I am solving the following equation: $$f(x)^2+\frac{f(x)^2}{f'(x)^2}=h(x)$$
for a known function $h(x)\geq 0, x\in \mathbb{R}$.
I'm asking for help with:
Identifying this ODE in terms of well-known ODE types (Ricatti equation, Bernoulli eqation, etc.).
Solving the equation.
Proving the existence and uniqueness for Cauchy problems (I think Peano's theorem applies here, but I would be grateful for confirmation).
Solving for special cases. So far with Mathematica I managed to get a solution for $h(x)=const$. and $h(x)=x^2$. If you can generate more solutions, I will be very grateful.
Thanks, Michał
EDIT:
E4.1 A solution for $h(x)=x^n$ seems achievable.
E2.1 I have been trying to solve this equation since 2 months now. One path that I did not explore fully is to factor the equation. Putting $h(x)=-g(x)^2$ we can write $$f(x)^2+\frac{f(x)^2}{f'(x)^2}+g(x)^2=0$$ Using quaternions we could factor this as $$(if(x)+j f(x)/f'(x)+k g(x))(-if(x)-j f(x)/f'(x)-k g(x))=0$$ Which would have the same solution as the following quaternion ODE (this step is "an uneducated guess") $$if(x)+j f(x)/f'(x)+k g(x)=0$$ and then to recover the solution I consider I simply take the real part. Does this make any sense for the quaternion experts?
E2.2 With some substitutions one can recover the form I posted here around 2 months ago (Almost Abel Form)