Functions such that $xf''(x) + f'(x) > 0$ I'm struggling on how to approach this question:

Find functions $f$ such that $xf''(x) + f'(x) &gt; 0$ with $x \in \mathbb{R}^+$

Any help?
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 $g'(x)&gt;0$ for all $x$, then define
$$
F(x)=\begin{cases}
g'(0) &amp; \text{if $x=0$}\\[3px]
\dfrac{g(x)}{x} &amp; \text{if $x\ne0$}
\end{cases}
$$
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 $g'(x)=xf'(x)+f''(x)&gt;0$ by assumption.

Now it's only a matter of choosing a function $g$: if $g(x)=x$, you surely have $g'(x)&gt;0$ and $g(0)=0$, so $F(x)=1$ and $f(x)=x$. But you can also choose
$$
g_1(x)=\arctan x,\qquad g_2(x)=e^x-1,\qquad g_3(x)=x^3+3x,\qquad \dots
$$
The LHS is equal to

$$\frac{d}{dt}\left(tf'(t)\right)&gt;0$$

Integrating from $0$ to $x$:

$$\int_{0}^x  \frac{d}{dt}\left(tf'(t)\right) dt=xf'(x)-0\cdot f'(0)&gt;0$$

How $x&gt;0$

$$f'(x)&gt;0$$

Then any function strictly inscreasing satisfy your equation.