Solution (a) For $y=f(x)=x^{4}-5x+1,$ we have $$\begin{equation*} y^{\prime} =4x^{3}-5\qquad y^{\prime \prime} =12x^{2} \end{equation*}$$

The function $f$ satisfies the differential equation $y^{\prime \prime} -12x^{2}=0$ and so $f$ is a solution.

(b) For $y=g(x)=x^{4}+C_{1}x+C_{2}$, we have $$\begin{equation*} y^{\prime} =4x^{3}+C_{1}\qquad y^{\prime \prime} =12x^{2} \end{equation*}$$

The function $g$ satisfies the differential equation $y^{\prime \prime} -12x^{2}=0$, so $g$ is a solution.

(c) Differentiate $x^{2}-x^{3}y+3y^{4}=C$ implicitly with respect to $x$ to find $\dfrac{dy}{dx}$. $$\begin{eqnarray*} 2x-x^{3}\dfrac{dy}{dx}-3x^{2}y+12y^{3}\dfrac{dy}{dx} &=&0 \\ ( 12y^{3}-x^{3}) \dfrac{dy}{dx} &=&3x^{2}y-2x \\[3pt] \dfrac{dy}{dx} &=&\dfrac{3x^{2}y-2x}{12y^{3}-x^{3}} \end{eqnarray*}$$

The function $y=f( x)$ defined by the equation $x^{2}-x^{3}y+3y^{4}=C$ satisfies the first-order differential equation and so is a solution.