ODE with functional argument

Question

I am trying to find a closed form solution for this ODE: $(a_0+a_1x)f(x)+(b_0x+b_1x^2)f^\prime (x) = (c_0+c_1x)f(d_0+d_1x)$. It is not a simple ODE due to the $f(d_0+d_1x)$ term. I also found a related subject as the delayed functional ODE that allows terms like $f(x-d_0)$. Is there any approach to find a closed form solution for this problem? Thanks for your advise.

Answer 1

By a linear change of variables, we may assume $d_0 = 0$. Then if we assume a formal series solution of the form $$ f(x) = \sum_{k=0}^\infty A_k x^{k+p} $$ we get a recursion for the coefficients $$ (a_0 + (k+p) b_0 - c_0 d_1^{k+p}) A_k + (a_1 + (k+p-1) b_1 - c_1 d_1^{k+p-1}) A_{k-1} = 0, \ k \ge 1 $$ with $(a_0 + p b_0 - c_0 d_1^p) A_0 = 0$. For a nonzero solution, we'll want to choose $p$ so that $a_0 + p b_0 - c_0 d_1^p = 0$. This can be solved using the Lambert W function:

$$p = -\frac{W\left(-\dfrac{\ln(d_1) c_0}{b_0} d_1^{-a_0/b_0}\right)}{\ln(d_1)} -{\frac {a_{{0}}}{b_{{0}}}} $$