Functions such that $xf''(x) + f'(x) > 0$ I'm struggling on how to approach this question:
Find functions $f$ such that $xf''(x) + f'(x) > 0$ with $x \in \mathbb{R}^+$
Any help?
 A: Consider $g(x)=xf'(x)$; then $g'(x)=f'(x)+xf''(x)$, so you want to find the function $g$ so that it's increasing and $g(0)=0$. If $g$ is any differentiable function with $g(0)=0$ and $g'(x)>0$ for all $x$, then define
$$
F(x)=\begin{cases}
g'(0) & \text{if $x=0$}\\[3px]
\dfrac{g(x)}{x} & \text{if $x\ne0$}
\end{cases}
$$
and
$$
f(x)=\int_{0}^x F(t)\,dt
$$
will satisfy your requests, because
$$
f'(x)=F(x)
$$
and
$$
xf'(x)=g(x)
$$
for all $x$, so $g'(x)=xf'(x)+f''(x)>0$ by assumption.
Now it's only a matter of choosing a function $g$: if $g(x)=x$, you surely have $g'(x)>0$ and $g(0)=0$, so $F(x)=1$ and $f(x)=x$. But you can also choose
$$
g_1(x)=\arctan x,\qquad g_2(x)=e^x-1,\qquad g_3(x)=x^3+3x,\qquad \dots
$$
A: \begin{equation}
xf''(x)+f'(x) = \frac{d}{dx}\left(xf'(x)\right)
\end{equation}
Which wlog
\begin{equation}
\implies xf'(x)=c
\end{equation}
where $c$ is a constant. since $x \in \mathbb{R^{+}}$, we write wlog
\begin{equation}
f'(x)=\frac{c}{x}
\end{equation}
You can maybe take it from here?
A: Solve the differential equation $(xf'(x))'=xf''(x)+f'(x)=1$ for example. You could also replace $1$ with $e^x$
A: Let $p(x)$ be a strictly positive function. Its antiderivative $P(x)$ is a strictly increasing function.
Integrating twice from $0$ to $x$,
$$xf''(x)+f'(x)=p(x),$$
$$xf'(x)=0f'(0)+\int_0^x p(x)\,dx+c=P(x)-P(0),$$
$$f'(x)=\frac{P(x)-P(0)}{x},$$
$$f(x)=f(0)+\int_0^x\frac{P(x)-P(0)}xdx.$$

For a nontrivial example, take $P(x)=-\dfrac1{x+1}$. Then,
$$f(x)=f(0)+\int_0^x\frac{dx}{x+1}=\ln(x+1)+c.$$
Check:
$$f'(x)=\frac1{x+1},$$
$$f''(x)=-\frac1{(x+1)^2},$$
$$xf''(x)+f'(x)=\frac1{(x+1)^2}>0.$$
A: The LHS is equal to
$$\frac{d}{dt}\left(tf'(t)\right)>0$$
Integrating from $0$ to $x$:
$$\int_{0}^x  \frac{d}{dt}\left(tf'(t)\right) dt=xf'(x)-0\cdot f'(0)>0$$
How $x>0$
$$f'(x)>0$$
Then any function strictly inscreasing satisfy your equation.
