What type of functional equation is this?

Question

I'm trying to solve the following functional equation

$f\left(x\right)=A\mbox{ exp}\left\{ \int\frac{1}{f\left(x\right)x^{2}+Bx}dx\right\}$

where $f\left(x\right):\mathbb{R}_{+}\rightarrow\mathbb{R}_{\geq0}$, A and B are constants in $\mathbb{R}$.

Does anybody recognize this type of functional equation so I can look up for the solution?

Alternatively, does anybody know how to solve it? Or a suggestion on how to start tackling this problem?

Thank you!

After the initial comments I realized the problem is equivalent to solving the following first-order nonlinear ODE: $f^{\prime}f=\frac{1}{x^{2}}\left(\left(Ax+Bx^{2}\right)f^{\prime}+Bf\right)$

This equation seems similar to an Abel differential equation of the second kind, i.e.

$ff^{\prime}=g\left(x\right)f+h\left(x\right)$

although it's not quite the same. If anybody has an idea how to deal with it I would appreciate it!

Answer 1

Differentiate and divide the result by the original expression (on second thought: simply take logarithms, then differentiate) to get a first-order nonlinear ODE :

$$ \ln(f(x))'=1/(f(x)x^2+Bx). $$

If $B=0$ you get an explicit solution. If $B>0$ there exists more and more slowly yet strictly increasing solutions for all $x\geq 0$

Answer 2

First, there is no concept about "indefinite integral equation" , so you should modify the question as $f(x)=Ae^{\int_k^x\frac{1}{f(x)x^2+Bx}dx}$

$\ln\dfrac{f(x)}{A}=\int_k^x\dfrac{1}{f(x)x^2+Bx}dx$

$\dfrac{1}{f(x)}\dfrac{df(x)}{dx}=\dfrac{1}{f(x)x^2+Bx}$ with $f(k)=A$

$f\dfrac{dx}{df}=fx^2+Bx$ with $x(A)=k$

Case $1$: $B=0$

Then $f\dfrac{dx}{df}=fx^2$ with $x(A)=k$

$df=\dfrac{dx}{x^2}$ with $f(k)=A$

$\int_A^fdf=\int_k^x\dfrac{dx}{x^2}$

$[f]_A^f=\left[-\dfrac{1}{x}\right]_k^x$

$f-A=\dfrac{1}{k}-\dfrac{1}{x}$

$f(x)=A+\dfrac{1}{k}-\dfrac{1}{x}$

Case $1$: $B=0$

Then $f\dfrac{dx}{df}=fx^2+Bx$ with $x(A)=k$

$\dfrac{dx}{df}-\dfrac{Bx}{f}=x^2$ with $x(A)=k$

Let $x=\dfrac{1}{y}$ ,

Then $\dfrac{dx}{df}=-\dfrac{1}{y^2}\dfrac{dy}{df}$

$\therefore-\dfrac{1}{y^2}\dfrac{dy}{df}-\dfrac{B}{fy}=\dfrac{1}{y^2}$ with $y(A)=\dfrac{1}{k}$

$\dfrac{dy}{df}+\dfrac{By}{f}=-1$ with $y(A)=\dfrac{1}{k}$

I.F. $=e^{\int\frac{B}{f}df}=e^{B\ln f}=f^B$

$\therefore\dfrac{d(f^By)}{df}=-f^B$

$f^By=-\int f^B~df$

$f^By=\begin{cases}-\dfrac{f^{B+1}}{B+1}+C&\text{when}~B\neq0,-1\\-\ln f+C&\text{when}~B=-1\end{cases}$

$y(A)=\dfrac{1}{k}$ :

$\dfrac{A^B}{k}=\begin{cases}-\dfrac{A^{B+1}}{B+1}+C&\text{when}~B\neq0,-1\\-\ln A+C&\text{when}~B=-1\end{cases}$

$C=\begin{cases}\dfrac{A^B}{k}+\dfrac{A^{B+1}}{B+1}&\text{when}~B\neq0,-1\\\dfrac{1}{kA}+\ln A&\text{when}~B=-1\end{cases}$

$\therefore f^By=\begin{cases}\dfrac{A^B}{k}+\dfrac{A^{B+1}}{B+1}-\dfrac{f^{B+1}}{B+1}&\text{when}~B\neq0,-1\\\dfrac{1}{kA}+\ln A-\ln f&\text{when}~B=-1\end{cases}$

$\dfrac{f^B}{x}=\begin{cases}\dfrac{A^B}{k}+\dfrac{A^{B+1}}{B+1}-\dfrac{f^{B+1}}{B+1}&\text{when}~B\neq0,-1\\\dfrac{1}{kA}+\ln A-\ln f&\text{when}~B=-1\end{cases}$

$x=\begin{cases}\dfrac{1}{\dfrac{A^B}{kf^B}+\dfrac{A^{B+1}}{(B+1)f^B}-\dfrac{f}{B+1}}&\text{when}~B\neq0,-1\\\dfrac{1}{\dfrac{f}{kA}+f\ln A-f\ln f}&\text{when}~B=-1\end{cases}$