Second order involving an exponential, Bessels Solution

This differential equation,

`y''+aexp(kx)y=0`

has the solution,

`y=c_1J_0(w)+c_2Y_0(w)` with `w=(2sqrt(a))/|k| exp((kx)/2)`

Where `J_0` and `Y_0` are the Bessel Functions of the First and Second Kind respectively.

Derivation

We use the substitution, `b=a/k^2` and we obtain,

`d/(dx) (d/(dx)y)+k^2bexp(kx)y=0`

Then use the substitution `z=exp(kx)`, which requires,

`(dz)/(dx)=kexp(kx)=kz` and `1/(dx)=kz 1/(dz)`

The equation becomes,

`kz d/(dz)  (kz d/(dz) y) +k^2bzy=0`

The constant k is moved outside the differentiation, and we divide by z,

`k^2 d/(dz) (z (dy)/(dz))  +k^2by=0`

Dividing then by `k^2` we obtain,

`d/(dz) (z (dy)/(dz))  +by=0`

We then apply the product rule, and arrive at a special case of the Bessel differential equation,

`(z y'_(zz))' +by=0=zy''_(zz)+y'_z+by`

Compare this to the Bessel Differential Equation, cf:

`x^2y''+xy'+(x^2-v^2)y=0`

On multipling our equation by z, we can see that `v=0`, which means the solution will involve Bessel function of the First and Second Kind, `J_0` and `Y_0`. The differential equation has the solution,

`y=c_1J_0(2sqrt(bz))+c_2Y_0(2sqrt(bz))`

Reversing the substitution for b, and then for z,

`y=c_1J_0(2sqrt(a/k^2exp(kx)))+c_2Y_0(2sqrt(a/k^2exp(kx)))`

Once simplified we obtain the result,

`y=c_1J_0((2sqrt(a))/|k|exp((kx)/2))+c_2Y_0((2sqrt(a))/|k|exp((kx)/2))`