Let $\mathcal{L}(x)=a(t)x^{\prime \prime \prime}+b(t)x^{\prime \prime}+c(t)x^{\prime}+d(t)x+e(t)$ where $a(t),b(t),c(t),d(t),e(t)\text{are functions in the variable $t$}$, determine if $\mathcal{L}(x)$ is linear.
The function isn´t linear since for $\alpha\in \mathbb{R}$ we have that $$\mathcal{L}(\alpha x)=a(t)(\alpha x)^{\prime \prime \prime}+b(t)(\alpha x)^{\prime \prime}+c(t)(\alpha x)^{\prime}+d(t)(\alpha x)+e(t)$$
$$=\alpha\left[a(t)x^{\prime \prime \prime}+b(t)x^{\prime \prime}+c(t)x^{\prime}+d(t)x\right]+e(t)$$ so $\mathcal{L}$ isn´t linear, now I think under why conditions $\mathcal{L}$ is linear and conlude that $\mathcal{L}$ is linear if and only if $e(t)=0$ for any $t\in \mathbb{R}$
The second equation is
$$\mathcal{H}(x)=x^{\prime}+ \sin x+x^2$$
Thus $\mathcal{H}(x)$ isn´t linear too since $\sin x$ isn´t linear with the sum and product by constant $\alpha$
The last one equation is
$$\mathcal{J}(x)=x^{\prime \prime}+|x|$$
Which isn´t linear because although $\mathcal{J}(\alpha x)=\alpha \mathcal{J}(x)$ don´t separate the sum because $|x+y| \neq |x|+|y|$.
I miss some?