# The product of the derivative and anti derivative is the function?

When is $$f(x) = f'(x)\int{f(x)}dx$$? I just though of the problem but couldn't solve it myself.

• $f(x)=0$ could work.. :-) – Simply Beautiful Art Mar 30 '17 at 22:11
• It might be better to write $g(x)=\int f(x)dx$; then this equation becomes $g'(x)=g''(x)g(x)$. – Steven Stadnicki Mar 30 '17 at 22:25
• Keep in mind that $\int f(x) \, dx$ is a family of functions. Are we permitted to choose any constant? Maybe it would be better to write it as $f'(x) = f''(x) f(x)$? (I see that Steven has beaten me to the punch!) – Brian Tung Mar 30 '17 at 22:32
• There appear to be no nice solutions: Maple gives a general implicit solution in terms of an elliptic integral function. On the other hand, if we replace the left-hand side with $f(x)^2$ the resulting equation is relatively easy to solve. – Travis Willse Mar 30 '17 at 22:51
• I was thinking of using the constant 0 but feel free if you have a general solution! – mtheorylord Mar 30 '17 at 22:55

We can write your equation as follows substituting $y+C_1=\int f(x)$ $$y'=(y+C_1)y''$$ $$y''=y'/(y+C_1)$$ Now integrating each side gives $$y' = \ln({y+C_1})+C_2$$ $$\frac{1}{\ln{(y+C_1)}+C_2}y'=1$$ And this is as far as we will go. $$\int \frac{1}{\ln{(y+C_1)}+C_2}dy=x$$ So, the crux of this matter comes to finding $\int \frac{1}{\ln{y}}dy$ for which there is no closed form. From here we would take the inverse of this non closed form function, and then take the derivative of that...