Prove monotonicity of a function by a given relationship and integration

Question

I have a differentiable function $f:\mathbb{R}\to \mathbb{R}$ for which is true that $$f''(x)=f'(x)\cdot f(x)\text{ }\forall x\in\mathbb{R}$$ and $\exists x_0\in\mathbb{R}$ for which is true that $$f(x_0)=f'(x_0)=k, \quad k\in(0,2).$$ I want to prove that $f$ is strictly increasing.

So I do the following: $$ \begin{split} f''(x)&=f'(x)\cdot f(x)\\ f''(x)+f'(x)&=f'(x)\cdot f(x)+f'(x)\\ (f(x)+f'(x))'&=\left[\frac{1}{2}(f(x)+1)^2\right]'\\ f(x)+f'(x)&=\frac{1}{2}\left(f(x)+1\right)^2+c \quad \forall x\in\mathbb{R} \end{split} $$ I prove that $c=-(k-1)^2$ but now I don't know how to continue. Any ideas?

Answer 1

From your last equation we can get $f'(x)=\frac{1}{2}(f(x))^2+\frac{1}{2}+c$, using the initial condition we obtain that $c=-\frac{1}{2}(k-1)^2$ and $c+\frac{1}{2} >0$ for $0< k < 2$. Thus, $f'(x)>0$

Answer 2

I would proceed a bit differently. Let us integrate the differential equation f''=f'f directly: $$\int_{x_0}^xf''(y) dy = \int_{x_0}^x f'fdy$$ Then you obtain $$f'(x)-f'(x_0) = \frac{1}{2} f(x)^2 - \frac{1}{2} f(x_0)^2$$ hence $$f'(x)= \frac{1}{2} f(x)^2 - \frac{1}{2}k^2+k> 0$$ if $k\in (0,2)$.