I'm trying to solve the following differential equation with Wolfram Alpha (I have a pro subscription):
y''+2*a*y'+(a^2-c^2-b/d*Exp(-x/d)-b^2*Exp(-2*x/d))*y=0
I want to assume that 'a', 'b', 'c', 'd' and 'x' are real positive numbers. How do I do this in Wolfram|Alpha?
Currently, by not assuming that, I'm getting analytic solutions but it contains associated Laguerre polynomials 'L': $$ y(x)=k_1 U\left(c d+1,2 c d+1,2 b d e^{-x/d}\right) \exp \left(d \left((a+c) \log \left(e^{-x/d}\right)-b e^{-x/d}\right)\right)+k_2 L_{-c d-1}^{2 c d}\left(2 b d e^{-x/d}\right) \exp \left(d \left((a+c) \log \left(e^{-x/d}\right)-b e^{-x/d}\right)\right) $$ where the n=-cd-1 is a not a real number.
Edit: Didn't know but Laguerre polynomials are defined for non-integer orders as well. My question still remains for the syntax of "assuming" in Wolfram Alpha though.
Wolfram Alpha can read Mathematica syntaxis, and so I would use the full correct expression with
Assumptions, for example:However, this seems too complicated for WA.
But your equation has too many unnecessary parameters.
Let's move to a new variable:
$$t=x \cdot d, \quad y(x)=f(t)$$
And new parameters:
$$ad=A, \quad bd=B, \quad cd=C$$
Now we have:
$$f''(t)+2 A f'(t)+ \left(A^2-\text{C}^2-B^2 \exp (-2 t)-B \exp (-t)\right)f(t)=0$$
This is not all, we can also move to another variable:
$$s=t-\log B, \quad f(t)=g(s)$$
$$g''(s)+2 A g'(s)+ \left(A^2-\text{C}^2-\exp (-2 t)-\exp (-t)\right)f(t)=0$$
Note that Mathematica has capital C reserved, so I will use
Cc:Or:
$$g(s)=c_1 U\left(\text{C}+1,2 \text{C}+1,2 e^{-s}\right) e^{(A+\text{C}) \log \left(e^{-s}\right)-e^{-s}}+c_2 L_{-\text{C}-1}^{2 \text{C}}\left(2 e^{-s}\right) e^{(A+\text{C}) \log \left(e^{-s}\right)-e^{-s}}$$
Wolfram Alpha still doesn't take this, however it doesn't matter as we still get the exact same solution you've got, in terms of generalized Laguerre polynomial and confluent hypergeometric function.