Intersects:
\begin{equation} f(x) = (x+c)^{2} \end{equation}
\begin{equation} h(x) = c x \end{equation}
Doesn’t Intersect:
\begin{equation} g(x) = c e^{\frac{x^{4}}{4}}} \end{equation}
\begin{align} &h_1(x)-h_2(x) = c_1x-c_2x \\ \Rightarrow\ & 0 = c_1x-c_2x \\ \Rightarrow\ & 0 = x(c_1-c_2) \end{align}
\begin{align} &g_1(x)-g_2(x) = c_1e^{\frac{x^{4}}{4}} - c_2e^{\frac{x^{4}}{4}} \\ \Rightarrow\ & 0 = \qty(c_1 - c_2)e^{\frac{x^{4}}{4}} \\ \Rightarrow\ & 0 = e^{\frac{x^{4}}{4}}(c_1-c_2) \end{align}
\begin{align} & f_1(x)-f_2(x)=(x+c_1)^{2}-(x+c_2)^{2} \\ \Rightarrow\ & 0 = (x+c_1)^{2}-(x+c_2)^{2} \\ \Rightarrow\ & 0 = 2x(c_1-c_2)+{c_1}^{2}+{c_2}^{2} \end{align}
\begin{equation} \dv{y}{x} + P’(x)y = Q’(x) \end{equation}
\begin{align} &y = e^{\int P’(x)\dd{x}} \int e^{\int P’(x)\dd{x}} Q’(x)\dd{x} \\ \Rightarrow\ & e^{-P(x)} \int e^{P(x)}Q’(x)\dd{x} \\ \Rightarrow\ & e^{-P(x)} (\dots+C) \\ \Rightarrow\ & e^{-P(x)}C + \dots \end{align}
\begin{equation} h(x) \in e^{-P(x)}C + \dots \end{equation}
\begin{equation} g(x) \in e^{-P(x)}C + \dots \end{equation}
\begin{align} &0 = (e^{-P(x)}C_1+\dots)-(e^{-P(x)}C_2 + \dots) \\ \Rightarrow\ & e^{-P(x)}C_1-e^{-P(x)}C_2 \\ \Rightarrow\ & e^{-P(x)}(C_1-C_2) = 0 \end{align}