Doubts about change of variable in ODE

[arroy_0205]

I was trying to check some calculations from a paper myself and I got
stuck at a differential equation. Can anybody help me with suggestions?
I am giving a simpler version of the differential eqn by setting all numerical
constants=1. Consider:
$$e^{-2y}\left(z''(y)-2z'(y)+\frac{3}{4}z(y)\right)+z(y)=0$$
We change variable to
$$u=e^y$$
then the ODE is
$$u^2z''(u)-uz'(u)+z(u)(u^2+\frac{3}{4})=0$$
The author writes the solution is:
$$z(y)=\frac{1}{u^{3/2}}z(u)=u\left(J_{\frac{1}{2}}(u)+c_1N_{\frac{1}{2}}(u) \right)$$
where J and N are Bessel functions. My problem is, I am getting the relation
$$z(u)=u\left(c_1 J_{\frac{1}{2}}(u)+c_2 N_{\frac{1}{2}}(u) \right)$$
But note that according to the author,
$$z(y)={u^{5/2}}\left(J_{\frac{1}{2}}(u)+c_1N_{\frac{1}{2}}(u) \right)$$
Or in other words though I can get the desired solution in terms of changed variable,
I am not able to get the relation:
$$z(y)=\frac{1}{u^{3/2}}z(u)$$
at all. How to get this relation? Also, note that the author uses one constant
in the solution but I think there should be two independent constants. What am I missing in the calculations?

 

[gtoguy97]

One approach to solve the differential equation is to use the method of variation of parameters. To do this, we need to first solve the homogeneous equation, which in this case is:e^{-2y}\left(z''(y)-2z'(y)+\frac{3}{4}z(y)\right)+z(y)=0The homogenous equation has two linearly independent solutions, which can be written as:z_1(y) = c_1 e^{-y/2} z_2(y) = c_2 e^{y/2}We then need to find a particular solution of the form:z_p(y) = u(y)z_1(y) + v(y)z_2(y)Substituting this into the original differential equation, we obtain:u'(y)z_1(y) - v'(y)z_2(y) + u(y)z_1'(y) + v(y)z_2'(y) + \frac{3}{4}[u(y)z_1(y) + v(y)z_2(y)] = 0Solving for u' and v', we obtain:u'(y) = \frac{3}{4}u(y) - \frac{1}{2}v(y)v'(y) = \frac{3}{4}v(y) + \frac{1}{2}u(y)We can then solve these equations for u and v, and thus we have the solution:z(y) = c_1 e^{-y/2}u(y) + c_2 e^{y/2}v(y)Using the change of variables u = e^y, we can rewrite this as:z(u) = u \left(c_1 u^{-3/2} J_{1/2}(u) + c_2 u^{3/2} N_{1/2}(u) \right)We then have the relation:z(y) = \frac{1}{u^{3/2