# When does an equation of this type have a closed-form solution?

Suppose I have an equation of the form

$f(x) + a x f'(x) + b = 0.$

Note: I am not looking for a general solution to this equation. I am not interested in solving an ODE of this type! Read below.

I am looking for functions $f(x)$ that let me solve for $x$ explicitly (they need not be true for all $x$, so we are not looking at a differential equation per se!).

So far I could just come up with

$f(x) = p x^r + q$, $\ \ \ \ \ \ r\neq 0,$

which has the trivial solution $x^* = \left(-\frac{b+q}{p(1+ar)}\right)^{\frac{1}{r}}$.

Another idea is to solve for the differential equations

$f(x) = \kappa * \left(x f'(x)\right)^2 + C\ \ \$ and

$f(x) = \kappa * \left(x f'(x)\right)^{1/2} + C$.

Inserting either function into the original equation will lead to a quadratic equation in $x f'(x)$ which implies two solutions for $x f'(x)$. With some "luck", these might be explicitly solvable for $x$.

However, I'm wondering whether there is a more systematic approach -- and more importantly, whether I have overlooked some other simple examples.

Many thanks in advance!

Edit: I just realized that $f(x) = p ln(qx) + r$ also allows for an explicit solution $x^*$.

• Wolfram Alpha finds a general solution of the differential equation. – lhf Aug 20 '13 at 10:00
• @Martin: Are you looking the explicit solution? As, I have found above you hadn't solved the OE. Am I right? – mrs Aug 20 '13 at 10:07
• @lhf: First, I think you should have entered $y + a x y' + b = 0$, which yields a much simpler general solution. However, as indicated in the text, I am not looking for a general solution to this type of equation! I am just looking for functions $f(x)$ that let me solve the equation explicitly for $x$ (as a function of the parameters $a$ and $b$). The equation need not be satisfied in general! – Martin Aug 20 '13 at 10:08
• @ Babak S.: As indicated in the text, I am not looking for a solution to the ODE. I am just interested in types of functions $f(x)$ that let me solve for $x$ explicitly (as a function of the parameters $a$ and $b$). – Martin Aug 20 '13 at 10:11

## 1 Answer

Let $y=f(x)$. The equation is rewritten as:

$$\frac{dy}{y+b}=\frac{dx}{-ax}$$ Therefore: $$Ln(y+b)=-\frac{1}{a}Ln(x)+C$$ $$y=e^{-\frac{1}{a}Ln(x)+C_{1}}-b$$ or: $$y=C_{2}x^{-\frac{1}{a}}-b$$ or: $$x=C_{3}(y+b)^{-a}$$

• Thanks for your effort! However, unless I'm misunderstanding something, this is not the problem I am facing and the answer I am looking for. Please compare with my original question (edited for emphasizing my problem). – Martin Aug 20 '13 at 10:21
• @MartinObradovits Ok, as you see the solution to ODE lets you to derive $x$ in terms of $y=f(x)$. You are free to choose any rule for $f(x)$ and put it in the last equation above. – Amir Kazemi Aug 20 '13 at 10:31
• I don't think this helps me much, as you still impose that the first equation has to be fulfilled for all x. I don't need this property. I am not interested in a solution to an ODE. I want properties on $f(x)$ (where $f(x)$ need not solve the first equation for all $x$!) that allow me to find an explicit solution $x^*(a,b)$. – Martin Aug 20 '13 at 11:47