$$g’’(x) + f^2*g’(x) + 4 g(x) = 0$$$$ g(0) = 0$$$$ g’(\pi/2) = 0$$ where $x \in\mathbb{R}$, $f$ is frequency.
If we use Laplace transform, the initial value of g'(pi/2) makes me cannot go further, but we don't have any ideas to do the Fourier transform. Can anyone give some comments?
Edit: Solution by Laplace transform
Thanks to @Mikako's clarification, the symbol "$*$" means scalar multiplication instead of convolution. As such, let us deal with \begin{align} g''+f^2g+4g&=0&&x\in\mathbb{R},\\ g&=0&&x=0,\\ g'&=0&&x=\pi/2. \end{align}
Apply Laplace transform to the governing ODE, we have $$ \left[s^2G(s)-sg(0)-g'(0)\right]+f^2\left[sG(s)-g(0)\right]+4G(s)=0, $$ where $G(s)=\mathcal{L}\{g\}(s)$. Note that $g(0)=0$, and the above equation reduces to $$ \left(s^2+f^2s+4\right)G(s)=\alpha, $$ where $\alpha=g'(0)$ will soon be determined by $g'(\pi/2)=0$. We have $$ G(s)=\frac{\alpha}{s^2+f^2s+4}=\frac{\alpha}{\left(s+f^2/2\right)^2+\left(4-f^4/4\right)}, $$ or more specifically, \begin{align} G(s)&=\frac{\alpha}{\sqrt{4-f^4/4}}\frac{\sqrt{4-f^4/4}}{\left(s+f^2/2\right)^2+\left(4-f^4/4\right)},&&\text{if }\left|f\right|<2,\\ G(s)&=\alpha\frac{1}{\left(s+f^2/2\right)^2},&&\text{if }\left|f\right|=2,\\ G(s)&=\frac{\alpha}{\sqrt{f^4/4-4}}\frac{\sqrt{f^4/4-4}}{\left(s+f^2/2\right)^2-\left(f^4/4-4\right)},&&\text{if }\left|f\right|>2. \end{align} Perform the Laplace inverse transform for each case, and we obtain \begin{align} g(x)&=\frac{\alpha}{\sqrt{4-f^4/4}}e^{-\left(f^2/2\right)x}\sin\left(x\sqrt{4-f^4/4}\right),&&\text{if }\left|f\right|<2,\\ g(x)&=\alpha xe^{-\left(f^2/2\right)x},&&\text{if }\left|f\right|=2,\\ g(x)&=\frac{\alpha}{\sqrt{f^4/4-4}}e^{-\left(f^2/2\right)x}\sinh\left(x\sqrt{f^4/4-4}\right),&&\text{if }\left|f\right|>2. \end{align} Finally, apply $g'(\pi/2)=0$ to each case, and their respective $\alpha$ could be determined accordingly. For $\left|f\right|\ge 2$, it is a must that $\alpha=0$.
Note that if one wishes to get non-trivial $g$, the value of $f$ shall not be arbitrary. In fact, for those $\left|f\right|\le 2$, if $f$ satisfies $$ \tan\left(\frac{\pi}{4}\sqrt{16-f^4}\right)=\frac{\sqrt{16-f^4}}{f^2}, $$ there would be no constraint on $\alpha$ (hence $g$ could be non-trivial).
Original post: Mistaken "*" as convolution
Suppose $g\in L^2(\mathbb{R})$. Apply Fourier transform to the equation with the convention \begin{align} \hat{f}(\xi)&=\int_{\mathbb{R}}f(x)e^{-2\pi ix\xi}{\rm d}x,\\ f(x)&=\int_{\mathbb{R}}\hat{f}(\xi)e^{2\pi ix\xi}{\rm d}\xi, \end{align} and the equation becomes $$ \widehat{g''}+\widehat{f^2*g'}+4\hat{g}=0. $$ Provided that $$ \widehat{g''}(\xi)=\left(2\pi i\xi\right)^2\hat{g}(\xi)=-4\pi^2\xi^2\hat{g}(\xi), $$ and that (by convolution theorem) $$ \widehat{f^2*g'}(\xi)=\widehat{f^2}(\xi)\widehat{g'}(\xi)=\widehat{f^2}(\xi)\left[\left(2\pi i\xi\right)\hat{g}(\xi)\right]=2\pi i\xi\widehat{f^2}(\xi)\hat{g}(\xi), $$ the governing equation becomes $$ \left(-4\pi^2\xi^2+2\pi i\xi\widehat{f^2}(\xi)+4\right)\hat{g}(\xi)=0. $$
Note that $$ -4\pi^2\xi^2+2\pi i\xi\widehat{f^2}(\xi)+4=0\iff\widehat{f^2}(\xi)=\frac{4\pi^2\xi^2-4}{2\pi i\xi}\notin L^2(\mathbb{R}). $$ Thus $-4\pi^2\xi^2+2\pi i\xi\widehat{f^2}(\xi)+4$ shall not be constantly zero, for which $$ \hat{g}(\xi)=0. $$ This result, as per the Fourier inverse transform, yields $$ g(x)=0. $$ Obviously, this $g$ satisfies both of the given conditions. Hence it is the only solution in $L^2(\mathbb{R})$ we are able to figure out.