1. ## Bessel Equation

I don't know how to solve the following problem!!

Prove that the general solution of,

$y'' + \frac{1-2a}{x}y' + \left[ (bcx^{c-1})^{2} = \frac{a^{2}-n^{2}c^{2}}{x^{2}}\right]y =0$

is given by $y=x^{a}[c_{1}J_{n}(bx^{c}) + c_{2}N_{n}(bx^{c}) ]$

Use this result to solve $y''+xy=0$

Where do I start?? what substitution do I make to make it look like a bessel equation???? Any suggestion?

2. Firstly I'm assuming you meant a - inplace of the equals sign in the brackets. Also, naturally, I'll assume that $a$, $b$ and $c$ are all constants.
To do this you simply use the substitution $y = x^a v(u)$ where $u=b x^c$ to transform the equation in terms of $u$, the function $v(u)$ and its derivatives wrt $u$. So to start you off you should find that

$y^{\prime} = ax^{a-1} \, v + x^a \, \frac{dv}{du} \cdot \frac{du}{dx}$

and, using Leibniz's differentiation of products rule),

$y^{\prime \prime} = a(a-1)x^{a-2} \, v + 2 a x^{a-1}\, \frac{dv}{du} \cdot \frac{du}{dx} + x^a \left( \left(\frac{du}{dx} \right)^2 \cdot \frac{d^2 v}{du^2} + \frac{dv}{du} \cdot \frac{d^2 u}{dx^2} \right)$ .

Further to this you only need to substitute for the derivatives of $u$ wrt $x$ using:

$\frac{du}{dx} = bc x^{c-1}$ and $\frac{d^2 u}{dx^2} = bc(c-1) x^{c-2}$.

Assuming your original equation was correct this should do the trick transforming it into the Bessel equation. Hope that helped!