}{du^2} + 4 u^{1/4} \dfrac {dt}{du}- \dfrac {1}{4} tu^{-3/4} + 4u^{5/4} t=0 $

Multiply throughout by $\dfrac {u^{3/4}}{4},$

$\therefore u^2 \dfrac {d^2 t}{du^2}+ u \dfrac {dt}{du} - \dfrac {1}{16} + u^2 t=0 $

$\therefore u^2 \dfrac {d^2 t}{du^2}+ u \dfrac {dt}{du} + \left [ u^2 - \left ( \dfrac {1}{4} \right )^2 \right ] t=0 \ \to (s)$

But, Bessel's Equation is $u^2 \dfrac {d^2 t}{du^2} + u\dfrac {dt}{du} + (u^2 - n^2) t=0 \to (9)$

And its solution is $AJ_n (u) + BJ_{-n}(w) \to (10)$

Comparing (8) and (9) we get, $n= \dfrac {1}{4}$

∴ From (10), $t=AJ_{1/4} (u) + BJ_{-1/4} (u)$

$\therefore \dfrac {y}{u^{1/4}} = AJ_{1/4}(u)+ BJ_{-1/4}(u) \ \ (From \ 5)$

$\therefore y=u^{1/4} \Big [ AJ_{1/4}(u) + BJ_{-1/4}(u) \Big ]$

$\therefore y(x^2)^{1/4} \Big [ AJ_{1/4}(x^2)+BJ_{-1/4}(x^2) \Big ] \ (From \ 1)$

$\therefore y=\sqrt{x} \Big [ AJ_{1/4}(x^2) + BJ_{-1/4}(x^2) \Big ]$

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